Mobile Numeric Keypad Problem

Difficulty Level : Hard
Last Updated : 24 Mar, 2023

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Given the mobile numeric keypad. You can only press buttons that are up, left, right or down to the current button. You are not allowed to press bottom row corner buttons (i.e. * and # ).

Mobile-keypad

Given a number N, find out the number of possible numbers of the given length.

Examples:

For N=1, number of possible numbers would be 10 (0, 1, 2, 3, …., 9) 
For N=2, number of possible numbers would be 36 
Possible numbers: 00,08 11,12,14 22,21,23,25 and so on. 
If we start with 0, valid numbers will be 00, 08 (count: 2) 
If we start with 1, valid numbers will be 11, 12, 14 (count: 3) 
If we start with 2, valid numbers will be 22, 21, 23,25 (count: 4) 
If we start with 3, valid numbers will be 33, 32, 36 (count: 3) 
If we start with 4, valid numbers will be 44,41,45,47 (count: 4) 
If we start with 5, valid numbers will be 55,54,52,56,58 (count: 5) 
.................................... 
....................................
We need to print the count of possible numbers.

N = 1 is trivial case, number of possible numbers would be 10 (0, 1, 2, 3, …., 9)
For N > 1, we need to start from some button, then move to any of the four direction (up, left, right or down) which takes to a valid button (should not go to *, #). Keep doing this until N length number is obtained (depth first traversal).

Recursive Solution:

Mobile Keypad is a rectangular grid of 4X3 (4 rows and 3 columns)
Lets say Count(i, j, N) represents the count of N length numbers starting from position (i, j)

If N = 1
  Count(i, j, N) = 10  
Else
  Count(i, j, N) = Sum of all Count(r, c, N-1) where (r, c) is new 
                   position after valid move of length 1 from current 
                   position (i, j)

Following is the implementation of above recursive formula.

C++

// A Naive Recursive C program to count number of possible numbers
// of given length
#include <stdio.h>
 
// left, up, right, down move from current location
int row[] = {0, 0, -1, 0, 1};
int col[] = {0, -1, 0, 1, 0};
 
// Returns count of numbers of length n starting from key position
// (i, j) in a numeric keyboard.
int getCountUtil(char keypad[][3], int i, int j, int n)
{
    if (keypad == NULL || n <= 0)
        return 0;
 
    // From a given key, only one number is possible of length 1
    if (n == 1)
        return 1;
 
    int k=0, move=0, ro=0, co=0, totalCount = 0;
 
    // move left, up, right, down from current location and if
    // new location is valid, then get number count of length
    // (n-1) from that new position and add in count obtained so far
    for (move=0; move<5; move++)
    {
        ro = i + row[move];
        co = j + col[move];
        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
           keypad[ro][co] != '*' && keypad[ro][co] != '#')
        {
            totalCount += getCountUtil(keypad, ro, co, n-1);
        }
    }
 
    return totalCount;
}
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
    // Base cases
    if (keypad == NULL || n <= 0)
        return 0;
    if (n == 1)
        return 10;
 
    int i=0, j=0, totalCount = 0;
    for (i=0; i<4; i++)  // Loop on keypad row
    {
        for (j=0; j<3; j++)   // Loop on keypad column
        {
            // Process for 0 to 9 digits
            if (keypad[i][j] != '*' && keypad[i][j] != '#')
            {
                // Get count when number is starting from key
                // position (i, j) and add in count obtained so far
                totalCount += getCountUtil(keypad, i, j, n);
            }
        }
    }
    return totalCount;
}
 
// Driver program to test above function
int main(int argc, char *argv[])
{
   char keypad[4][3] = {{'1','2','3'},
                        {'4','5','6'},
                        {'7','8','9'},
                        {'*','0','#'}};
   printf("Count for numbers of length %d: %dn \n", 1, getCount(keypad, 1));
   printf("Count for numbers of length %d: %dn \n", 2, getCount(keypad, 2));
   printf("Count for numbers of length %d: %dn \n", 3, getCount(keypad, 3));
   printf("Count for numbers of length %d: %dn \n", 4, getCount(keypad, 4));
   printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));
 
   return 0;
}

Java

// A Naive Recursive Java program 
// to count number of possible 
// numbers of given length
import java.util.*;
import java.io.*;
 
class GfG
{
 
// left, up, right, down 
// move from current location
static int row[] = {0, 0, -1, 0, 1};
static int col[] = {0, -1, 0, 1, 0};
 
// Returns count of numbers of length
// n starting from key position
// (i, j) in a numeric keyboard.
static int getCountUtil(char keypad[][], 
                        int i, int j, int n)
{
    if (keypad == null || n <= 0)
        return 0;
 
    // From a given key, only one 
    // number is possible of length 1
    if (n == 1)
        return 1;
 
    int k = 0, move = 0, ro = 0, co = 0, totalCount = 0;
 
    // move left, up, right, down
    // from current location and if
    // new location is valid, then 
    // get number count of length
    // (n-1) from that new position 
    // and add in count obtained so far
    for (move=0; move<5; move++)
    {
        ro = i + row[move];
        co = j + col[move];
        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
        keypad[ro][co] != '*' && keypad[ro][co] != '#')
        {
            totalCount += getCountUtil(keypad, ro, co, n - 1);
        }
    }
    return totalCount;
}
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
static int getCount(char keypad[][], int n)
{
    // Base cases
    if (keypad == null || n <= 0)
        return 0;
    if (n == 1)
        return 10;
 
    int i = 0, j = 0, totalCount = 0;
    for (i = 0; i < 4; i++) // Loop on keypad row
    {
        for (j=0; j<3; j++) // Loop on keypad column
        {
            // Process for 0 to 9 digits
            if (keypad[i][j] != '*' && keypad[i][j] != '#')
            {
                // Get count when number is starting from key
                // position (i, j) and add in count obtained so far
                totalCount += getCountUtil(keypad, i, j, n);
            }
        }
    }
    return totalCount;
}
 
// Driver code
public static void main(String[] args) 
{
    char keypad[][] = {{'1','2','3'},
                        {'4','5','6'},
                        {'7','8','9'},
                        {'*','0','#'}};
    System.out.printf("Count for numbers of"+
                    " length %d: %d", 1, getCount(keypad, 1));
    System.out.printf("\nCount for numbers of" + 
                    " length %d: %d", 2, getCount(keypad, 2));
    System.out.printf("\nCount for numbers of" + 
                    " length %d: %d", 3, getCount(keypad, 3));
    System.out.printf("\nCount for numbers of" + 
                    " length %d: %d", 4, getCount(keypad, 4));
    System.out.printf("\nCount for numbers of" + 
                    " length %d: %d", 5, getCount(keypad, 5));
}
}
 
// This code has been contributed by 29AjayKumar

Python3

# A Naive Recursive Python program to count number of possible numbers
# of given length
 
# left, up, right, down move from current location
row = [0, 0, -1, 0, 1]
col = [0, -1, 0, 1, 0]
 
# Returns count of numbers of length n starting from key position
# (i, j) in a numeric keyboard.
def getCountUtil(keypad, i, j, n):
    if (keypad == None or n <= 0):
        return 0
 
    # From a given key, only one number is possible of length 1
    if (n == 1):
        return 1
    k = 0
    move = 0
    ro = 0
    co = 0
    totalCount = 0
 
    # move left, up, right, down from current location and if
    # new location is valid, then get number count of length
    # (n-1) from that new position and add in count obtained so far
    for move in range(5):
        ro = i + row[move]
        co = j + col[move]
        if (ro >= 0 and ro <= 3 and co >= 0 and co <= 2 and
                keypad[ro][co] != '*' and keypad[ro][co] != '#'):
            totalCount += getCountUtil(keypad, ro, co, n - 1)
    return totalCount
 
# Return count of all possible numbers of length n
# in a given numeric keyboard
def getCount(keypad, n):
     
    # Base cases
    if (keypad == None or n <= 0):
        return 0
    if (n == 1):
        return 10
    i = 0
    j = 0
    totalCount = 0
    for i in range(4):  # Loop on keypad row
        for j in range(3):   # Loop on keypad column
             
            # Process for 0 to 9 digits
            if (keypad[i][j] != '*' and keypad[i][j] != '#'):
 
              # Get count when number is starting from key
                # position (i, j) and add in count obtained so far
                totalCount += getCountUtil(keypad, i, j, n)
    return totalCount
 
# Driver code
keypad = [['1', '2', '3'],
          ['4', '5', '6'],
          ['7', '8', '9'],
          ['*', '0', '#']]
print("Count for numbers of length 1:", getCount(keypad, 1))
print("Count for numbers of length 2:", getCount(keypad, 2))
print("Count for numbers of length 3:", getCount(keypad, 3))
print("Count for numbers of length 4:", getCount(keypad, 4))
print("Count for numbers of length 5:", getCount(keypad, 5))
 
# This code is contributed by subhammahato348

C#

// A Naive Recursive C# program 
// to count number of possible 
// numbers of given length
using System;
 
class GfG
{
 
// left, up, right, down 
// move from current location
static int []row = {0, 0, -1, 0, 1};
static int []col = {0, -1, 0, 1, 0};
 
// Returns count of numbers of length
// n starting from key position
// (i, j) in a numeric keyboard.
static int getCountUtil(char [,]keypad, 
                        int i, int j, int n)
{
    if (keypad == null || n <= 0)
        return 0;
 
    // From a given key, only one 
    // number is possible of length 1
    if (n == 1)
        return 1;
 
    int k = 0, move = 0, ro = 0, co = 0, totalCount = 0;
 
    // move left, up, right, down
    // from current location and if
    // new location is valid, then 
    // get number count of length
    // (n-1) from that new position 
    // and add in count obtained so far
    for (move=0; move<5; move++)
    {
        ro = i + row[move];
        co = j + col[move];
        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
        keypad[ro,co] != '*' && keypad[ro,co] != '#')
        {
            totalCount += getCountUtil(keypad, ro, co, n - 1);
        }
    }
    return totalCount;
}
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
static int getCount(char [,]keypad, int n)
{
    // Base cases
    if (keypad == null || n <= 0)
        return 0;
    if (n == 1)
        return 10;
 
    int i = 0, j = 0, totalCount = 0;
    for (i = 0; i < 4; i++) // Loop on keypad row
    {
        for (j = 0; j < 3; j++) // Loop on keypad column
        {
            // Process for 0 to 9 digits
            if (keypad[i, j] != '*' && keypad[i, j] != '#')
            {
                // Get count when number is starting from key
                // position (i, j) and add in count obtained so far
                totalCount += getCountUtil(keypad, i, j, n);
            }
        }
    }
    return totalCount;
}
 
// Driver code
public static void Main() 
{
    char [,]keypad = {{'1','2','3'},
                        {'4','5','6'},
                        {'7','8','9'},
                        {'*','0','#'}};
    Console.Write("Count for numbers of"+
                    " length {0}: {1}", 1, getCount(keypad, 1));
    Console.Write("\nCount for numbers of" + 
                    "length {0}: {1}", 2, getCount(keypad, 2));
    Console.Write("\nCount for numbers of" + 
                    "length {0}: {1}", 3, getCount(keypad, 3));
    Console.Write("\nCount for numbers of" + 
                    "length {0}: {1}", 4, getCount(keypad, 4));
    Console.Write("\nCount for numbers of" + 
                    "length {0}: {1}", 5, getCount(keypad, 5));
}
}
 
/* This code contributed by PrinciRaj1992 */

PHP

<?php
// A Naive Recursive PHP program 
// to count number of possible 
// numbers of given length
// left, up, right, down 
// move from current location
// static $row = array(0, 0, -1, 0, 1);
//static $col = array(0, -1, 0, 1, 0);
 
// Returns count of numbers of length
// n starting from key position
// (i, j) in a numeric keyboard.
function getCountUtil($keypad, 
                        $i, $j, $n)
{
    static $row= array(0,0,-1,0,1);
    static $col= array(0,-1,0,1,0);
    if ($keypad == null || $n <= 0)
        return 0;
 
    // From a given key, only one 
    // number is possible of length 1
    if ($n == 1)
        return 1;
 
    $k = 0; $move = 0; $ro = 0; $co = 0; $totalCount = 0;
 
    // move left, up, right, down
    // from current location and if
    // new location is valid, then 
    // get number count of length
    // (n-1) from that new position 
    // and add in count obtained so far
    for ($move = 0; $move < 5; $move++)
    {
        $ro = $i + $row[$move];
        $co = $j + $col[$move];
        if ($ro >= 0 && $ro <= 3 && $co >=0 && $co <= 2 &&
        $keypad[$ro][$co] != '*' && $keypad[$ro][$co] != '#')
        {
            $totalCount += getCountUtil($keypad, $ro, $co, $n - 1);
        }
    }
    return $totalCount;
}
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
function getCount($keypad, $n)
{
    // Base cases
    if ($keypad == null || $n <= 0)
        return 0;
    if ($n == 1)
        return 10;
 
    $i = 0; $j = 0; $totalCount = 0;
    for ($i = 0; $i < 4; $i++) // Loop on keypad row
    {
        for ($j = 0; $j < 3; $j++) // Loop on keypad column
        {
            // Process for 0 to 9 digits
            if ($keypad[$i][$j] != '*' && $keypad[$i][$j] != '#')
            {
                // Get count when number is starting from key
                // position (i, j) and add in count obtained so far
                $totalCount += getCountUtil($keypad, $i, $j, $n);
            }
        }
    }
    return $totalCount;
}
 
// Driver code
{
    $keypad = array(array('1','2','3'),
                        array('4','5','6'),
                        array('7','8','9'),
                        array('*','0','#'));
    echo("Count for numbers of"." length". getCount($keypad, 1)); 
    echo("\nCount for numbers of" . 
                    " length ". getCount($keypad, 2));
    echo("\nCount for numbers of" . 
                    " length ".getCount($keypad, 3));
    echo("\nCount for numbers of" .
                    " length ".getCount($keypad, 4));
    echo("\nCount for numbers of" . 
                    " length ".getCount($keypad, 5));
}
 
// This code has been contributed by Code_Mech.

Javascript

<script>
 
// A Naive Recursive Javascript program
// to count number of possible
// numbers of given length
 
     
    // left, up, right, down
    // move from current location
    let row=[0, 0, -1, 0, 1];
    let col=[0, -1, 0, 1, 0];
     
    // Returns count of numbers of length
    // n starting from key position
    // (i, j) in a numeric keyboard.
    function getCountUtil(keypad,i,j,n)
    {
        if (keypad == null || n <= 0)
        {return 0;}
         
        // From a given key, only one
        // number is possible of length 1
        if (n == 1)
            return 1;
         
        let k = 0, move = 0, ro = 0, co = 0, totalCount = 0;
         
        // move left, up, right, down
        // from current location and if
        // new location is valid, then
        // get number count of length
        // (n-1) from that new position
        // and add in count obtained so far
        for (move=0; move<5; move++)
        {
            ro = i + row[move];
            co = j + col[move];
            if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
            keypad[ro][co] != '*' && keypad[ro][co] != '#')
            {
                totalCount += getCountUtil(keypad, ro, co, n - 1);
            }
        }
        return totalCount;
    }
     
    // Return count of all possible numbers of length n
    // in a given numeric keyboard  
    function getCount(keypad,n)
    {
        // Base cases
        if (keypad == null || n <= 0)
            return 0;
        if (n == 1)
            return 10;
      
        let i = 0, j = 0, totalCount = 0;
        for (i = 0; i < 4; i++) // Loop on keypad row
        {
            for (j=0; j<3; j++) // Loop on keypad column
            {
                // Process for 0 to 9 digits
                if (keypad[i][j] != '*' && keypad[i][j] != '#')
                {
                    // Get count when number is starting from key
                    // position (i, j) and add in count obtained so far
                    totalCount += getCountUtil(keypad, i, j, n);
                }
            }
        }
        return totalCount;
    }
     
    // Driver code
    let keypad=[['1','2','3'],['4','5','6'],['7','8','9'],['*','0','#']];
    document.write("Count for numbers of"+
                    " length ", 1,": ", getCount(keypad, 1));
    document.write("<br>Count for numbers of" +
                    "length ", 2,": ", getCount(keypad, 2));
    document.write("<br>Count for numbers of" +
                    "length ", 3,": ", getCount(keypad, 3));
    document.write("<br>Count for numbers of" +
                    "length ", 4,": ", getCount(keypad, 4));
    document.write("<br>Count for numbers of" +
                    "length ", 5,": ", getCount(keypad, 5));
                     
    // This code is contributed by avanitrachhadiya2155
     
</script>

Output

Count for numbers of length 1: 10n 
Count for numbers of length 2: 36n 
Count for numbers of length 3: 138n 
Count for numbers of length 4: 532n 
Count for numbers of length 5: 2062n

Time Complexity: O(n*m) where n and m are row and column of keypad .
Auxiliary Space: O(n*m) where n and m are row and column of keypad .

Dynamic Programming

There are many repeated traversal on smaller paths (traversal for smaller N) to find all possible longer paths (traversal for bigger N). See following two diagrams for example. In this traversal, for N = 4 from two starting positions (buttons ‘4’ and ‘8’), we can see there are few repeated traversals for N = 2 (e.g. 4 -> 1, 6 -> 3, 8 -> 9, 8 -> 7 etc).

mobile2

mobile3

Since the problem has both properties: Optimal Substructure and Overlapping Subproblems, it can be efficiently solved using dynamic programming.

Following is the program for dynamic programming implementation.

C++

// A Dynamic Programming based C program to count number of
// possible numbers of given length
#include <stdio.h>
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
    if(keypad == NULL || n <= 0)
        return 0;
    if(n == 1)
        return 10;
 
    // left, up, right, down move from current location
    int row[] = {0, 0, -1, 0, 1};
    int col[] = {0, -1, 0, 1, 0};
 
    // taking n+1 for simplicity - count[i][j] will store
    // number count starting with digit i and length j
    int count[10][n+1];
    int i=0, j=0, k=0, move=0, ro=0, co=0, num = 0;
    int nextNum=0, totalCount = 0;
 
    // count numbers starting with digit i and of lengths 0 and 1
    for (i=0; i<=9; i++)
    {
        count[i][0] = 0;
        count[i][1] = 1;
    }
 
    // Bottom up - Get number count of length 2, 3, 4, ... , n
    for (k=2; k<=n; k++)
    {
        for (i=0; i<4; i++)  // Loop on keypad row
        {
            for (j=0; j<3; j++)   // Loop on keypad column
            {
                // Process for 0 to 9 digits
                if (keypad[i][j] != '*' && keypad[i][j] != '#')
                {
                    // Here we are counting the numbers starting with
                    // digit keypad[i][j] and of length k keypad[i][j]
                    // will become 1st digit, and we need to look for
                    // (k-1) more digits
                    num = keypad[i][j] - '0';
                    count[num][k] = 0;
 
                    // move left, up, right, down from current location
                    // and if new location is valid, then get number
                    // count of length (k-1) from that new digit and
                    // add in count we found so far
                    for (move=0; move<5; move++)
                    {
                        ro = i + row[move];
                        co = j + col[move];
                        if (ro >= 0 && ro <= 3 && co >=0 && co <= 2 &&
                           keypad[ro][co] != '*' && keypad[ro][co] != '#')
                        {
                            nextNum = keypad[ro][co] - '0';
                            count[num][k] += count[nextNum][k-1];
                        }
                    }
                }
            }
        }
    }
 
    // Get count of all possible numbers of length "n" starting
    // with digit 0, 1, 2, ..., 9
    totalCount = 0;
    for (i=0; i<=9; i++)
        totalCount += count[i][n];
    return totalCount;
}
 
// Driver program to test above function
int main(int argc, char *argv[])
{
   char keypad[4][3] = {{'1','2','3'},
                        {'4','5','6'},
                        {'7','8','9'},
                        {'*','0','#'}};
   printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));
   printf("\nCount for numbers of length %d: %dn", 2, getCount(keypad, 2));
   printf("\nCount for numbers of length %d: %dn", 3, getCount(keypad, 3));
   printf("\nCount for numbers of length %d: %dn", 4, getCount(keypad, 4));
   printf("\nCount for numbers of length %d: %dn", 5, getCount(keypad, 5));
 
   return 0;
}

Java

// A Dynamic Programming based Java program to 
// count number of possible numbers of given length
import java.util.*;
import java.io.*;
 
class GFG
{
     
// Return count of all possible numbers of length n
// in a given numeric keyboard
static int getCount(char keypad[][], int n)
{
    if(keypad == null || n <= 0)
        return 0;
    if(n == 1)
        return 10;
 
    // left, up, right, down move from current location
    int row[] = {0, 0, -1, 0, 1};
    int col[] = {0, -1, 0, 1, 0};
 
    // taking n+1 for simplicity - count[i][j] will store
    // number count starting with digit i and length j
    int [][]count = new int[10][n + 1];
    int i = 0, j = 0, k = 0, move = 0, 
             ro = 0, co = 0, num = 0;
    int nextNum = 0, totalCount = 0;
 
    // count numbers starting with digit i 
    // and of lengths 0 and 1
    for (i = 0; i <= 9; i++)
    {
        count[i][0] = 0;
        count[i][1] = 1;
    }
 
    // Bottom up - Get number count of length 2, 3, 4, ... , n
    for (k = 2; k <= n; k++)
    {
        for (i = 0; i < 4; i++) // Loop on keypad row
        {
            for (j = 0; j < 3; j++) // Loop on keypad column
            {
                // Process for 0 to 9 digits
                if (keypad[i][j] != '*' && 
                    keypad[i][j] != '#')
                {
                    // Here we are counting the numbers starting with
                    // digit keypad[i][j] and of length k keypad[i][j]
                    // will become 1st digit, and we need to look for
                    // (k-1) more digits
                    num = keypad[i][j] - '0';
                    count[num][k] = 0;
 
                    // move left, up, right, down from current location
                    // and if new location is valid, then get number
                    // count of length (k-1) from that new digit and
                    // add in count we found so far
                    for (move = 0; move < 5; move++)
                    {
                        ro = i + row[move];
                        co = j + col[move];
                        if (ro >= 0 && ro <= 3 && co >= 0 && 
                            co <= 2 && keypad[ro][co] != '*' && 
                                       keypad[ro][co] != '#')
                        {
                            nextNum = keypad[ro][co] - '0';
                            count[num][k] += count[nextNum][k - 1];
                        }
                    }
                }
            }
        }
    }
 
    // Get count of all possible numbers of length "n" 
    // starting with digit 0, 1, 2, ..., 9
    totalCount = 0;
    for (i = 0; i <= 9; i++)
        totalCount += count[i][n];
    return totalCount;
}
 
// Driver Code
public static void main(String[] args)
{
    char keypad[][] = {{'1','2','3'},
                       {'4','5','6'},
                       {'7','8','9'},
                       {'*','0','#'}};
    System.out.printf("Count for numbers of length %d: %d\n", 1, 
                                           getCount(keypad, 1));
    System.out.printf("Count for numbers of length %d: %d\n", 2,
                                           getCount(keypad, 2));
    System.out.printf("Count for numbers of length %d: %d\n", 3, 
                                           getCount(keypad, 3));
    System.out.printf("Count for numbers of length %d: %d\n", 4, 
                                           getCount(keypad, 4));
    System.out.printf("Count for numbers of length %d: %d\n", 5,
                                           getCount(keypad, 5));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# A Dynamic Programming based C program to count number of
# possible numbers of given length
 
# Return count of all possible numbers of length n
# in a given numeric keyboard
def getCount(keypad, n):
    if (keypad == None or n <= 0):
        return 0
    if (n == 1):
        return 10
 
    # left, up, right, down move from current location
    row = [0, 0, -1, 0, 1]
    col = [0, -1, 0, 1, 0]
 
    # taking n+1 for simplicity - count[i][j] will store
    # number count starting with digit i and length j
    # count[10][n+1]
    count = [[0]*(n + 1)]*10
    i = 0
    j = 0
    k = 0
    move = 0
    ro = 0
    co = 0
    num = 0
    nextNum = 0
    totalCount = 0
 
    # count numbers starting with 
    # digit i and of lengths 0 and 1
    for i in range(10):
        count[i][0] = 0
        count[i][1] = 1
 
    # Bottom up - Get number 
    # count of length 2, 3, 4, ... , n
    for k in range(2, n + 1):
        for i in range(4):  # Loop on keypad row
            for j in range(3):   # Loop on keypad column
               
                # Process for 0 to 9 digits
                if (keypad[i][j] != '*' and keypad[i][j] != '#'):
                   
                    # Here we are counting the numbers starting with
                    # digit keypad[i][j] and of length k keypad[i][j]
                    # will become 1st digit, and we need to look for
                    # (k-1) more digits
                    num = ord(keypad[i][j]) - 48
                    count[num][k] = 0
 
                    # move left, up, right, down from current location
                    # and if new location is valid, then get number
                    # count of length (k-1) from that new digit and
                    # add in count we found so far
                    for move in range(5):
                        ro = i + row[move]
                        co = j + col[move]
                        if (ro >= 0 and ro <= 3 and co >= 0 and co <= 2 and
                                keypad[ro][co] != '*' and keypad[ro][co] != '#'):
                            nextNum = ord(keypad[ro][co]) - 48
                            count[num][k] += count[nextNum][k - 1]
 
    # Get count of all possible numbers of length "n" starting
    # with digit 0, 1, 2, ..., 9
    totalCount = 0
    for i in range(10):
        totalCount += count[i][n]
    return totalCount
 
# Driver code
if __name__ == "__main__":
    keypad = [['1','2','3'],
            ['4','5','6'],
            ['7','8','9'],
            ['*','0','#']]
     
    print("Count for numbers of length", 1, ":", getCount(keypad, 1))
    print("Count for numbers of length", 2, ":", getCount(keypad, 2))
    print("Count for numbers of length", 3, ":", getCount(keypad, 3))
    print("Count for numbers of length", 4, ":", getCount(keypad, 4))
    print("Count for numbers of length", 5, ":", getCount(keypad, 5))
 
    # This code is contributed by subhammahato348

C#

// A Dynamic Programming based C# program to 
// count number of possible numbers of given Length
using System;
 
class GFG
{
     
// Return count of all possible numbers of Length n
// in a given numeric keyboard
static int getCount(char [,]keypad, int n)
{
    if(keypad == null || n <= 0)
        return 0;
    if(n == 1)
        return 10;
 
    // left, up, right, down move
    // from current location
    int []row = {0, 0, -1, 0, 1};
    int []col = {0, -1, 0, 1, 0};
 
    // taking n+1 for simplicity - count[i,j] 
    // will store number count starting with
    // digit i and.Length j
    int [,]count = new int[10,n + 1];
    int i = 0, j = 0, k = 0, move = 0, 
              ro = 0, co = 0, num = 0;
    int nextNum = 0, totalCount = 0;
 
    // count numbers starting with digit i 
    // and of.Lengths 0 and 1
    for (i = 0; i <= 9; i++)
    {
        count[i, 0] = 0;
        count[i, 1] = 1;
    }
 
    // Bottom up - Get number count of
    // Length 2, 3, 4, ... , n
    for (k = 2; k <= n; k++)
    {
        for (i = 0; i < 4; i++) // Loop on keypad row
        {
            for (j = 0; j < 3; j++) // Loop on keypad column
            {
                // Process for 0 to 9 digits
                if (keypad[i, j] != '*' && 
                    keypad[i, j] != '#')
                {
                    // Here we are counting the numbers starting with
                    // digit keypad[i,j] and of.Length k keypad[i,j]
                    // will become 1st digit, and we need to look for
                    // (k-1) more digits
                    num = keypad[i, j] - '0';
                    count[num, k] = 0;
 
                    // move left, up, right, down from current location
                    // and if new location is valid, then get number
                    // count of.Length (k-1) from that new digit and
                    //.Add in count we found so far
                    for (move = 0; move < 5; move++)
                    {
                        ro = i + row[move];
                        co = j + col[move];
                        if (ro >= 0 && ro <= 3 && co >= 0 && 
                            co <= 2 && keypad[ro, co] != '*' && 
                                       keypad[ro, co] != '#')
                        {
                            nextNum = keypad[ro, co] - '0';
                            count[num, k] += count[nextNum, k - 1];
                        }
                    }
                }
            }
        }
    }
 
    // Get count of all possible numbers of.Length "n" 
    // starting with digit 0, 1, 2, ..., 9
    totalCount = 0;
    for (i = 0; i <= 9; i++)
        totalCount += count[i, n];
    return totalCount;
}
 
// Driver Code
public static void Main(String[] args)
{
    char [,]keypad = {{'1', '2', '3'},
                      {'4', '5', '6'},
                      {'7', '8', '9'},
                      {'*', '0', '#'}};
    Console.Write("Count for numbers of.Length {0}: {1}\n", 1, 
                                        getCount(keypad, 1));
    Console.Write("Count for numbers of.Length {0}: {1}\n", 2,
                                        getCount(keypad, 2));
    Console.Write("Count for numbers of.Length {0}: {1}\n", 3, 
                                        getCount(keypad, 3));
    Console.Write("Count for numbers of.Length {0}: {1}\n", 4, 
                                        getCount(keypad, 4));
    Console.Write("Count for numbers of.Length {0}: {1}\n", 5,
                                        getCount(keypad, 5));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// A Dynamic Programming based Javascript program to
// count number of possible numbers of given length
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
function getCount(keypad, n)
{
    if (keypad == null || n <= 0)
        return 0;
         
    if (n == 1)
        return 10;
     
    // left, up, right, down move
    // from current location    
    let row = [ 0, 0, -1, 0, 1 ];
    let col = [ 0, -1, 0, 1, 0 ];
     
    // Taking n+1 for simplicity - count[i][j]
    // will store number count starting with 
    // digit i and length j
    let count = new Array(10);
    for(let i = 0; i < 10; i++)
    {
        count[i] = new Array(n + 1);
        for(let j = 0; j < n + 1; j++)
        {
            count[i][j] = 0;
        }
    }
     
    let i = 0, j = 0, k = 0, move = 0,
         ro = 0, co = 0, num = 0;
    let nextNum = 0, totalCount = 0;
     
    // count numbers starting with digit i
    // and of lengths 0 and 1
    for(i = 0; i <= 9; i++)
    {
        count[i][0] = 0;
        count[i][1] = 1;
    }
  
    // Bottom up - Get number count 
    // of length 2, 3, 4, ... , n
    for(k = 2; k <= n; k++)
    {
         
        // Loop on keypad row
        for(i = 0; i < 4; i++) 
        {
             
            // Loop on keypad column
            for(j = 0; j < 3; j++) 
            {
                 
                // Process for 0 to 9 digits
                if (keypad[i][j] != '*' &&
                    keypad[i][j] != '#')
                {
                     
                    // Here we are counting the numbers starting with
                    // digit keypad[i][j] and of length k keypad[i][j]
                    // will become 1st digit, and we need to look for
                    // (k-1) more digits
                    num = keypad[i][j].charCodeAt(0) - 
                                   '0'.charCodeAt(0);
                    count[num][k] = 0;
  
                    // Move left, up, right, down from current location
                    // and if new location is valid, then get number
                    // count of length (k-1) from that new digit and
                    // add in count we found so far
                    for(move = 0; move < 5; move++)
                    {
                        ro = i + row[move];
                        co = j + col[move];
                        if (ro >= 0 && ro <= 3 && co >= 0 &&
                            co <= 2 && keypad[ro][co] != '*' &&
                                       keypad[ro][co] != '#')
                        {
                            nextNum = keypad[ro][co].charCodeAt(0) - 
                                                 '0'.charCodeAt(0);
                            count[num][k] += count[nextNum][k - 1];
                        }
                    }
                }
            }
        }
    }
  
    // Get count of all possible numbers of length "n"
    // starting with digit 0, 1, 2, ..., 9
    totalCount = 0;
    for(i = 0; i <= 9; i++)
        totalCount += count[i][n];
         
    return totalCount;
     
}
 
// Driver Code
let keypad = [ [ '1','2','3' ],
               [ '4','5','6' ],
               [ '7','8','9' ],
               [ '*','0','#' ] ];
                
document.write("Count for numbers of length " +
         1 + " : " + getCount(keypad, 1) + "<br>")
document.write("Count for numbers of length " +
         2 + " : " + getCount(keypad, 2) + "<br>")
document.write("Count for numbers of length " +
         3 + " : " + getCount(keypad, 3) + "<br>")
document.write("Count for numbers of length " + 
         4 + " : " + getCount(keypad, 4) + "<br>")
document.write("Count for numbers of length " + 
         5 + " : " + getCount(keypad, 5) + "<br>")
 
// This code is contributed by rag2127
 
</script>

Output

Count for numbers of length 1: 10n
Count for numbers of length 2: 36n
Count for numbers of length 3: 138n
Count for numbers of length 4: 532n
Count for numbers of length 5: 2062n

A Space Optimized Solution:

The above dynamic programming approach also runs in O(n) time and requires O(n) auxiliary space, as only one for loop runs n times, other for loops runs for constant time. We can see that nth iteration needs data from (n-1)th iteration only, so we need not keep the data from older iterations. We can have a space efficient dynamic programming approach with just two arrays of size 10. Thanks to Nik for suggesting this solution.

C++

// A Space Optimized C++ program to count number of possible numbers
// of given length
#include <bits/stdc++.h>
using namespace std;
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
    if (keypad == NULL || n <= 0)
        return 0;
    if (n == 1)
        return 10;
 
    // odd[i], even[i] arrays represent count of numbers starting
    // with digit i for any length j
    int odd[10], even[10];
    int i = 0, j = 0, useOdd = 0, totalCount = 0;
 
    for (i = 0; i <= 9; i++)
        odd[i] = 1; // for j = 1
 
    for (j = 2; j <= n; j++) // Bottom Up calculation from j = 2 to n
    {
        useOdd = 1 - useOdd;
 
        // Here we are explicitly writing lines for each number 0
        // to 9. But it can always be written as DFS on 4X3 grid
        // using row, column array valid moves
        if (useOdd == 1)
        {
            even[0] = odd[0] + odd[8];
            even[1] = odd[1] + odd[2] + odd[4];
            even[2] = odd[2] + odd[1] + odd[3] + odd[5];
            even[3] = odd[3] + odd[2] + odd[6];
            even[4] = odd[4] + odd[1] + odd[5] + odd[7];
            even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];
            even[6] = odd[6] + odd[3] + odd[5] + odd[9];
            even[7] = odd[7] + odd[4] + odd[8];
            even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];
            even[9] = odd[9] + odd[6] + odd[8];
        }
        else
        {
            odd[0] = even[0] + even[8];
            odd[1] = even[1] + even[2] + even[4];
            odd[2] = even[2] + even[1] + even[3] + even[5];
            odd[3] = even[3] + even[2] + even[6];
            odd[4] = even[4] + even[1] + even[5] + even[7];
            odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];
            odd[6] = even[6] + even[3] + even[5] + even[9];
            odd[7] = even[7] + even[4] + even[8];
            odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];
            odd[9] = even[9] + even[6] + even[8];
        }
    }
 
    // Get count of all possible numbers of length "n" starting
    // with digit 0, 1, 2, ..., 9
    totalCount = 0;
    if (useOdd == 1)
    {
        for (i = 0; i <= 9; i++)
            totalCount += even[i];
    }
    else
    {
        for (i = 0; i <= 9; i++)
            totalCount += odd[i];
    }
    return totalCount;
}
 
// Driver program to test above function
int main()
{
    char keypad[4][3] = {{'1', '2', '3'},
                         {'4', '5', '6'},
                         {'7', '8', '9'},
                         {'*', '0', '#'}};
    cout << "Count for numbers of length 1: " << getCount(keypad, 1) << endl;
    cout << "Count for numbers of length 2: " << getCount(keypad, 2) << endl;
    cout << "Count for numbers of length 3: " << getCount(keypad, 3) << endl;
    cout << "Count for numbers of length 4: " << getCount(keypad, 4) << endl;
    cout << "Count for numbers of length 5: " << getCount(keypad, 5) << endl;
 
    return 0;
}
 
//This code is contributed by Mayank Tyagi

C

// A Space Optimized C program to count number of possible numbers
// of given length
#include <stdio.h>
 
// Return count of all possible numbers of length n
// in a given numeric keyboard
int getCount(char keypad[][3], int n)
{
    if(keypad == NULL || n <= 0)
        return 0;
    if(n == 1)
        return 10;
 
    // odd[i], even[i] arrays represent count of numbers starting
    // with digit i for any length j
    int odd[10], even[10];
    int i = 0, j = 0, useOdd = 0, totalCount = 0;
 
    for (i=0; i<=9; i++)
        odd[i] = 1;  // for j = 1
 
    for (j=2; j<=n; j++) // Bottom Up calculation from j = 2 to n
    {
        useOdd = 1 - useOdd;
 
        // Here we are explicitly writing lines for each number 0
        // to 9. But it can always be written as DFS on 4X3 grid
        // using row, column array valid moves
        if(useOdd == 1)
        {
            even[0] = odd[0] + odd[8];
            even[1] = odd[1] + odd[2] + odd[4];
            even[2] = odd[2] + odd[1] + odd[3] + odd[5];
            even[3] = odd[3] + odd[2] + odd[6];
            even[4] = odd[4] + odd[1] + odd[5] + odd[7];
            even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6];
            even[6] = odd[6] + odd[3] + odd[5] + odd[9];
            even[7] = odd[7] + odd[4] + odd[8];
            even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9];
            even[9] = odd[9] + odd[6] + odd[8];
        }
        else
        {
            odd[0] = even[0] + even[8];
            odd[1] = even[1] + even[2] + even[4];
            odd[2] = even[2] + even[1] + even[3] + even[5];
            odd[3] = even[3] + even[2] + even[6];
            odd[4] = even[4] + even[1] + even[5] + even[7];
            odd[5] = even[5] + even[2] + even[4] + even[8] + even[6];
            odd[6] = even[6] + even[3] + even[5] + even[9];
            odd[7] = even[7] + even[4] + even[8];
            odd[8] = even[8] + even[0] + even[5] + even[7] + even[9];
            odd[9] = even[9] + even[6] + even[8];
        }
    }
 
    // Get count of all possible numbers of length "n" starting
    // with digit 0, 1, 2, ..., 9
    totalCount = 0;
    if(useOdd == 1)
    {
        for (i=0; i<=9; i++)
            totalCount += even[i];
    }
    else
    {
        for (i=0; i<=9; i++)
            totalCount += odd[i];
    }
    return totalCount;
}
 
// Driver program to test above function
int main()
{
    char keypad[4][3] = {{'1','2','3'},
        {'4','5','6'},
        {'7','8','9'},
        {'*','0','#'}
    };
    printf("Count for numbers of length %d: %dn", 1, getCount(keypad, 1));
    printf("Count for numbers of length %d: %dn", 2, getCount(keypad, 2));
    printf("Count for numbers of length %d: %dn", 3, getCount(keypad, 3));
    printf("Count for numbers of length %d: %dn", 4, getCount(keypad, 4));
    printf("Count for numbers of length %d: %dn", 5, getCount(keypad, 5));
 
    return 0;
}

Java

// A Space Optimized Java program to 
// count number of possible numbers 
// of given length
import java.util.*;
import java.io.*;
 
class GFG 
{
 
// Return count of all possible numbers of 
// length n in a given numeric keyboard
static int getCount(char keypad[][], int n)
{
    if(keypad == null || n <= 0)
        return 0;
    if(n == 1)
        return 10;
 
    // odd[i], even[i] arrays represent count of 
    // numbers starting with digit i for any length j
    int []odd = new int[10];
    int []even = new int[10];
    int i = 0, j = 0, useOdd = 0, totalCount = 0;
 
    for (i = 0; i <= 9; i++)
        odd[i] = 1; // for j = 1
     
    // Bottom Up calculation from j = 2 to n
    for (j = 2; j <= n; j++) 
    {
        useOdd = 1 - useOdd;
 
        // Here we are explicitly writing lines 
        // for each number 0 to 9. But it can always be 
        // written as DFS on 4X3 grid using row, 
        // column array valid moves
        if(useOdd == 1)
        {
            even[0] = odd[0] + odd[8];
            even[1] = odd[1] + odd[2] + odd[4];
            even[2] = odd[2] + odd[1] + 
                      odd[3] + odd[5];
            even[3] = odd[3] + odd[2] + odd[6];
            even[4] = odd[4] + odd[1] + 
                      odd[5] + odd[7];
            even[5] = odd[5] + odd[2] + odd[4] + 
                               odd[8] + odd[6];
            even[6] = odd[6] + odd[3] + 
                      odd[5] + odd[9];
            even[7] = odd[7] + odd[4] + odd[8];
            even[8] = odd[8] + odd[0] + odd[5] + 
                               odd[7] + odd[9];
            even[9] = odd[9] + odd[6] + odd[8];
        }
        else
        {
            odd[0] = even[0] + even[8];
            odd[1] = even[1] + even[2] + even[4];
            odd[2] = even[2] + even[1] + 
                     even[3] + even[5];
            odd[3] = even[3] + even[2] + even[6];
            odd[4] = even[4] + even[1] + 
                     even[5] + even[7];
            odd[5] = even[5] + even[2] + even[4] + 
                               even[8] + even[6];
            odd[6] = even[6] + even[3] + 
                     even[5] + even[9];
            odd[7] = even[7] + even[4] + even[8];
            odd[8] = even[8] + even[0] + even[5] + 
                               even[7] + even[9];
            odd[9] = even[9] + even[6] + even[8];
        }
    }
 
    // Get count of all possible numbers of 
    // length "n" starting with digit 0, 1, 2, ..., 9
    totalCount = 0;
    if(useOdd == 1)
    {
        for (i = 0; i <= 9; i++)
            totalCount += even[i];
    }
    else
    {
        for (i = 0; i <= 9; i++)
            totalCount += odd[i];
    }
    return totalCount;
}
 
// Driver Code
public static void main(String[] args) 
{
    char keypad[][] = {{'1','2','3'},
                       {'4','5','6'},
                       {'7','8','9'},
                       {'*','0','#'}};
    System.out.printf("Count for numbers of length %d: %d\n", 1, 
                                           getCount(keypad, 1));
    System.out.printf("Count for numbers of length %d: %d\n", 2, 
                                           getCount(keypad, 2));
    System.out.printf("Count for numbers of length %d: %d\n", 3,
                                           getCount(keypad, 3));
    System.out.printf("Count for numbers of length %d: %d\n", 4,
                                           getCount(keypad, 4));
    System.out.printf("Count for numbers of length %d: %d\n", 5,
                                           getCount(keypad, 5));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# A Space Optimized Python program to count 
# number of possible numbers
# of given length
 
# Return count of all possible numbers 
# of length n
# in a given numeric keyboard
def getCount(keypad, n):
 
    if(not keypad or n <= 0):
        return 0
    if(n == 1):
        return 10
 
    # odd[i], even[i] arrays represent 
    # count of numbers starting
    # with digit i for any length j
    odd = [0]*10
    even = [0]*10
    i = 0
    j = 0
    useOdd = 0
    totalCount = 0
 
    for i in range(10):
        odd[i] = 1 # for j = 1
 
    for j in range(2,n+1): # Bottom Up calculation from j = 2 to n
     
        useOdd = 1 - useOdd
 
        # Here we are explicitly writing lines for each number 0
        # to 9. But it can always be written as DFS on 4X3 grid
        # using row, column array valid moves
        if(useOdd == 1):
         
            even[0] = odd[0] + odd[8]
            even[1] = odd[1] + odd[2] + odd[4]
            even[2] = odd[2] + odd[1] + odd[3] + odd[5]
            even[3] = odd[3] + odd[2] + odd[6]
            even[4] = odd[4] + odd[1] + odd[5] + odd[7]
            even[5] = odd[5] + odd[2] + odd[4] + odd[8] + odd[6]
            even[6] = odd[6] + odd[3] + odd[5] + odd[9]
            even[7] = odd[7] + odd[4] + odd[8]
            even[8] = odd[8] + odd[0] + odd[5] + odd[7] + odd[9]
            even[9] = odd[9] + odd[6] + odd[8]
         
        else:
         
            odd[0] = even[0] + even[8]
            odd[1] = even[1] + even[2] + even[4]
            odd[2] = even[2] + even[1] + even[3] + even[5]
            odd[3] = even[3] + even[2] + even[6]
            odd[4] = even[4] + even[1] + even[5] + even[7]
            odd[5] = even[5] + even[2] + even[4] + even[8] + even[6]
            odd[6] = even[6] + even[3] + even[5] + even[9]
            odd[7] = even[7] + even[4] + even[8]
            odd[8] = even[8] + even[0] + even[5] + even[7] + even[9]
            odd[9] = even[9] + even[6] + even[8]
 
    # Get count of all possible numbers of length "n" starting
    # with digit 0, 1, 2, ..., 9
    totalCount = 0
    if(useOdd == 1):
        for i in range(10):
            totalCount += even[i]
     
    else:
        for i in range(10):
            totalCount += odd[i]
 
    return totalCount
 
# Driver program to test above function
if __name__ == "__main__":
    keypad = [['1','2','3'],
            ['4','5','6'],
            ['7','8','9'],
            ['*','0','#']]
     
    print("Count for numbers of length ",1,": ", getCount(keypad, 1))
    print("Count for numbers of length ",2,": ", getCount(keypad, 2))
    print("Count for numbers of length ",3,": ", getCount(keypad, 3))
    print("Count for numbers of length ",4,": ", getCount(keypad, 4))
    print("Count for numbers of length ",5,": ", getCount(keypad, 5))
     
# This code is contributed by
# ChitraNayal

C#

// A Space Optimized C# program to 
// count number of possible numbers 
// of given length
using System;
     
class GFG 
{
 
// Return count of all possible numbers of 
// length n in a given numeric keyboard
static int getCount(char [,]keypad, int n)
{
    if(keypad == null || n <= 0)
        return 0;
    if(n == 1)
        return 10;
 
    // odd[i], even[i] arrays represent count of 
    // numbers starting with digit i for any length j
    int []odd = new int[10];
    int []even = new int[10];
    int i = 0, j = 0, useOdd = 0, totalCount = 0;
 
    for (i = 0; i <= 9; i++)
        odd[i] = 1; // for j = 1
     
    // Bottom Up calculation from j = 2 to n
    for (j = 2; j <= n; j++) 
    {
        useOdd = 1 - useOdd;
 
        // Here we are explicitly writing lines 
        // for each number 0 to 9. But it can always be 
        // written as DFS on 4X3 grid using row, 
        // column array valid moves
        if(useOdd == 1)
        {
            even[0] = odd[0] + odd[8];
            even[1] = odd[1] + odd[2] + odd[4];
            even[2] = odd[2] + odd[1] + 
                      odd[3] + odd[5];
            even[3] = odd[3] + odd[2] + odd[6];
            even[4] = odd[4] + odd[1] + 
                      odd[5] + odd[7];
            even[5] = odd[5] + odd[2] + odd[4] + 
                               odd[8] + odd[6];
            even[6] = odd[6] + odd[3] + 
                      odd[5] + odd[9];
            even[7] = odd[7] + odd[4] + odd[8];
            even[8] = odd[8] + odd[0] + odd[5] + 
                               odd[7] + odd[9];
            even[9] = odd[9] + odd[6] + odd[8];
        }
        else
        {
            odd[0] = even[0] + even[8];
            odd[1] = even[1] + even[2] + even[4];
            odd[2] = even[2] + even[1] + 
                     even[3] + even[5];
            odd[3] = even[3] + even[2] + even[6];
            odd[4] = even[4] + even[1] + 
                     even[5] + even[7];
            odd[5] = even[5] + even[2] + even[4] + 
                               even[8] + even[6];
            odd[6] = even[6] + even[3] + 
                     even[5] + even[9];
            odd[7] = even[7] + even[4] + even[8];
            odd[8] = even[8] + even[0] + even[5] + 
                               even[7] + even[9];
            odd[9] = even[9] + even[6] + even[8];
        }
    }
 
    // Get count of all possible numbers of 
    // length "n" starting with digit 0, 1, 2, ..., 9
    totalCount = 0;
    if(useOdd == 1)
    {
        for (i = 0; i <= 9; i++)
            totalCount += even[i];
    }
    else
    {
        for (i = 0; i <= 9; i++)
            totalCount += odd[i];
    }
    return totalCount;
}
 
// Driver Code
public static void Main(String[] args) 
{
    char [,]keypad = {{'1','2','3'},
                      {'4','5','6'},
                      {'7','8','9'},
                      {'*','0','#'}};
    Console.Write("Count for numbers of length {0}: {1}\n", 1, 
                                        getCount(keypad, 1));
    Console.Write("Count for numbers of length {0}: {1}\n", 2, 
                                        getCount(keypad, 2));
    Console.Write("Count for numbers of length {0}: {1}\n", 3,
                                        getCount(keypad, 3));
    Console.Write("Count for numbers of length {0}: {1}\n", 4,
                                        getCount(keypad, 4));
    Console.Write("Count for numbers of length {0}: {1}\n", 5,
                                        getCount(keypad, 5));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// A Space Optimized javascript program to 
// count number of possible numbers 
// of given length
 
// Return count of all possible numbers of 
// length n in a given numeric keyboard
function getCount(keypad , n)
{
    if(keypad == null || n <= 0)
        return 0;
    if(n == 1)
        return 10;
 
    // odd[i], even[i] arrays represent count of 
    // numbers starting with digit i for any length j
    var odd = Array.from({length: 10}, (_, i) => 0);
    var even = Array.from({length: 10}, (_, i) => 0);
    var i = 0, j = 0, useOdd = 0, totalCount = 0;
 
    for (i = 0; i <= 9; i++)
        odd[i] = 1; // for j = 1
     
    // Bottom Up calculation from j = 2 to n
    for (j = 2; j <= n; j++) 
    {
        useOdd = 1 - useOdd;
 
        // Here we are explicitly writing lines 
        // for each number 0 to 9. But it can always be 
        // written as DFS on 4X3 grid using row, 
        // column array valid moves
        if(useOdd == 1)
        {
            even[0] = odd[0] + odd[8];
            even[1] = odd[1] + odd[2] + odd[4];
            even[2] = odd[2] + odd[1] + 
                      odd[3] + odd[5];
            even[3] = odd[3] + odd[2] + odd[6];
            even[4] = odd[4] + odd[1] + 
                      odd[5] + odd[7];
            even[5] = odd[5] + odd[2] + odd[4] + 
                               odd[8] + odd[6];
            even[6] = odd[6] + odd[3] + 
                      odd[5] + odd[9];
            even[7] = odd[7] + odd[4] + odd[8];
            even[8] = odd[8] + odd[0] + odd[5] + 
                               odd[7] + odd[9];
            even[9] = odd[9] + odd[6] + odd[8];
        }
        else
        {
            odd[0] = even[0] + even[8];
            odd[1] = even[1] + even[2] + even[4];
            odd[2] = even[2] + even[1] + 
                     even[3] + even[5];
            odd[3] = even[3] + even[2] + even[6];
            odd[4] = even[4] + even[1] + 
                     even[5] + even[7];
            odd[5] = even[5] + even[2] + even[4] + 
                               even[8] + even[6];
            odd[6] = even[6] + even[3] + 
                     even[5] + even[9];
            odd[7] = even[7] + even[4] + even[8];
            odd[8] = even[8] + even[0] + even[5] + 
                               even[7] + even[9];
            odd[9] = even[9] + even[6] + even[8];
        }
    }
 
    // Get count of all possible numbers of 
    // length "n" starting with digit 0, 1, 2, ..., 9
    totalCount = 0;
    if(useOdd == 1)
    {
        for (i = 0; i <= 9; i++)
            totalCount += even[i];
    }
    else
    {
        for (i = 0; i <= 9; i++)
            totalCount += odd[i];
    }
    return totalCount;
}
 
// Driver Code
 
var keypad = [['1','2','3'],
                   ['4','5','6'],
                   ['7','8','9'],
                   ['*','0','#']];
                    
document.write("Count for numbers of length "+ 1+": "+ 
                                       getCount(keypad, 1));
document.write("<br>Count for numbers of length  "+ 2+": "+ 
                                       getCount(keypad, 2));
document.write("<br>Count for numbers of length "+ 3+": "+ 
                                       getCount(keypad, 3));
document.write("<br>Count for numbers of length "+ 4+": "+ 
                                       getCount(keypad, 4));
document.write("<br>Count for numbers of length "+ 5+": "+ 
                                       getCount(keypad, 5));
 
 
// This code is contributed by 29AjayKumar 
 
</script>

Output

Count for numbers of length 1: 10
Count for numbers of length 2: 36
Count for numbers of length 3: 138
Count for numbers of length 4: 532
Count for numbers of length 5: 2062

Time complexity :- O(N)
Auxiliary Space: O(1)

My Personal Notes

1. Mobile Numeric Keypad Problem | Set 2

2. Convert a Mobile Numeric Keypad sequence to equivalent sentence

3. Convert a sentence into its equivalent mobile numeric keypad sequence

4. Count of buttons pressed in a keypad mobile

5. Count of distinct Numbers that can be formed by chess knight in N moves on a mobile keypad

6. Print all n digit patterns formed by mobile Keypad

7. Remove uppercase, lowercase, special, numeric, and non-numeric characters from a String

8. Check whether the string can be printed using same row of qwerty keypad

9. Nuts & Bolts Problem (Lock & Key problem) using Quick Sort

10. Nuts & Bolts Problem (Lock & Key problem) using Hashmap

Palindrome Partitioning | DP-17

Longest Common Subsequence with at most k changes allowed

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Data Structures & Algorithms in Python - Self Paced

Python Programming Foundation -Self Paced

Data Structures and Algorithms - Self Paced

Master Java Programming - Complete Beginner to Advanced

Complete Test Series for Product-Based Companies

Competitive Programming - Live

JavaScript Foundation - Self Paced

Complete Interview Preparation - Self Paced

Master C++ Programming - Complete Beginner to Advanced

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Mastering Data Analytics

System Design - Live

Python Backend Development with Django - Live

Complete Machine Learning & Data Science Program

Full Stack Development with React & Node JS - Live

JAVA Backend Development - Live

Complete Test Series for Service-Based Companies

Start Your Coding Journey Now!