Sum of Digits of the Good Strings

• Last Updated : 25 May, 2021

A string is called good if it is made with only digits 0 to 9 and adjacent elements are different. The task is to find the sum of the digits of all possible good strings of length X that end with the given digit Y. The answer could be large, so print the answer modulo 109 + 7.

Examples:

Input: X = 2, Y = 2
Output: 61
All possible strings of length 2 that end with 2 are:
02, 12, 32, 42, 52, 62, 72, 82, 92.
Now, ((0 + 2) + (1 + 2) + (3 + 2) + (4 + 2) + (5 + 2)
+ (6 + 2) + (7 + 2) + (8 + 2) + (9 + 2)) = 61

Input: X = 6, Y = 4
Output: 1567751

Approach: This problem can be solved by using dynamic programming. Let’s define the following states:

1. dp[i][j]: Sum of the digits of all possible good strings of length i that end with j.
2. cnt[i][j]: Count the good strings of length i that end with j.

The value of the previous state will have to be used to compute the value for the current state as the adjacent digits have to be compared to whether they are equal or not. Now, the recurrence relation will be:

dp[i][j] = dp[i][j] + dp[i – 1][k] + cnt[i – 1][k] * j

Here, dp[i – 1][k] is the sum of the digits of good strings of length (i – 1) that end with k and k != j
cnt[i -1][k] is the count of good strings of length (i – 1) that end with k and k != j
So for position i, (cnt(i – 1)[k] * j) has to be added as j is being put at index i and the count of possible strings that have length (i – 1) is cnt[i – 1][k].
Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;   #define DIGITS 10 #define MAX 10000 #define MOD 1000000007   // To store the states of the dp long dp[MAX][DIGITS], cnt[MAX][DIGITS];   // Function to fill the dp table void precompute() {       // dp[i][j] : Sum of the digits of all     // possible good strings of length     // i that end with j     // cnt[i][j] : Count of the good strings     // of length i that end with j       // Sum of digits of the string of length     // 1 is i as i is only number in that string     // and count of good strings of length 1     // that end with i is also 1     for (int i = 0; i < DIGITS; i++)         dp[i] = i, cnt[i] = 1;       for (int i = 2; i < MAX; i++) {         for (int j = 0; j < DIGITS; j++) {             for (int k = 0; k < DIGITS; k++) {                   // Adjacent digits are different                 if (j != k) {                     dp[i][j] = dp[i][j]                                + (dp[i - 1][k] + (cnt[i - 1][k] * j) % MOD)                                      % MOD;                     dp[i][j] %= MOD;                       // Increment the count as digit at                     // (i - 1)'th index is k and count                     // of good strings is equal to this                     // because at the end of the strings of                     // length (i - 1) we are just                     // putting digit j as the last digit                     cnt[i][j] += cnt[i - 1][k];                     cnt[i][j] %= MOD;                 }             }         }     } }   // Driver code int main() {     long long int x = 6, y = 4;       precompute();       cout << dp[x][y];       return 0; }

Java

 // Java implementation of the approach class GFG {     final static int DIGITS = 10;     final static int MAX = 10000;     final static int MOD = 1000000007;           // To store the states of the dp     static int dp[][] = new int[MAX][DIGITS];     static int cnt[][] = new int[MAX][DIGITS];           // Function to fill the dp table     static void precompute()     {               // dp[i][j] : Sum of the digits of all         // possible good strings of length         // i that end with j         // cnt[i][j] : Count of the good strings         // of length i that end with j               // Sum of digits of the string of length         // 1 is i as i is only number in that string         // and count of good strings of length 1         // that end with i is also 1         for (int i = 0; i < DIGITS; i++)         {             dp[i] = i;             cnt[i] = 1;         }               for (int i = 2; i < MAX; i++)         {             for (int j = 0; j < DIGITS; j++)             {                 for (int k = 0; k < DIGITS; k++)                 {                           // Adjacent digits are different                     if (j != k)                     {                         dp[i][j] = dp[i][j] + (dp[i - 1][k] +                                              (cnt[i - 1][k] * j) % MOD)                                                                  % MOD;                         dp[i][j] %= MOD;                               // Increment the count as digit at                         // (i - 1)'th index is k and count                         // of good strings is equal to this                         // because at the end of the strings of                         // length (i - 1) we are just                         // putting digit j as the last digit                         cnt[i][j] += cnt[i - 1][k];                         cnt[i][j] %= MOD;                     }                 }             }         }     }           // Driver code     public static void main (String[] args)     {         int x = 6, y = 4;               precompute();               System.out.println(dp[x][y]);     } }   // This code is contributed by AnkitRai01

Python3

 # Python3 implementation of the approach DIGITS = 10; MAX = 10000; MOD = 1000000007;   # To store the states of the dp dp = [[0 for i in range(DIGITS)]          for i in range(MAX)]; cnt = [[0 for i in range(DIGITS)]           for i in range(MAX)];   # Function to fill the dp table def precompute():       # dp[i][j] : Sum of the digits of all     # possible good strings of length     # i that end with j     # cnt[i][j] : Count of the good strings     # of length i that end with j       # Sum of digits of the string of length     # 1 is i as i is only number in that string     # and count of good strings of length 1     # that end with i is also 1     for i in range(DIGITS):               dp[i] = i;         cnt[i] = 1;           for i in range(2, MAX):         for j in range(DIGITS):             for k in range(DIGITS):                                   # Adjacent digits are different                 if (j != k):                                       dp[i][j] = dp[i][j] + (dp[i - 1][k] +\                                          (cnt[i - 1][k] * j) % MOD) % MOD;                     dp[i][j] %= MOD;                       # Increment the count as digit at                     # (i - 1)'th index is k and count                     # of good strings is equal to this                     # because at the end of the strings of                     # length (i - 1) we are just                     # putting digit j as the last digit                     cnt[i][j] += cnt[i - 1][k];                     cnt[i][j] %= MOD;   # Driver code x = 6; y = 4;   precompute();   print(dp[x][y]);   # This code is contributed by 29AjayKumar

C#

 // C# implementation of the approach using System;       class GFG {     readonly static int DIGITS = 10;     readonly static int MAX = 10000;     readonly static int MOD = 1000000007;           // To store the states of the dp     static int [,]dp = new int[MAX, DIGITS];     static int [,]cnt = new int[MAX, DIGITS];           // Function to fill the dp table     static void precompute()     {               // dp[i][j] : Sum of the digits of all         // possible good strings of length         // i that end with j         // cnt[i][j] : Count of the good strings         // of length i that end with j               // Sum of digits of the string of length         // 1 is i as i is only number in that string         // and count of good strings of length 1         // that end with i is also 1         for (int i = 0; i < DIGITS; i++)         {             dp[1, i] = i;             cnt[1, i] = 1;         }               for (int i = 2; i < MAX; i++)         {             for (int j = 0; j < DIGITS; j++)             {                 for (int k = 0; k < DIGITS; k++)                 {                           // Adjacent digits are different                     if (j != k)                     {                         dp[i, j] = dp[i, j] + (dp[i - 1, k] +                                              (cnt[i - 1, k] * j) % MOD)                                                                  % MOD;                         dp[i, j] %= MOD;                               // Increment the count as digit at                         // (i - 1)'th index is k and count                         // of good strings is equal to this                         // because at the end of the strings of                         // length (i - 1) we are just                         // putting digit j as the last digit                         cnt[i, j] += cnt[i - 1, k];                         cnt[i, j] %= MOD;                     }                 }             }         }     }           // Driver code     public static void Main (String[] args)     {         int x = 6, y = 4;               precompute();               Console.WriteLine(dp[x,y]);     } }   // This code is contributed by Rajput-Ji

Javascript



Output:

1567751

Time Complexity: O(N)

My Personal Notes arrow_drop_up
Recommended Articles
Page :