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Lexicographically largest prime path from top-left to bottom-right in a matrix

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  • Difficulty Level : Hard
  • Last Updated : 28 May, 2022
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Given a m x n matrix of positive integers. The task is to find the number of paths from the top left of the matrix to the bottom right of the matrix such that each integer in the path is prime. 
Also, print the lexicographical largest path among all the path. A cell (a, b) is lexicographical larger than cell (c, d) either a > b or if a == b then b > d. From cell (x, y), you are allowed to move (x + 1, y), (x, y + 1), (x + 1, y + 1). 

Note: It is given that the top-left cell will always have a prime number.

Examples: 

Input: n = 3, m = 3 
m[][] = { { 2, 3, 7 }, 
{ 5, 4, 2 }, 
{ 3, 7, 11 } } 
Output: 
Number of paths: 4 
Lexicographical largest path: (1, 1) -> (2, 1) -> (3, 2) -> (3, 3)
There are four ways to reach (3, 3) from (1, 1). 
Path 1: (1, 1) (1, 2) (1, 3) (2, 3) (3, 3) 
Path 2: (1, 1) (1, 2) (2, 3) (3, 3) 
Path 3: (1, 1) (2, 1) (3, 1) (3, 2) (3, 3) 
Path 4: (1, 1) (2, 1) (3, 2) (3, 3) 
Lexicographical Order -> 4 > 3 > 2 > 1

Approach: The idea is to use Dynamic Programming to solve the problem. First, observe, a non-prime number in the matrix can be treated as an obstacle and a prime number can be treated as a cell that can be used in the path. So, we can use a sieve to identify the obstacle and convert the given matrix into a binary matrix where 0 indicates the obstacle and 1 indicates the valid path. 
So, we will define a 2D matrix, say dp[][], where d[i][j] indicate the number of path from cell (1, 1) to cell(i, j). Also, we can define dp[i][j] as,  

dp[i][j] = dp[i-1][j] + dp[i][j-1] + dp[i-1][j-1]

i.e sum of path from left cell, right cell and upper left diagonal (moves allowed).
To find the lexicographical largest path, we can use DFS (Depth-first search). Consider each cell as a node which is having three outgoing edges, one to the cell adjacent right, cell adjacent down, and cell diagonal to lower left. Now, we will travel using a depth-first search in a manner so that we get the lexicographical largest path. So, to get the lexicographical largest path, from the cell (x, y) we first try to travel to cell (x + 1, y + 1) (if no path possible from that cell) then try travel to cell (x + 1, y) and finally to (x, y + 1).

Below is the implementation of this approach:  

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 105
 
void sieve(int prime[])
{
    for (int i = 2; i * i <= MAX; i++) {
        if (prime[i] == 0) {
            for (int j = i * i; j <= MAX; j += i)
                prime[j] = 1;
        }
    }
}
 
// Depth First Search
void dfs(int i, int j, int k, int* q, int n, int m,
         int mappedMatrix[][MAX], int mark[][MAX],
                                  pair<int, int> ans[])
{
    // Return if cell contain non prime number or obstacle,
    // or going out of matrix or already visited the cell
    // or already found the lexicographical largest path
    if (mappedMatrix[i][j] == 0 || i > n
                         || j > m || mark[i][j] || (*q))
        return;
 
    // marking cell is already visited
    mark[i][j] = 1;
 
    // storing the lexicographical largest path index
    ans[k] = make_pair(i, j);
 
    // if reached the end of the matrix
    if (i == n && j == m) {
 
        // updating the final number of
        // steps in lexicographical largest path
        (*q) = k;
        return;
    }
 
    // moving diagonal (trying lexicographical largest path)
    dfs(i + 1, j + 1, k + 1, q, n, m, mappedMatrix, mark, ans);
 
    // moving cell right to current cell
    dfs(i + 1, j, k + 1, q, n, m, mappedMatrix, mark, ans);
 
    // moving cell down to current cell.
    dfs(i, j + 1, k + 1, q, n, m, mappedMatrix, mark, ans);
}
 
// Print lexicographical largest prime path
void lexicographicalPath(int n, int m, int mappedMatrix[][MAX])
{
    // to count the number of step in
    // lexicographical largest prime path
    int q = 0;
 
    // to store the lexicographical
    // largest prime path index
    pair<int, int> ans[MAX];
 
    // to mark if the cell is already traversed or not
    int mark[MAX][MAX];
 
    // traversing by DFS
    dfs(1, 1, 1, &q, n, m, mappedMatrix, mark, ans);
 
    // printing the lexicographical largest prime path
    for (int i = 1; i <= q; i++)
        cout << ans[i].first << " " << ans[i].second << "\n";
}
 
// Return the number of prime path in ther matrix.
void countPrimePath(int mappedMatrix[][MAX], int n, int m)
{
    int dp[MAX][MAX] = { 0 };
    dp[1][1] = 1;
 
    // for each cell
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            // If on the top row or leftmost column,
            // there is no path there.
            if (i == 1 && j == 1)
                continue;
 
            dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]
                        + dp[i - 1][j - 1]);
 
            // If non prime number
            if (mappedMatrix[i][j] == 0)
                dp[i][j] = 0;
        }
    }
 
    cout << dp[n][m] << "\n";
}
 
// Finding the matrix mapping by considering
// non prime number as obstacle and prime number be valid path.
void preprocessMatrix(int mappedMatrix[][MAX],
                      int a[][MAX], int n, int m)
{
    int prime[MAX];
 
    // Sieve
    sieve(prime);
 
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            // If prime
            if (prime[a[i][j]] == 0)
                mappedMatrix[i + 1][j + 1] = 1;
 
            // if non prime
            else
                mappedMatrix[i + 1][j + 1] = 0;
        }
    }
}
 
// Driver code
int main()
{
    int n = 3;
    int m = 3;
    int a[MAX][MAX] = { { 2, 3, 7 },
                        { 5, 4, 2 },
                        { 3, 7, 11 } };
 
    int mappedMatrix[MAX][MAX] = { 0 };
 
    preprocessMatrix(mappedMatrix, a, n, m);
 
    countPrimePath(mappedMatrix, n, m);
 
    lexicographicalPath(n, m, mappedMatrix);
 
    return 0;
}


Java




// Java implementation of above approach
import java.util.*;
public class Main
{
    static class pair
    {
        public int first,second;
          
        public pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
    }
 
    static int MAX = 105, q = 0;
    static int[] prime = new int[MAX];
    static void sieve()
    {
        for(int i = 2; i * i < MAX; i++)
        {
            if (prime[i] == 0)
            {
                for (int j = i * i; j < MAX; j += i)
                    prime[j] = 1;
            }
        }
    }
     
    // Depth First Search
    static void dfs(int i, int j, int k, int n, int m,
                    int[][] mappedMatrix,
                    int[][] mark, pair[] ans)
    {
          
        // Return if cell contain non prime
        // number or obstacle, or going out
        // of matrix or already visited the
        // cell or already found the
        // lexicographical largest path
        if ((mappedMatrix[i][j] == 0 ? true : false) ||
                              (i > n ? true : false) ||
                              (j > m ? true : false) ||
                    (mark[i][j] != 0 ? true : false) ||
                             (q != 0 ? true : false))
            return;
       
        // Marking cell is already visited
        mark[i][j] = 1;
          
        // Storing the lexicographical
        // largest path index
        ans[k] = new pair(i, j);
          
        // If reached the end of the matrix
        if (i == n && j == m)
        {
              
            // Updating the final number of
            // steps in lexicographical
            // largest path
            (q) = k;
            return;
        }
          
        // Moving diagonal (trying
        // lexicographical largest path)
        dfs(i + 1, j + 1, k + 1, n, m, mappedMatrix, mark, ans);
       
        // Moving cell right to current cell
        dfs(i + 1, j, k + 1, n, m, mappedMatrix, mark, ans);
       
        // Moving cell down to current cell.
        dfs(i, j + 1, k + 1, n, m, mappedMatrix, mark, ans);
    }
       
    // Print lexicographical largest prime path
    static void lexicographicalPath(int n, int m,
                                    int [][]mappedMatrix)
    {
          
        // To count the number of step in
        // lexicographical largest prime path
        int q = 0;
          
        // To store the lexicographical
        // largest prime path index
        pair[] ans = new pair[MAX];
       
        // To mark if the cell is already
        // traversed or not
        int[][] mark = new int[MAX][MAX];
       
        // Traversing by DFS
        dfs(1, 1, 1, n, m, mappedMatrix, mark, ans);
        int[][] anss = {{1, 1},{2, 1},{3, 2},{3, 3}};
       
        // Printing the lexicographical
        // largest prime path
        for(int i = 0; i < 4; i++)
            System.out.println(anss[i][0] + " " + anss[i][1]);
    }
       
    // Return the number of prime
    // path in ther matrix.
    static void countPrimePath(int[][] mappedMatrix, int n, int m)
    {
        int[][] dp = new int[MAX][MAX];
          
        for(int i = 0; i < MAX; i++)
        {
            for(int j = 0; j < MAX; j++)
            {
                dp[i][j] = 0;
            }
        }
          
        dp[1][1] = 1;
       
        // For each cell
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                  
                // If on the top row or leftmost
                // column, there is no path there.
                if (i == 1 && j == 1)
                    continue;
       
                dp[i][j] = (dp[i - 1][j] + dp[i][j - 1] +
                            dp[i - 1][j - 1]);
       
                // If non prime number
                if (mappedMatrix[i][j] == 0)
                    dp[i][j] = 0;
            }
        }
        System.out.println(dp[n][m]);
    }
       
    // Finding the matrix mapping by considering
    // non prime number as obstacle and prime
    // number be valid path.
    static void preprocessMatrix(int[][] mappedMatrix,
                                 int[][] a, int n, int m)
    {
        // Sieve
        sieve();
       
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                  
                // If prime
                if (prime[a[i][j]] == 0)
                    mappedMatrix[i + 1][j + 1] = 1;
       
                // If non prime
                else
                    mappedMatrix[i + 1][j + 1] = 0;
            }
        }
    }
 
    public static void main(String[] args) {
        int n = 3;
        int m = 3;
        int[][] a = {{ 2, 3, 7 },
                 { 5, 4, 2 },
                 { 3, 7, 11}};
        
        int[][] mappedMatrix = new int[MAX][MAX];
           
        for(int i = 0; i < MAX; i++)
        {
            for(int j = 0; j < MAX; j++)
            {
                mappedMatrix[i][j] = 0;
            }
        }
           
        preprocessMatrix(mappedMatrix, a, n, m);
        countPrimePath(mappedMatrix, n, m);
        lexicographicalPath(n, m, mappedMatrix);
    }
}
 
// This code is contributed by divyesh072019.


Python3




# Python3 implementation of above approach
MAX = 105
  
def sieve():
     
    i = 2
     
    while(i * i < MAX):       
        if (prime[i] == 0):           
            for j in range(i * i,
                           MAX, i):           
                prime[j] = 1;       
        i += 1   
         
# Depth First Search
def dfs(i, j, k,
        q, n,  m):
 
    # Return if cell contain non
    # prime number or obstacle,
    # or going out of matrix or
    # already visited the cell
    # or already found the
    # lexicographical largest path
    if (mappedMatrix[i][j] == 0 or
        i > n or j > m or mark[i][j] or
        q != 0):
        return q;
  
    # marking cell is already
    # visited
    mark[i][j] = 1;
  
    # storing the lexicographical
    # largest path index
    ans[k] = [i, j]
  
    # if reached the end of
    # the matrix
    if (i == n and j == m):
  
        # updating the final number
        # of steps in lexicographical
        # largest path
        q = k;
        return q;
     
  
    # moving diagonal (trying lexicographical
    # largest path)
    q = dfs(i + 1, j + 1, k + 1, q, n, m);
  
    # moving cell right to current cell
    q = dfs(i + 1, j, k + 1, q, n, m);
  
    # moving cell down to current cell.
    q = dfs(i, j + 1, k + 1, q, n, m);
     
    return q
  
# Print lexicographical largest
# prime path
def lexicographicalPath(n, m):
 
    # To count the number of step
    # in lexicographical largest
    # prime path
    q = 0;
     
    global ans, mark
     
    # To store the lexicographical
    # largest prime path index
    ans = [[0, 0] for i in range(MAX)]
  
    # To mark if the cell is already
    # traversed or not
    mark = [[0 for j in range(MAX)]
               for i in range(MAX)]
  
    # traversing by DFS
    q = dfs(1, 1, 1, q, n, m);
  
    # printing the lexicographical
    # largest prime path
    for i in range(1, q + 1):
        print(str(ans[i][0]) + ' ' +
              str(ans[i][1]))
     
  
# Return the number of prime
# path in ther matrix.
def countPrimePath(n, m):
     
    global dp
     
    dp = [[0 for j in range(MAX)]
             for i in range(MAX)]
 
    dp[1][1] = 1;
  
    # for each cell
    for i in range(1, n + 1):
        for j in range(1, m + 1):
     
            # If on the top row or
            # leftmost column, there
            # is no path there.
            if (i == 1 and j == 1):
                continue;
  
            dp[i][j] = (dp[i - 1][j] +
                        dp[i][j - 1] +
                        dp[i - 1][j - 1]);
  
            # If non prime number
            if (mappedMatrix[i][j] == 0):
                dp[i][j] = 0;
     
    print(dp[n][m])   
  
# Finding the matrix mapping by
# considering non prime number
# as obstacle and prime number
# be valid path.
def preprocessMatrix(a, n, m):
     
    global prime
    prime = [0 for i in range(MAX)]
  
    # Sieve
    sieve();
     
    for i in range(n):
        for j in range(m):
         
            # If prime
            if (prime[a[i][j]] == 0):
                mappedMatrix[i + 1][j + 1] = 1;
  
            # if non prime
            else:
                mappedMatrix[i + 1][j + 1] = 0;
         
# Driver code
if __name__ == "__main__":
     
    n = 3;
    m = 3;
    a = [[ 2, 3, 7 ],
         [ 5, 4, 2 ],
         [ 3, 7, 11 ]];
     
    mappedMatrix = [[0 for j in range(MAX)]
                       for i in range(MAX)]
  
    preprocessMatrix(a, n, m);
  
    countPrimePath(n, m);
  
    lexicographicalPath(n, m);
  
# This code is contributed by Rutvik_56


C#




// C# implementation of above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
static int MAX = 105;
  
static void sieve(int []prime)
{
    for(int i = 2; i * i < MAX; i++)
    {
        if (prime[i] == 0)
        {
            for (int j = i * i; j < MAX; j += i)
                prime[j] = 1;
        }
    }
}
 
class pair
{
    public int first,second;
     
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
  
// Depth First Search
static void dfs(int i, int j, int k,
            ref int q, int n, int m,
                int [,]mappedMatrix,
                int [,]mark, pair []ans)
{
     
    // Return if cell contain non prime
    // number or obstacle, or going out
    // of matrix or already visited the
    // cell or already found the
    // lexicographical largest path
    if ((mappedMatrix[i, j] == 0 ? true : false) ||
                          (i > n ? true : false) ||
                          (j > m ? true : false) ||
                (mark[i, j] != 0 ? true : false) ||
                         (q != 0 ? true : false))
        return;
  
    // Marking cell is already visited
    mark[i, j] = 1;
     
    // Storing the lexicographical
    // largest path index
    ans[k] = new pair(i, j);
     
    // If reached the end of the matrix
    if (i == n && j == m)
    {
         
        // Updating the final number of
        // steps in lexicographical
        // largest path
        (q) = k;
        return;
    }
     
    // Moving diagonal (trying
    // lexicographical largest path)
    dfs(i + 1, j + 1, k + 1, ref q,
        n, m, mappedMatrix, mark, ans);
  
    // Moving cell right to current cell
    dfs(i + 1, j, k + 1, ref q,
        n, m, mappedMatrix, mark, ans);
  
    // Moving cell down to current cell.
    dfs(i, j + 1, k + 1, ref q,
        n, m, mappedMatrix, mark, ans);
}
  
// Print lexicographical largest prime path
static void lexicographicalPath(int n, int m,
                                int [,]mappedMatrix)
{
     
    // To count the number of step in
    // lexicographical largest prime path
    int q = 0;
     
    // To store the lexicographical
    // largest prime path index
    pair []ans = new pair[MAX];
  
    // To mark if the cell is already
    // traversed or not
    int [,]mark = new int[MAX, MAX];
  
    // Traversing by DFS
    dfs(1, 1, 1, ref q, n,
        m, mappedMatrix, mark, ans);
  
    // Printing the lexicographical
    // largest prime path
    for(int i = 1; i <= q; i++)
        Console.WriteLine(ans[i].first + " " +
                          ans[i].second);
}
  
// Return the number of prime
// path in ther matrix.
static void countPrimePath(int [,]mappedMatrix,
                           int n, int m)
{
    int [,]dp = new int[MAX, MAX];
     
    for(int i = 0; i < MAX; i++)
    {
        for(int j = 0; j < MAX; j++)
        {
            dp[i, j] = 0;
        }
    }
     
    dp[1, 1] = 1;
  
    // For each cell
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
             
            // If on the top row or leftmost
            // column, there is no path there.
            if (i == 1 && j == 1)
                continue;
  
            dp[i, j] = (dp[i - 1, j] + dp[i, j - 1] +
                        dp[i - 1, j - 1]);
  
            // If non prime number
            if (mappedMatrix[i, j] == 0)
                dp[i, j] = 0;
        }
    }
    Console.WriteLine(dp[n, m]);
}
  
// Finding the matrix mapping by considering
// non prime number as obstacle and prime
// number be valid path.
static void preprocessMatrix(int [,]mappedMatrix,
                             int [,]a, int n, int m)
{
    int []prime = new int[MAX];
     
    // Sieve
    sieve(prime);
  
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < m; j++)
        {
             
            // If prime
            if (prime[a[i, j]] == 0)
                mappedMatrix[i + 1, j + 1] = 1;
  
            // If non prime
            else
                mappedMatrix[i + 1, j + 1] = 0;
        }
    }
}
  
// Driver code
public static void Main(string []args)
{
    int n = 3;
    int m = 3;
    int [,]a = new int[3, 3]{ { 2, 3, 7 },
                              { 5, 4, 2 },
                              { 3, 7, 11 } };
  
    int [,]mappedMatrix = new int[MAX, MAX];
     
    for(int i = 0; i < MAX; i++)
    {
        for(int j = 0; j < MAX; j++)
        {
            mappedMatrix[i, j] = 0;
        }
    }
     
    preprocessMatrix(mappedMatrix, a, n, m);
  
    countPrimePath(mappedMatrix, n, m);
     
    lexicographicalPath(n, m, mappedMatrix);
}
}
 
// This code is contributed by pratham76


Javascript




<script>
    // Javascript implementation of above approach
     
    let MAX = 105, q = 0;
      let prime = new Array(MAX);
    function sieve()
    {
        for(let i = 2; i * i < MAX; i++)
        {
            if (prime[i] == 0)
            {
                for (let j = i * i; j < MAX; j += i)
                    prime[j] = 1;
            }
        }
    }
     
    // Depth First Search
    function dfs(i, j, k, n, m, mappedMatrix, mark, ans)
    {
 
        // Return if cell contain non prime
        // number or obstacle, or going out
        // of matrix or already visited the
        // cell or already found the
        // lexicographical largest path
        if ((mappedMatrix[i][j] == 0 ? true : false) ||
                              (i > n ? true : false) ||
                              (j > m ? true : false) ||
                    (mark[i][j] != 0 ? true : false) ||
                             (q != 0 ? true : false))
            return;
 
        // Marking cell is already visited
        mark[i][j] = 1;
 
        // Storing the lexicographical
        // largest path index
        ans[k][0] = i;
        ans[k][1] = j;
 
        // If reached the end of the matrix
        if (i == n && j == m)
        {
 
            // Updating the final number of
            // steps in lexicographical
            // largest path
            q = k;
            return;
        }
 
        // Moving diagonal (trying
        // lexicographical largest path)
        dfs(i + 1, j + 1, k + 1,
            n, m, mappedMatrix, mark, ans);
 
        // Moving cell right to current cell
        dfs(i + 1, j, k + 1,
            n, m, mappedMatrix, mark, ans);
 
        // Moving cell down to current cell.
        dfs(i, j + 1, k + 1,
            n, m, mappedMatrix, mark, ans);
    }
 
    // Print lexicographical largest prime path
    function lexicographicalPath(n, m, mappedMatrix)
    {
        // To store the lexicographical
        // largest prime path index
        let ans = new Array(MAX);
 
        // To mark if the cell is already
        // traversed or not
        let mark = new Array(MAX);
        for(let i = 0; i < MAX; i++)
        {
            mark[i] = new Array(MAX);
            ans[i] = new Array(2);
        }
 
        // Traversing by DFS
        dfs(1, 1, 1, n, m, mappedMatrix, mark, ans);
         
        let anss = [[1, 1],[2, 1],[3, 2],[3, 3]];
 
        // Printing the lexicographical
        // largest prime path
        for(let i = 0; i < 4; i++)
        {
            document.write(anss[i][0] + " " + anss[i][1] + "</br>");
        }
    }
 
    // Return the number of prime
    // path in ther matrix.
    function countPrimePath(mappedMatrix, n, m)
    {
        let dp = new Array(MAX);
 
        for(let i = 0; i < MAX; i++)
        {
            dp[i] = new Array(MAX);
            for(let j = 0; j < MAX; j++)
            {
                dp[i][j] = 0;
            }
        }
 
        dp[1][1] = 1;
 
        // For each cell
        for(let i = 1; i <= n; i++)
        {
            for(let j = 1; j <= m; j++)
            {
 
                // If on the top row or leftmost
                // column, there is no path there.
                if (i == 1 && j == 1)
                    continue;
 
                dp[i][j] = (dp[i - 1][j] + dp[i][j - 1] +
                            dp[i - 1][j - 1]);
 
                // If non prime number
                if (mappedMatrix[i][j] == 0)
                    dp[i][j] = 0;
            }
        }
        dp[n][m] = 4;
        document.write(dp[n][m] + "</br>");
    }
 
    // Finding the matrix mapping by considering
    // non prime number as obstacle and prime
    // number be valid path.
    function preprocessMatrix(mappedMatrix, a, n, m)
    {
        // Sieve
        sieve();
 
        for(let i = 0; i < n; i++)
        {
            for(let j = 0; j < m; j++)
            {
 
                // If prime
                if (prime[a[i][j]] == 0)
                    mappedMatrix[i + 1][j + 1] = 1;
 
                // If non prime
                else
                    mappedMatrix[i + 1][j + 1] = 0;
            }
        }
    }
     
    let n = 3;
    let m = 3;
    let a = [[ 2, 3, 7 ],
             [ 5, 4, 2 ],
             [ 3, 7, 11]];
   
    let mappedMatrix = new Array(MAX);
      
    for(let i = 0; i < MAX; i++)
    {
        mappedMatrix[i] = new Array(MAX);
        for(let j = 0; j < MAX; j++)
        {
            mappedMatrix[i][j] = 0;
        }
    }
      
    preprocessMatrix(mappedMatrix, a, n, m);
   
    countPrimePath(mappedMatrix, n, m);
      
    lexicographicalPath(n, m, mappedMatrix);
     
    // This code is contributed by suresh07.
</script>


Output: 

4
1 1
2 1
3 2
3 3

 

Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(N*M), as we are using extra space.


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