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 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 prime number can be treated as a cell that can be used in the path. So, we can use sieve to identify the obstacle and convert the given matrix into a binary matrix where 0 indicate the obstacle and 1 indicate 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 you 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 depth-first search in a manner so that we get lexicographical largest path. So, to get lexicographical largest path, from 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++ 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 larget 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

4 1 1 2 1 3 2 3 3

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