# Sum of cost of all paths to reach a given cell in a Matrix

Given a matrix grid[][] and two integers M and N, the task is to find the sum of cost of all possible paths from the (0, 0) to (M, N) by moving a cell down or right. Cost of each path is defined as the sum of values of the cells visited in the path.

Examples:

Input: M = 1, N = 1, grid[][] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Output: 18
Explanation:
There are only 2 ways to reach (1, 1)
Path 1: (0, 0) => (0, 1) => (1, 1)
Path cost = 1 + 2 + 5 = 8
Path 2: (0, 0) => (1, 0) => (1, 1)
Path cost = 1 + 4 + 5 = 10
Total Path Sum = 8 + 10 = 18

Input: M = 2, N = 2, grid = { {1, 1, 1}, {1, 1, 1}, {1, 1, 1} }
Output: 30
Explanation:
Sum of path cost of all path is 30.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to find the contribution of each cell of the matrix for reaching (M, N), that is, the contribution of the every i and j, where 0 <= i <= M and 0 <= j <= N.
Below is the illustration of the contribution of each cell to all paths from (0, 0) to (M, N) through the respective cells:

Number of ways to reach (M, N) from (0, 0) = Number of ways to reach (M, N) from (0, 0) via (i, j) = Therefore, Contribution of each grid (i, j) is = Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the ` `// sum of cost of all paths ` `// to reach (M, N) ` ` `  `#include ` `using` `namespace` `std; ` ` `  `const` `int` `Col = 3; ` `int` `fact(``int` `n); ` ` `  `// Function for computing ` `// combination ` `int` `nCr(``int` `n, ``int` `r) ` `{ ` `    ``return` `fact(n) / (fact(r) ` `                      ``* fact(n - r)); ` `} ` ` `  `// Function to find the ` `// factorial of N ` `int` `fact(``int` `n) ` `{ ` `    ``int` `res = 1; ` ` `  `    ``// Loop to find the factorial ` `    ``// of a given number ` `    ``for` `(``int` `i = 2; i <= n; i++) ` `        ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function for coumputing the ` `// sum of all path cost ` `int` `sumPathCost(``int` `grid[][Col], ` `                ``int` `m, ``int` `n) ` `{ ` `    ``int` `sum = 0, count; ` ` `  `    ``// Loop to find the contribution ` `    ``// of each (i, j) in the all possible ` `    ``// ways ` `    ``for` `(``int` `i = 0; i <= m; i++) { ` `        ``for` `(``int` `j = 0; j <= n; j++) { ` ` `  `            ``// Count number of ` `            ``// times (i, j) visited ` `            ``count ` `                ``= nCr(i + j, i) ` `                  ``* nCr(m + n - i - j, m - i); ` ` `  `            ``// Add the contribution of ` `            ``// grid[i][j] in the result ` `            ``sum += count * grid[i][j]; ` `        ``} ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `m = 2; ` `    ``int` `n = 2; ` `    ``int` `grid[][Col] = { { 1, 2, 3 }, ` `                        ``{ 4, 5, 6 }, ` `                        ``{ 7, 8, 9 } }; ` ` `  `    ``// Function Call ` `    ``cout << sumPathCost(grid, m, n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the ` `// sum of cost of all paths ` `// to reach (M, N) ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `static` `int` `Col = ``3``; ` ` `  `// Function for computing ` `// combination ` `static` `int` `nCr(``int` `n, ``int` `r) ` `{ ` `    ``return` `fact(n) / (fact(r) *  ` `                      ``fact(n - r)); ` `} ` ` `  `// Function to find the ` `// factorial of N ` `static` `int` `fact(``int` `n) ` `{ ` `    ``int` `res = ``1``; ` ` `  `    ``// Loop to find the factorial ` `    ``// of a given number ` `    ``for``(``int` `i = ``2``; i <= n; i++) ` `       ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function for coumputing the ` `// sum of all path cost ` `static` `int` `sumPathCost(``int` `grid[][], ` `                       ``int` `m, ``int` `n) ` `{ ` `    ``int` `sum = ``0``, count; ` ` `  `    ``// Loop to find the contribution ` `    ``// of each (i, j) in the all possible ` `    ``// ways ` `    ``for``(``int` `i = ``0``; i <= m; i++) ` `    ``{ ` `       ``for``(``int` `j = ``0``; j <= n; j++) ` `       ``{ ` `           `  `          ``// Count number of ` `          ``// times (i, j) visited ` `          ``count = nCr(i + j, i) *  ` `                  ``nCr(m + n - i - j, m - i); ` `           `  `          ``// Add the contribution of ` `          ``// grid[i][j] in the result ` `          ``sum += count * grid[i][j]; ` `       ``} ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `m = ``2``; ` `    ``int` `n = ``2``; ` `    ``int` `grid[][] = { { ``1``, ``2``, ``3` `}, ` `                     ``{ ``4``, ``5``, ``6` `}, ` `                     ``{ ``7``, ``8``, ``9` `} }; ` ` `  `    ``// Function Call ` `    ``System.out.println(sumPathCost(grid, m, n)); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 implementation to find the sum  ` `# of cost of all paths to reach (M, N) ` ` `  `Col ``=` `3``; ` ` `  `# Function for computing ` `# combination ` `def` `nCr(n, r): ` `     `  `    ``return` `fact(n) ``/` `(fact(r) ``*` `                      ``fact(n ``-` `r)); ` ` `  `# Function to find the ` `# factorial of N ` `def` `fact(n): ` `     `  `    ``res ``=` `1``; ` ` `  `    ``# Loop to find the factorial ` `    ``# of a given number ` `    ``for` `i ``in` `range``(``2``, n ``+` `1``): ` `        ``res ``=` `res ``*` `i; ` `    ``return` `res; ` ` `  `# Function for coumputing the ` `# sum of all path cost ` `def` `sumPathCost(grid, m, n): ` `     `  `    ``sum` `=` `0``; ` `    ``count ``=` `0``; ` ` `  `    ``# Loop to find the contribution ` `    ``# of each (i, j) in the all possible ` `    ``# ways ` `    ``for` `i ``in` `range``(``0``, m ``+` `1``): ` `        ``for` `j ``in` `range``(``0``, n ``+` `1``): ` `             `  `            ``# Count number of ` `            ``# times (i, j) visited ` `            ``count ``=` `(nCr(i ``+` `j, i) ``*`  `                     ``nCr(m ``+` `n ``-` `i ``-` `j, m ``-` `i)); ` ` `  `            ``# Add the contribution of ` `            ``# grid[i][j] in the result ` `            ``sum` `+``=` `count ``*` `grid[i][j]; ` ` `  `    ``return` `sum``; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``m ``=` `2``; ` `    ``n ``=` `2``; ` `    ``grid ``=` `[ [ ``1``, ``2``, ``3` `],  ` `             ``[ ``4``, ``5``, ``6` `], ` `             ``[ ``7``, ``8``, ``9` `] ]; ` ` `  `    ``# Function Call ` `    ``print``(``int``(sumPathCost(grid, m, n))); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# implementation to find the ` `// sum of cost of all paths ` `// to reach (M, N) ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function for computing ` `// combination ` `static` `int` `nCr(``int` `n, ``int` `r) ` `{ ` `    ``return` `fact(n) / (fact(r) *  ` `                      ``fact(n - r)); ` `} ` ` `  `// Function to find the ` `// factorial of N ` `static` `int` `fact(``int` `n) ` `{ ` `    ``int` `res = 1; ` ` `  `    ``// Loop to find the factorial ` `    ``// of a given number ` `    ``for``(``int` `i = 2; i <= n; i++) ` `       ``res = res * i; ` `    ``return` `res; ` `} ` ` `  `// Function for coumputing the ` `// sum of all path cost ` `static` `int` `sumPathCost(``int` `[,]grid, ` `                       ``int` `m, ``int` `n) ` `{ ` `    ``int` `sum = 0, count; ` ` `  `    ``// Loop to find the contribution ` `    ``// of each (i, j) in the all possible ` `    ``// ways ` `    ``for``(``int` `i = 0; i <= m; i++) ` `    ``{ ` `       ``for``(``int` `j = 0; j <= n; j++) ` `       ``{ ` `            `  `          ``// Count number of ` `          ``// times (i, j) visited ` `          ``count = nCr(i + j, i) *  ` `                  ``nCr(m + n - i - j, m - i); ` `                   `  `          ``// Add the contribution of ` `          ``// grid[i][j] in the result ` `          ``sum += count * grid[i, j]; ` `       ``} ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `m = 2; ` `    ``int` `n = 2; ` `    ``int` `[, ]grid = { { 1, 2, 3 }, ` `                     ``{ 4, 5, 6 }, ` `                     ``{ 7, 8, 9 } }; ` ` `  `    ``// Function Call ` `    ``Console.Write(sumPathCost(grid, m, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```150
```

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Improved By : offbeat, Code_Mech, 29AjayKumar