Related Articles

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

• Difficulty Level : Hard
• Last Updated : 03 May, 2021

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.

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`

## Javascript

 ``
Output:
`150`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up