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 =
![grid[i][j] * \binom{m+n-i-j}{m-i} * \binom{i+j}{i}](https://neveropen.co.za/wp-content/uploads/2023/10/neveropen_quicklatex.com-010913adcdf5e09bf69b8691d400b292_l3.png "Rendered by QuickLaTeX.com")
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 <iostream>using namespace std;const int Col = 3;int fact(int n);// Function for computing// combinationint nCr(int n, int r){ return fact(n) / (fact(r) * fact(n - r));}// Function to find the// factorial of Nint 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 costint 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 Codeint 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// combinationstatic int nCr(int n, int r){ return fact(n) / (fact(r) * fact(n - r));}// Function to find the// factorial of Nstatic 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 coststatic 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 codepublic 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# combinationdef nCr(n, r): return fact(n) / (fact(r) * fact(n - r));# Function to find the# factorial of Ndef 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 costdef 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 codeif __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// combinationstatic int nCr(int n, int r){ return fact(n) / (fact(r) * fact(n - r));}// Function to find the// factorial of Nstatic 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 coststatic 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 codepublic 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
<script>// Javascript implementation to find the// sum of cost of all paths// to reach (M, N)var Col = 3;// Function for computing// combinationfunction nCr(n, r){ return fact(n) / (fact(r) * fact(n - r));}// Function to find the// factorial of Nfunction fact(n){ var res = 1; // Loop to find the factorial // of a given number for (var i = 2; i <= n; i++) res = res * i; return res;}// Function for coumputing the// sum of all path costfunction sumPathCost(grid, m, n){ var sum = 0, count; // Loop to find the contribution // of each (i, j) in the all possible // ways for (var i = 0; i <= m; i++) { for (var 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 Codevar m = 2;var n = 2;var grid = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ];// Function Calldocument.write( sumPathCost(grid, m, n));</script> |
Output:
150
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