Given a square matrix of size N*N, where each cell is associated with a specific cost. A path is defined as a specific sequence of cells which starts from top left cell move only right or down and ends on bottom right cell. We want to find a path with maximum average over all existing paths. Average is computed as total cost divided by number of cells visited in path.

Examples:

Input : Matrix = [1, 2, 3 4, 5, 6 7, 8, 9] Output : 5.8 Path with maximum average is, 1 -> 4 -> 7 -> 8 -> 9 Sum of the path is 29 and average is 29/5 = 5.8

One interesting observation is, the only allowed moves are down and right, we need N-1 down moves and N-1 right moves to reach destination (bottom rightmost). So any path from from top left corner to bottom right corner requires 2N – 1 cells. In average value, denominator is fixed and we need to just maximize numerator. Therefore we basically need to to find maximum sum path. Calculating maximum sum of path is a classic dynamic programming problem, if dp[i][j] represents maximum sum till cell (i, j) from (0, 0) then at each cell (i, j), we update dp[i][j] as below,

for all i, 1 <= i <= N dp[i][0] = dp[i-1][0] + cost[i][0]; for all j, 1 <= j <= j dp[0][j] = dp[0][j-1] + cost[0][j]; otherwise dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + cost[i][j];

Once we get maximum sum of all paths we will divide this sum by (2N – 1) and we will get our maximum average.

## C++

// C/C++ program to find maximum average cost path #include <bits/stdc++.h> using namespace std; // Maximum number of rows and/or columns const int M = 100; // method returns maximum average of all path of // cost matrix double maxAverageOfPath(int cost[M][M], int N) { int dp[N+1][N+1]; dp[0][0] = cost[0][0]; /* Initialize first column of total cost(dp) array */ for (int i = 1; i < N; i++) dp[i][0] = dp[i-1][0] + cost[i][0]; /* Initialize first row of dp array */ for (int j = 1; j < N; j++) dp[0][j] = dp[0][j-1] + cost[0][j]; /* Construct rest of the dp array */ for (int i = 1; i < N; i++) for (int j = 1; j <= N; j++) dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + cost[i][j]; // divide maximum sum by constant path // length : (2N - 1) for getting average return (double)dp[N-1][N-1] / (2*N-1); } /* Driver program to test above functions */ int main() { int cost[M][M] = { {1, 2, 3}, {6, 5, 4}, {7, 3, 9} }; printf("%f", maxAverageOfPath(cost, 3)); return 0; }

## Java

// JAVA Code for Path with maximum average // value class GFG { // method returns maximum average of all // path of cost matrix public static double maxAverageOfPath(int cost[][], int N) { int dp[][] = new int[N+1][N+1]; dp[0][0] = cost[0][0]; /* Initialize first column of total cost(dp) array */ for (int i = 1; i < N; i++) dp[i][0] = dp[i-1][0] + cost[i][0]; /* Initialize first row of dp array */ for (int j = 1; j < N; j++) dp[0][j] = dp[0][j-1] + cost[0][j]; /* Construct rest of the dp array */ for (int i = 1; i < N; i++) for (int j = 1; j < N; j++) dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]) + cost[i][j]; // divide maximum sum by constant path // length : (2N - 1) for getting average return (double)dp[N-1][N-1] / (2 * N - 1); } /* Driver program to test above function */ public static void main(String[] args) { int cost[][] = {{1, 2, 3}, {6, 5, 4}, {7, 3, 9}}; System.out.println(maxAverageOfPath(cost, 3)); } } // This code is contributed by Arnav Kr. Mandal.

Output:

5.2

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