 Open in App
Not now

# Path with maximum average value

• Difficulty Level : Easy
• Last Updated : 30 Dec, 2022

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 that starts from the top-left cell move only right or down and ends on bottom right cell. We want to find a path with the maximum average over all existing paths. Average is computed as total cost divided by the number of cells visited in the 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 the destination (bottom rightmost). So any path from top left corner to bottom right corner requires 2N – 1 cells. In average value, the denominator is fixed and we need to just maximize numerator. Therefore we basically need to find the 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] = dp[i-1] + cost[i];
for all j, 1 <= j <= N
dp[j] = dp[j-1] + cost[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.

Implementation:

## C++

 `//C/C++ program to find maximum average cost path``#include ``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 = cost;` `    ``/* Initialize first column of total cost(dp) array */``    ``for` `(``int` `i = 1; i < N; i++)``        ``dp[i] = dp[i-1] + cost[i];` `    ``/* Initialize first row of dp array */``    ``for` `(``int` `j = 1; j < N; j++)``        ``dp[j] = dp[j-1] + cost[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``import` `java.io.*;` `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.`

## Python3

 `# Python program to find``# maximum average cost path` `# Maximum number of rows``# and/or columns``M ``=` `100` `# method returns maximum average of``# all path of cost matrix``def` `maxAverageOfPath(cost, N):``    ` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(N ``+` `1``)] ``for` `j ``in` `range``(N ``+` `1``)]``    ``dp[``0``][``0``] ``=` `cost[``0``][``0``]` `    ``# Initialize first column of total cost(dp) array``    ``for` `i ``in` `range``(``1``, N):``        ``dp[i][``0``] ``=` `dp[i ``-` `1``][``0``] ``+` `cost[i][``0``]` `    ``# Initialize first row of dp array``    ``for` `j ``in` `range``(``1``, N):``        ``dp[``0``][j] ``=` `dp[``0``][j ``-` `1``] ``+` `cost[``0``][j]` `    ``# Construct rest of the dp array``    ``for` `i ``in` `range``(``1``, N):``        ``for` `j ``in` `range``(``1``, N):``            ``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` `dp[N ``-` `1``][N ``-` `1``] ``/` `(``2` `*` `N ``-` `1``)` `# Driver program to test above function``cost ``=` `[[``1``, ``2``, ``3``],``        ``[``6``, ``5``, ``4``],``        ``[``7``, ``3``, ``9``]]` `print``(maxAverageOfPath(cost, ``3``))` `# This code is contributed by Soumen Ghosh.`

## C#

 `// C# Code for Path with maximum average``// value``using` `System;``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 Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[,]cost = {{1, 2, 3},``                       ``{6, 5, 4},``                       ``{7, 3, 9}};``                ` `        ``Console.Write(maxAverageOfPath(cost, 3));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``

Output

`5.200000`

Time complexity: O(N2) for  given input N
Auxiliary Space: O(N2) for given input N.

Method – 2: Without using Extra N*N space

We can use input cost array as a dp to store the ans. so this way we don’t need an extra dp array or no that extra space.\

One observation is that the only allowed moves are down and right, we need N-1 down moves and N-1 right moves to reach the destination (bottom rightmost). So any path from top left corner to bottom right corner requires 2N – 1 cell. In average value, the denominator is fixed and we need to just maximize numerator. Therefore we basically need to find the maximum sum path. Calculating maximum sum of path is a classic dynamic programming problem, also we don’t need any prev cost[i][j] value after calculating dp[i][j] so we can modifies the cost[i][j] value such that we don’t need extra space for dp[i][j].

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

Below is the implementation of the above approach:

## C++

 `// C++ program to find maximum average cost path``#include ``using` `namespace` `std;` `// Method returns maximum average of all path of cost matrix``double` `maxAverageOfPath(vector>cost)``{``    ``int` `N = cost.size();` `    ``// Initialize first column of total cost array``    ``for` `(``int` `i = 1; i < N; i++)``        ``cost[i] = cost[i] + cost[i - 1];` `    ``// Initialize first row of array``    ``for` `(``int` `j = 1; j < N; j++)``        ``cost[j] = cost[j - 1] + cost[j];` `    ``// Construct rest of the array``    ``for` `(``int` `i = 1; i < N; i++)``        ``for` `(``int` `j = 1; j <= N; j++)``            ``cost[i][j] = max(cost[i - 1][j], cost[i][j - 1]) + cost[i][j];` `    ``// divide maximum sum by constant path``    ``// length : (2N - 1) for getting average``    ``return` `(``double``)cost[N - 1][N - 1] / (2 * N - 1);``}` `// Driver program``int` `main()``{``    ``vector> cost = {{1, 2, 3},``        ``{6, 5, 4},``        ``{7, 3, 9}``    ``};``    ``cout << maxAverageOfPath(cost);``    ``return` `0;``}`

## Java

 `// Java program to find maximum average cost path``import` `java.io.*;` `class` `GFG {` `  ``// Method returns maximum average of all path of cost``  ``// matrix``  ``static` `double` `maxAverageOfPath(``int``[][] cost)``  ``{``    ``int` `N = cost.length;` `    ``// Initialize first column of total cost array``    ``for` `(``int` `i = ``1``; i < N; i++)``      ``cost[i][``0``] = cost[i][``0``] + cost[i - ``1``][``0``];` `    ``// Initialize first row of array``    ``for` `(``int` `j = ``1``; j < N; j++)``      ``cost[``0``][j] = cost[``0``][j - ``1``] + cost[``0``][j];` `    ``// Construct rest of the array``    ``for` `(``int` `i = ``1``; i < N; i++)``      ``for` `(``int` `j = ``1``; j < N; j++)``        ``cost[i][j] = Math.max(cost[i - ``1``][j],``                              ``cost[i][j - ``1``])``        ``+ cost[i][j];` `    ``// divide maximum sum by constant path``    ``// length : (2N - 1) for getting average``    ``return` `(``double``)cost[N - ``1``][N - ``1``] / (``2` `* N - ``1``);``  ``}` `  ``// Driver program``  ``public` `static` `void` `main(String[] args)``  ``{``    ``int``[][] cost``      ``= { { ``1``, ``2``, ``3` `}, { ``6``, ``5``, ``4` `}, { ``7``, ``3``, ``9` `} };``    ``System.out.println(maxAverageOfPath(cost));``  ``}``}` `// This code is contributed by karandeep1234`

## Python3

 `# Python program to find maximum average cost path``from` `typing ``import` `List` `def` `maxAverageOfPath(cost: ``List``[``List``[``int``]]) ``-``> ``float``:``    ``N ``=` `len``(cost)` `    ``# Initialize first column of total cost array``    ``for` `i ``in` `range``(``1``, N):``        ``cost[i][``0``] ``=` `cost[i][``0``] ``+` `cost[i ``-` `1``][``0``]` `    ``# Initialize first row of array``    ``for` `j ``in` `range``(``1``, N):``        ``cost[``0``][j] ``=` `cost[``0``][j ``-` `1``] ``+` `cost[``0``][j]` `    ``# Construct rest of the array``    ``for` `i ``in` `range``(``1``, N):``        ``for` `j ``in` `range``(``1``, N):``            ``cost[i][j] ``=` `max``(cost[i ``-` `1``][j], cost[i][j ``-` `1``]) ``+` `cost[i][j]` `    ``# divide maximum sum by constant path``    ``# length : (2N - 1) for getting average``    ``return` `cost[N ``-` `1``][N ``-` `1``] ``/` `(``2` `*` `N ``-` `1``)` `# Driver program``def` `main():``    ``cost ``=` `[[``1``, ``2``, ``3``],``            ``[``6``, ``5``, ``4``],``            ``[``7``, ``3``, ``9``]]``    ``print``(maxAverageOfPath(cost))` `if` `__name__ ``=``=` `'__main__'``:``    ``main()`

## C#

 `// C# program to find maximum average cost path``using` `System;``class` `GFG {` `  ``// Method returns maximum average of all path of cost``  ``// matrix``  ``static` `double` `maxAverageOfPath(``int``[, ] cost)``  ``{``    ``int` `N = cost.GetLength(0);` `    ``// Initialize first column of total cost array``    ``for` `(``int` `i = 1; i < N; i++)``      ``cost[i, 0] = cost[i, 0] + cost[i - 1, 0];` `    ``// Initialize first row of array``    ``for` `(``int` `j = 1; j < N; j++)``      ``cost[0, j] = cost[0, j - 1] + cost[0, j];` `    ``// Construct rest of the array``    ``for` `(``int` `i = 1; i < N; i++)``      ``for` `(``int` `j = 1; j < N; j++)``        ``cost[i, j] = Math.Max(cost[i - 1, j],``                              ``cost[i, j - 1])``        ``+ cost[i, j];` `    ``// divide maximum sum by constant path``    ``// length : (2N - 1) for getting average``    ``return` `(``double``)cost[N - 1, N - 1] / (2 * N - 1);``  ``}` `  ``// Driver program``  ``static` `void` `Main(``string``[] args)``  ``{``    ``int``[, ] cost``      ``= { { 1, 2, 3 }, { 6, 5, 4 }, { 7, 3, 9 } };``    ``Console.WriteLine(maxAverageOfPath(cost));``  ``}``}` `// This code is contributed by karandeep1234`

## Javascript

 `// Method returns maximum average of all path of cost matrix``function` `maxAverageOfPath(cost)``{``    ``let N = cost.length;` `    ``// Initialize first column of total cost array``    ``for` `(let i = 1; i < N; i++)``        ``cost[i] = cost[i] + cost[i - 1];` `    ``// Initialize first row of array``    ``for` `(let j = 1; j < N; j++)``        ``cost[j] = cost[j - 1] + cost[j];` `    ``// Construct rest of the array``    ``for` `(let i = 1; i < N; i++)``        ``for` `(let j = 1; j <= N; j++)``            ``cost[i][j] = Math.max(cost[i - 1][j], cost[i][j - 1]) + cost[i][j];` `    ``// divide maximum sum by constant path``    ``// length : (2N - 1) for getting average``    ``return` `(cost[N - 1][N - 1]) / (2.0 * N - 1);``}` `// Driver program``let cost = [[1, 2, 3],``        ``[6, 5, 4],``        ``[7, 3, 9]];``console.log(maxAverageOfPath(cost))` `// This code is contributed by karandeep1234.`

Output

`5.2`

Time Complexity: O(N*N)
Auxiliary Space: O(1)

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up