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Maximum sum path in a matrix from top to bottom and back
• Difficulty Level : Hard
• Last Updated : 24 May, 2019

Given a matrix of dimension N * M. The task is find the maximum sum of path from arr to arr[N – 1][M – 1] and back from arr[N – 1][M – 1] to arr.

On the path from arr to arr[N – 1][M – 1], you can traverse in down and right directions and on the path from arr[N – 1][M – 1] to arr, you can traverse in up and left directions.

Note: Both the path must not be equal i.e. there has to be at least one cell arr[i][j] which is not common in both the paths.

Examples:

```Input:
mat[][]= {{1, 0, 3, -1},
{3, 5, 1, -2},
{-2, 0, 1, 1},
{2, 1, -1, 1}}
Output: 16 Maximum sum on path from arr to arr
= 1 + 3 + 5 + 1 + 1 + 1 + 1 = 13
Maximum sum on path from arr to arr = 3
Total path sum = 13 + 3 = 16

Input:
mat[][]= {{1, 0},
{1, 1}}
Output: 3
```

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

Approach: This problem is somewhat similar to Minimum Cost Path problem except that in the present problem, two paths with maximum sum are to be found. Also, we need to take care that cells on both paths contribute only once to the sum.
First thing to notice is that path from arr[N – 1][M – 1] to arr is nothing but another path from arr to arr[N – 1][M – 1]. So, we have to find two paths from arr to arr[N – 1][M – 1] with maximum sum.
Approaching in a similar way as Minimum Cost Path problem, we start both paths from arr together and recur to neighbouring cells of the matrix till we reach arr[N – 1][M – 1]. To make sure that a cell doesn’t contribute more than once, we check if current cell on both path are the same or not. If they are same, it is added to the answer only once.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Input matrix``int` `n = 4, m = 4;``int` `arr = { { 1, 0, 3, -1 },``                  ``{ 3, 5, 1, -2 },``                  ``{ -2, 0, 1, 1 },``                  ``{ 2, 1, -1, 1 } };`` ` `// DP matrix``int` `cache;`` ` `// Function to return the sum of the cells``// arr[i1][j1] and arr[i2][j2]``int` `sum(``int` `i1, ``int` `j1, ``int` `i2, ``int` `j2)``{``    ``if` `(i1 == i2 && j1 == j2) {``        ``return` `arr[i1][j1];``    ``}``    ``return` `arr[i1][j1] + arr[i2][j2];``}`` ` `// Recursive function to return the``// required maximum cost path``int` `maxSumPath(``int` `i1, ``int` `j1, ``int` `i2)``{`` ` `    ``// Column number of second path``    ``int` `j2 = i1 + j1 - i2;`` ` `    ``// Base Case``    ``if` `(i1 >= n || i2 >= n || j1 >= m || j2 >= m) {``        ``return` `0;``    ``}`` ` `    ``// If already calculated, return from DP matrix``    ``if` `(cache[i1][j1][i2] != -1) {``        ``return` `cache[i1][j1][i2];``    ``}``    ``int` `ans = INT_MIN;`` ` `    ``// Recurring for neighbouring cells of both paths together``    ``ans = max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));``    ``ans = max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));``    ``ans = max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));``    ``ans = max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));`` ` `    ``// Saving result to the DP matrix for current state``    ``cache[i1][j1][i2] = ans;`` ` `    ``return` `ans;``}`` ` `// Driver code``int` `main()``{``    ``memset``(cache, -1, ``sizeof``(cache));``    ``cout << maxSumPath(0, 0, 0);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;`` ` `class` `GFG``{ ``     ` `// Input matrix``static` `int` `n = ``4``, m = ``4``;``static` `int` `arr[][] = { { ``1``, ``0``, ``3``, -``1` `},``                ``{ ``3``, ``5``, ``1``, -``2` `},``                ``{ -``2``, ``0``, ``1``, ``1` `},``                ``{ ``2``, ``1``, -``1``, ``1` `} };`` ` `// DP matrix``static` `int` `cache[][][] = ``new` `int``[``5``][``5``][``5``];`` ` `// Function to return the sum of the cells``// arr[i1][j1] and arr[i2][j2]``static` `int` `sum(``int` `i1, ``int` `j1, ``int` `i2, ``int` `j2)``{``    ``if` `(i1 == i2 && j1 == j2) ``    ``{``        ``return` `arr[i1][j1];``    ``}``    ``return` `arr[i1][j1] + arr[i2][j2];``}`` ` `// Recursive function to return the``// required maximum cost path``static` `int` `maxSumPath(``int` `i1, ``int` `j1, ``int` `i2)``{`` ` `    ``// Column number of second path``    ``int` `j2 = i1 + j1 - i2;`` ` `    ``// Base Case``    ``if` `(i1 >= n || i2 >= n || j1 >= m || j2 >= m) ``    ``{``        ``return` `0``;``    ``}`` ` `    ``// If already calculated, return from DP matrix``    ``if` `(cache[i1][j1][i2] != -``1``) ``    ``{``        ``return` `cache[i1][j1][i2];``    ``}``    ``int` `ans = Integer.MIN_VALUE;`` ` `    ``// Recurring for neighbouring cells of both paths together``    ``ans = Math.max(ans, maxSumPath(i1 + ``1``, j1, i2 + ``1``) + sum(i1, j1, i2, j2));``    ``ans = Math.max(ans, maxSumPath(i1, j1 + ``1``, i2) + sum(i1, j1, i2, j2));``    ``ans = Math.max(ans, maxSumPath(i1, j1 + ``1``, i2 + ``1``) + sum(i1, j1, i2, j2));``    ``ans = Math.max(ans, maxSumPath(i1 + ``1``, j1, i2) + sum(i1, j1, i2, j2));`` ` `    ``// Saving result to the DP matrix for current state``    ``cache[i1][j1][i2] = ans;`` ` `    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `main(String args[])``{``    ``//set initial value``    ``for``(``int` `i=``0``;i<``5``;i++)``    ``for``(``int` `i1=``0``;i1<``5``;i1++)``    ``for``(``int` `i2=``0``;i2<``5``;i2++)``    ``cache[i][i1][i2]=-``1``;``     ` `    ``System.out.println( maxSumPath(``0``, ``0``, ``0``));``}``}`` ` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python 3 implementation of the approach``import` `sys`` ` `# Input matrix``n ``=` `4``m ``=` `4``arr ``=` `[[``1``, ``0``, ``3``, ``-``1``],``    ``[``3``, ``5``, ``1``, ``-``2``],``    ``[``-``2``, ``0``, ``1``, ``1``],``    ``[``2``, ``1``, ``-``1``, ``1``]]`` ` `# DP matrix``cache ``=` `[[[``-``1` `for` `i ``in` `range``(``5``)] ``for` `j ``in` `range``(``5``)] ``for` `k ``in` `range``(``5``)]`` ` `# Function to return the sum of the cells``# arr[i1][j1] and arr[i2][j2]``def` `sum``(i1, j1, i2, j2):``    ``if` `(i1 ``=``=` `i2 ``and` `j1 ``=``=` `j2):``        ``return` `arr[i1][j1]``    ``return` `arr[i1][j1] ``+` `arr[i2][j2]`` ` `# Recursive function to return the``# required maximum cost path``def` `maxSumPath(i1, j1, i2):``     ` `    ``# Column number of second path``    ``j2 ``=` `i1 ``+` `j1 ``-` `i2`` ` `    ``# Base Case``    ``if` `(i1 >``=` `n ``or` `i2 >``=` `n ``or` `j1 >``=` `m ``or` `j2 >``=` `m):``        ``return` `0`` ` `    ``# If already calculated, return from DP matrix``    ``if` `(cache[i1][j1][i2] !``=` `-``1``):``        ``return` `cache[i1][j1][i2]``    ``ans ``=` `-``sys.maxsize``-``1`` ` `    ``# Recurring for neighbouring cells of both paths together``    ``ans ``=` `max``(ans, maxSumPath(i1 ``+` `1``, j1, i2 ``+` `1``) ``+` `sum``(i1, j1, i2, j2))``    ``ans ``=` `max``(ans, maxSumPath(i1, j1 ``+` `1``, i2) ``+` `sum``(i1, j1, i2, j2))``    ``ans ``=` `max``(ans, maxSumPath(i1, j1 ``+` `1``, i2 ``+` `1``) ``+` `sum``(i1, j1, i2, j2))``    ``ans ``=` `max``(ans, maxSumPath(i1 ``+` `1``, j1, i2) ``+` `sum``(i1, j1, i2, j2))`` ` `    ``# Saving result to the DP matrix for current state``    ``cache[i1][j1][i2] ``=` `ans`` ` `    ``return` `ans`` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``print``(maxSumPath(``0``, ``0``, ``0``))`` ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the approach ``using` `System;`` ` `class` `GFG``{ ``     ` `// Input matrix``static` `int` `n = 4, m = 4;``static` `int` `[,]arr = { { 1, 0, 3, -1 },``                ``{ 3, 5, 1, -2 },``                ``{ -2, 0, 1, 1 },``                ``{ 2, 1, -1, 1 } };`` ` `// DP matrix``static` `int` `[,,]cache = ``new` `int``[5, 5, 5];`` ` `// Function to return the sum of the cells``// arr[i1][j1] and arr[i2][j2]``static` `int` `sum(``int` `i1, ``int` `j1, ``int` `i2, ``int` `j2)``{``    ``if` `(i1 == i2 && j1 == j2) ``    ``{``        ``return` `arr[i1, j1];``    ``}``    ``return` `arr[i1, j1] + arr[i2, j2];``}`` ` `// Recursive function to return the``// required maximum cost path``static` `int` `maxSumPath(``int` `i1, ``int` `j1, ``int` `i2)``{`` ` `    ``// Column number of second path``    ``int` `j2 = i1 + j1 - i2;`` ` `    ``// Base Case``    ``if` `(i1 >= n || i2 >= n || j1 >= m || j2 >= m) ``    ``{``        ``return` `0;``    ``}`` ` `    ``// If already calculated, return from DP matrix``    ``if` `(cache[i1, j1, i2] != -1) ``    ``{``        ``return` `cache[i1, j1, i2];``    ``}``    ``int` `ans = ``int``.MinValue;`` ` `    ``// Recurring for neighbouring cells of both paths together``    ``ans = Math.Max(ans, maxSumPath(i1 + 1, j1, i2 + 1) + sum(i1, j1, i2, j2));``    ``ans = Math.Max(ans, maxSumPath(i1, j1 + 1, i2) + sum(i1, j1, i2, j2));``    ``ans = Math.Max(ans, maxSumPath(i1, j1 + 1, i2 + 1) + sum(i1, j1, i2, j2));``    ``ans = Math.Max(ans, maxSumPath(i1 + 1, j1, i2) + sum(i1, j1, i2, j2));`` ` `    ``// Saving result to the DP matrix for current state``    ``cache[i1, j1, i2] = ans;`` ` `    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``//set initial value``    ``for``(``int` `i = 0; i < 5; i++)``        ``for``(``int` `i1 = 0; i1 < 5; i1++)``            ``for``(``int` `i2 = 0; i2 < 5; i2++)``                ``cache[i,i1,i2]=-1;``     ` `    ``Console.WriteLine( maxSumPath(0, 0, 0));``}``}`` ` `// This code contributed by Rajput-Ji`
Output:
```16
```

Time Complexity: O((N2) * M)

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