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# Maximum path sum in matrix

Given a matrix of N * M. Find the maximum path sum in matrix. The maximum path is sum of all elements from first row to last row where you are allowed to move only down or diagonally to left or right. You can start from any element in first row.

Examples:

```Input : mat[][] = 10 10  2  0 20  4
1  0  0 30  2  5
0 10  4  0  2  0
1  0  2 20  0  4
Output : 74
The maximum sum path is 20-30-4-20.

Input : mat[][] = 1 2 3
9 8 7
4 5 6
Output : 17
The maximum sum path is 3-8-6.```

We are given a matrix of N * M. To find max path sum first we have to find max value in first row of matrix. Store this value in res. Now for every element in matrix update element with max value which can be included in max path. If the value is greater then res then update res. In last return res which consists of max path sum value.

Implementation:

## C++

 `// CPP program for finding max path in matrix``#include ``#define N 4``#define M 6``using` `namespace` `std;` `// To calculate max path in matrix``int` `findMaxPath(``int` `mat[][M])``{` `    ``for` `(``int` `i = 1; i < N; i++) {``        ``for` `(``int` `j = 0; j < M; j++) {` `            ``// When all paths are possible``            ``if` `(j > 0 && j < M - 1)``                ``mat[i][j] += max(mat[i - 1][j],``                             ``max(mat[i - 1][j - 1],``                             ``mat[i - 1][j + 1]));` `            ``// When diagonal right is not possible``            ``else` `if` `(j > 0)``                ``mat[i][j] += max(mat[i - 1][j],``                            ``mat[i - 1][j - 1]);` `            ``// When diagonal left is not possible``            ``else` `if` `(j < M - 1)``                ``mat[i][j] += max(mat[i - 1][j],``                            ``mat[i - 1][j + 1]);` `            ``// Store max path sum``        ``}``    ``}``    ``int` `res = 0;``    ``for` `(``int` `j = 0; j < M; j++)``        ``res = max(mat[N-1][j], res);``    ``return` `res;``}` `// Driver program to check findMaxPath``int` `main()``{``    ` `    ``int` `mat1[N][M] = { { 10, 10, 2, 0, 20, 4 },``                    ``{ 1, 0, 0, 30, 2, 5 },``                    ``{ 0, 10, 4, 0, 2, 0 },``                    ``{ 1, 0, 2, 20, 0, 4 } };``            ` `    ``cout << findMaxPath(mat1) << endl;``    ``return` `0;``}`

## Java

 `// Java program for finding max path in matrix` `import` `static` `java.lang.Math.max;` `class` `GFG``{``    ``public` `static` `int` `N = ``4``, M = ``6``;``    ` `    ``// Function to calculate max path in matrix``    ``static` `int` `findMaxPath(``int` `mat[][])``    ``{``        ``// To find max val in first row``        ``int` `res = -``1``;``        ``for` `(``int` `i = ``0``; i < M; i++)``            ``res = max(res, mat[``0``][i]);` `        ``for` `(``int` `i = ``1``; i < N; i++)``        ``{``            ``res = -``1``;``            ``for` `(``int` `j = ``0``; j < M; j++)``            ``{``                ``// When all paths are possible``                ``if` `(j > ``0` `&& j < M - ``1``)``                    ``mat[i][j] += max(mat[i - ``1``][j],``                                 ``max(mat[i - ``1``][j - ``1``],``                                    ``mat[i - ``1``][j + ``1``]));` `                ``// When diagonal right is not possible``                ``else` `if` `(j > ``0``)``                    ``mat[i][j] += max(mat[i - ``1``][j],``                                    ``mat[i - ``1``][j - ``1``]);` `                ``// When diagonal left is not possible``                ``else` `if` `(j < M - ``1``)``                    ``mat[i][j] += max(mat[i - ``1``][j],``                                ``mat[i - ``1``][j + ``1``]);` `                ``// Store max path sum``                ``res = max(mat[i][j], res);``            ``}``        ``}``        ``return` `res;``    ``}``    ` `    ``// driver program``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `mat[][] = { { ``10``, ``10``, ``2``, ``0``, ``20``, ``4` `},``                        ``{ ``1``, ``0``, ``0``, ``30``, ``2``, ``5` `},``                        ``{ ``0``, ``10``, ``4``, ``0``, ``2``, ``0` `},``                        ``{ ``1``, ``0``, ``2``, ``20``, ``0``, ``4` `}``                    ``};` `        ``System.out.println(findMaxPath(mat));``    ``}``}` `// Contributed by Pramod Kumar`

## Python3

 `# Python 3 program for finding max path in matrix``# To calculate max path in matrix` `def` `findMaxPath(mat):``  ` `    ``for` `i ``in` `range``(``1``, N):`` ` `        ``res ``=` `-``1``        ``for` `j ``in` `range``(M):`` ` `            ``# When all paths are possible``            ``if` `(j > ``0` `and` `j < M ``-` `1``):``                ``mat[i][j] ``+``=` `max``(mat[i ``-` `1``][j],``                                 ``max``(mat[i ``-` `1``][j ``-` `1``],``                                     ``mat[i ``-` `1``][j ``+` `1``]))`` ` `            ``# When diagonal right is not possible``            ``else` `if` `(j > ``0``):``                ``mat[i][j] ``+``=` `max``(mat[i ``-` `1``][j],``                                 ``mat[i ``-` `1``][j ``-` `1``])`` ` `            ``# When diagonal left is not possible``            ``else` `if` `(j < M ``-` `1``):``                ``mat[i][j] ``+``=` `max``(mat[i ``-` `1``][j],``                                 ``mat[i ``-` `1``][j ``+` `1``])`` ` `            ``# Store max path sum``            ``res ``=` `max``(mat[i][j], res)``    ``return` `res` `# Driver program to check findMaxPath``N``=``4``M``=``6``mat ``=` `([[ ``10``, ``10``, ``2``, ``0``, ``20``, ``4` `],``        ``[ ``1``, ``0``, ``0``, ``30``, ``2``, ``5` `],``        ``[ ``0``, ``10``, ``4``, ``0``, ``2``, ``0` `],``        ``[ ``1``, ``0``, ``2``, ``20``, ``0``, ``4` `]])``              ` `print``(findMaxPath(mat))` `# This code is contributed by Azkia Anam.`

## C#

 `// C# program for finding``// max path in matrix``using` `System;` `class` `GFG``{``    ``static` `int` `N = 4, M = 6;``    ` `    ``// find the max element``    ``static` `int` `max(``int` `a, ``int` `b)``    ``{``        ``if``(a > b)``        ``return` `a;``        ``else``        ``return` `b;``    ``}``    ` `    ``// Function to calculate``    ``// max path in matrix``    ``static` `int` `findMaxPath(``int` `[,]mat)``    ``{``        ``// To find max val``        ``// in first row``        ``int` `res = -1;``        ``for` `(``int` `i = 0; i < M; i++)``            ``res = max(res, mat[0, i]);` `        ``for` `(``int` `i = 1; i < N; i++)``        ``{``            ``res = -1;``            ``for` `(``int` `j = 0; j < M; j++)``            ``{``                ``// When all paths are possible``                ``if` `(j > 0 && j < M - 1)``                    ``mat[i, j] += max(mat[i - 1, j],``                                 ``max(mat[i - 1, j - 1],``                                     ``mat[i - 1, j + 1]));` `                ``// When diagonal right``                ``// is not possible``                ``else` `if` `(j > 0)``                    ``mat[i, j] += max(mat[i - 1, j],``                                     ``mat[i - 1, j - 1]);` `                ``// When diagonal left``                ``// is not possible``                ``else` `if` `(j < M - 1)``                    ``mat[i, j] += max(mat[i - 1, j],``                                 ``mat[i - 1, j + 1]);` `                ``// Store max path sum``                ``res = max(mat[i, j], res);``            ``}``        ``}``        ``return` `res;``    ``}``    ` `    ``// Driver code``    ``static` `public` `void` `Main (String[] args)``    ``{``        ``int``[,] mat = {{10, 10, 2, 0, 20, 4},``                      ``{1, 0, 0, 30, 2, 5},``                      ``{0, 10, 4, 0, 2, 0},``                      ``{1, 0, 2, 20, 0, 4}};` `        ``Console.WriteLine(findMaxPath(mat));``    ``}``}` `// This code is contributed``// by Arnab Kundu`

## PHP

 ` 0 && ``\$j` `< (``\$M` `- 1))``                ``\$mat``[``\$i``][``\$j``] += max(``\$mat``[``\$i` `- 1][``\$j``],``                                ``max(``\$mat``[``\$i` `- 1][``\$j` `- 1],``                                    ``\$mat``[``\$i` `- 1][``\$j` `+ 1]));` `            ``// When diagonal right is``            ``// not possible``            ``else` `if` `(``\$j` `> 0)``                ``\$mat``[``\$i``][``\$j``] += max(``\$mat``[``\$i` `- 1][``\$j``],``                                    ``\$mat``[``\$i` `- 1][``\$j` `- 1]);` `            ``// When diagonal left is``            ``// not possible``            ``else` `if` `(``\$j` `< (``\$M` `- 1))``                ``\$mat``[``\$i``][``\$j``] += max(``\$mat``[``\$i` `- 1][``\$j``],``                                    ``\$mat``[``\$i` `- 1][``\$j` `+ 1]);` `            ``// Store max path sum``        ``}``    ``}``    ` `    ``\$res` `= 0;``    ``for` `(``\$j` `= 0; ``\$j` `< ``\$M``; ``\$j``++)``        ``\$res` `= max(``\$mat``[``\$N` `- 1][``\$j``], ``\$res``);``    ``return` `\$res``;``}` `// Driver Code``\$mat1` `= ``array``( ``array``( 10, 10, 2, 0, 20, 4 ),``               ``array``( 1, 0, 0, 30, 2, 5 ),``               ``array``( 0, 10, 4, 0, 2, 0 ),``               ``array``( 1, 0, 2, 20, 0, 4 ));``        ` `echo` `findMaxPath(``\$mat1``),``"\n"``;` `// This code is contributed by Sach_Code``?>`

## Javascript

 ``

Output

`74`

Time Complexity: O(N*M), where N and M are the dimensions of the matrix
Space Complexity: O(1), since no extra space has been taken.

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