# Maximum path sum that starting with any cell of 0-th row and ending with any cell of (N-1)-th row

Given a N X N matrix Mat[N][N] of positive integers. There are only three possible moves from a cell (i, j)

1. (i+1, j)
2. (i+1, j-1)
3. (i+1, j+1)

Starting from any column in row 0, return the largest sum of any of the paths up to row N-1.

Examples:

```Input : mat[4][4] = { {4, 2, 3, 4},
{2, 9, 1, 10},
{15, 1, 3, 0},
{16, 92, 41, 44} };
Output :120
path : 4 + 9 + 15 + 92 = 120
```

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

The above problem can be recursively defined.

Let initial position be MaximumPathSum(N-1, j), where j varies from 0 to N-1. We return maximum value between all path that we start traversing (N-1, j) [ where j varies from 0 to N-1]

```i = N-1, j = 0 to N -1
int MaximumPath(Mat[][N], I, j)

// IF we reached to first row of
// matrix then return value of that
// element
IF ( i == 0 && j = 0 )
return    Mat[i][j]

// out of matrix bound
IF( i = N || j < 0 )
return 0;

// call all rest position that we reached
// from current position and find maximum
// between them and add current value in
// that path
return max(MaximumPath(Mat, i-1, j),
MaximumPath(Mat, i-1, j-1),
MaximumPath(Mat, i-1, j+1)))
+ Mat[i][j];
```

If we draw recursion tree of above recursive solution, we can observe overlapping subproblems. Since the problem has overlapping subproblems, we can solve it efficiently using Dynamic Programming. Below is Dynamic Programming based solution.

```// C++ program to find Maximum path sum
// start any column in row '0' and ends
// up to any column in row 'n-1'
#include<bits/stdc++.h>
using namespace std;
#define N 4

// function find maximum sum path
int MaximumPath(int Mat[][N])
{
int result = 0 ;

// creat 2D matrix to store the sum
// of the path
int dp[N][N+2];

// initialize all dp matrix as '0'
memset(dp, 0, sizeof(dp));

// copy all element of first column into
// 'dp' first column
for (int i = 0 ; i < N ; i++)
dp[0][i+1] = Mat[0][i];

for (int i = 1 ; i < N ; i++)
for (int j = 1 ; j <= N ; j++)
dp[i][j] = max(dp[i-1][j-1],
max(dp[i-1][j],
dp[i-1][j+1])) +
Mat[i][j-1] ;

// Find maximum path sum that end ups
// at  any column of last row 'N-1'
for (int i=0; i<=N; i++)
result = max(result, dp[N-1][i]);

// return maximum sum path
return result ;
}

// driver program to test above function
int main()
{
int Mat[4][4] = { { 4, 2 , 3 , 4 },
{ 2 , 9 , 1 , 10},
{ 15, 1 , 3 , 0 },
{ 16 ,92, 41, 44 }
};

cout << MaximumPath ( Mat ) <<endl ;
return 0;
}
```

Output:

```120
```

Time complexity : O(N2)

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

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