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++

 `// C++ program to find Maximum path sum ` `// start any column in row '0' and ends ` `// up to any column in row 'n-1' ` `#include ` `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 ) <

Java

 `// Java program to find Maximum path sum ` `// start any column in row '0' and ends ` `// up to any column in row 'n-1' ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `    ``static` `int` `N = ``4``; ` ` `  `    ``// function find maximum sum path ` `    ``static` `int` `MaximumPath(``int` `Mat[][]) ` `    ``{ ` `        ``int` `result = ``0``; ` ` `  `        ``// creat 2D matrix to store the sum ` `        ``// of the path ` `        ``int` `dp[][] = ``new` `int``[N][N + ``2``]; ` ` `  `        ``// initialize all dp matrix as '0' ` `        ``for` `(``int``[] rows : dp) ` `            ``Arrays.fill(rows, ``0``); ` ` `  `        ``// 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] = Math.max(dp[i - ``1``][j - ``1``], ` `                                    ``Math.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 = Math.max(result, dp[N - ``1``][i]); ` ` `  `        ``// return maximum sum path ` `        ``return` `result; ` `    ``} ` `     `  `    ``// driver code ` `    ``public` `static` `void` `main(String arg[]) ` `    ``{ ` `        ``int` `Mat[][] = { { ``4``, ``2``, ``3``, ``4` `}, ` `                        ``{ ``2``, ``9``, ``1``, ``10` `}, ` `                        ``{ ``15``, ``1``, ``3``, ``0` `}, ` `                        ``{ ``16``, ``92``, ``41``, ``44` `} }; ` ` `  `        ``System.out.println(MaximumPath(Mat)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

Python3

 `# Python3 program to find ` `# Maximum path sum ` `# start any column in ` `# row '0' and ends ` `# up to any column in row 'n-1' ` ` `  `N ``=` `4` ` `  `# function find maximum sum path ` `def` `MaximumPath(Mat): ` ` `  `    ``result ``=` `0` ` `  `    ``# create 2D matrix to store the sum ` `    ``# of the path ` `    ``# initialize all dp matrix as '0' ` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(N``+``2``)] ``for` `j ``in` `range``(N)] ` ` `  `    ``# copy all element of first column into ` `    ``# dp first column ` `    ``for` `i ``in` `range``(N): ` `        ``for` `j ``in` `range``(``1``, N``+``1``): ` `            ``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` `i ``in` `range``(N``+``1``): ` `        ``result ``=` `max``(result, dp[N``-``1``][i]) ` ` `  `    ``# return maximum sum path ` `    ``return` `result ` ` `  `# driver program to test above function ` `Mat ``=` `[[``4``, ``2``, ``3``, ``4``], ` `       ``[``2``, ``9``, ``1``, ``10``], ` `       ``[``15``, ``1``, ``3``, ``0``], ` `       ``[``16``, ``92``, ``41``, ``44``]] ` ` `  `print``(MaximumPath(Mat)) ` ` `  `# This code is contributed by Soumen Ghosh. `

C#

 `// C# program to find Maximum path sum  ` `// start any column in row '0' and ends  ` `// up to any column in row 'n-1'  ` `using` `System; ` ` `  `class` `GFG {  ` `     `  `    ``static` `int` `N = 4;  ` ` `  `    ``// function find maximum sum path  ` `    ``static` `int` `MaximumPath(``int` `[,] Mat)  ` `    ``{  ` `        ``int` `result = 0;  ` ` `  `        ``// creat 2D matrix to store the sum  ` `        ``// of the path  ` `        ``int` `[,]dp = ``new` `int``[N,N + 2];  ` ` `  `        ``// initialize all dp matrix as '0'  ` `        ``//for (int[] rows : dp)  ` `        ``//    Arrays.fill(rows, 0);  ` ` `  `        ``// 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] = Math.Max(dp[i - 1,j - 1],  ` `                                    ``Math.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 = Math.Max(result, dp[N - 1,i]);  ` ` `  `        ``// return maximum sum path  ` `        ``return` `result;  ` `    ``}  ` `     `  `    ``// driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[,]Mat = { { 4, 2, 3, 4 },  ` `                        ``{ 2, 9, 1, 10 },  ` `                        ``{ 15, 1, 3, 0 },  ` `                        ``{ 16, 92, 41, 44 } };  ` ` `  `        ``Console.WriteLine(MaximumPath(Mat));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga.  `

PHP

 ` `

Output:

```120
```

Time complexity : O(N2)

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