# Maximum sum of a path in a Right Number Triangle

Given a right triangle of numbers, find the largest of the sum of numbers that appear on the paths starting from the top towards the base, so that on each path the next number is located directly below or below-and-one-place-to-the-right.

Examples :

```Input : 1
1 2
4 1 2
2 3 1 1
Output : 9
Explanation : 1 + 1 + 4 + 3

Input : 2
4 1
1 2 7
Output : 10
Explanation : 2 + 1 + 7
```

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

The idea is to find largest sum ending at every cell of last row and return maximum of these sums. We can recursively compute these sums by recursively considering above two cells. Since there are overlapping subproblems, we use dynamic programming to find the maximum sum ending at particular cell of last row.

Below is the implementation of above idea.

## C++

 `// C++ program to print maximum sum ` `// in a right triangle of numbers ` `#include ` `using` `namespace` `std; ` ` `  `// function to find maximum sum path ` `int` `maxSum(``int` `tri[], ``int` `n) ` `{ ` `    ``// Adding the element of row 1 to both the ` `    ``// elements of row 2 to reduce a step from ` `    ``// the loop ` `    ``if` `(n > 1) ` `        ``tri = tri + tri; ` `        ``tri = tri + tri; ` ` `  `    ``// Traverse remaining rows ` `    ``for``(``int` `i = 2; i < n; i++) { ` `        ``tri[i] = tri[i] + tri[i-1]; ` `        ``tri[i][i] = tri[i][i] + tri[i-1][i-1]; ` ` `  `        ``//Loop to traverse columns ` `        ``for` `(``int` `j = 1; j < i; j++){ ` ` `  `            ``// Checking the two conditions,  ` `            ``// directly below and below right. ` `            ``// Considering the greater one ` `             `  `            ``// tri[i] would store the possible  ` `            ``// combinations of sum of the paths ` `            ``if` `(tri[i][j] + tri[i-1][j-1] >=  ` `                            ``tri[i][j] + tri[i-1][j]) ` `                 `  `                ``tri[i][j] = tri[i][j] + tri[i-1][j-1]; ` `            ``else` `                ``tri[i][j] = tri[i][j]+tri[i-1][j]; ` `        ``} ` `    ``} ` `     `  `    ``// array at n-1 index (tri[i]) stores  ` `    ``// all possible adding combination, finding  ` `    ``// the maximum one out of them ` `    ``int` `max=tri[n-1]; ` `     `  `    ``for``(``int` `i=1;i

## Java

 `// Java program to print maximum sum ` `// in a right triangle of numbers ` `class` `GFG ` `{ ` `     `  `    ``// function to find maximum sum path ` `    ``static` `int` `maxSum(``int` `tri[][], ``int` `n) ` `    ``{ ` `         `  `        ``// Adding the element of row 1 to both the ` `        ``// elements of row 2 to reduce a step from ` `        ``// the loop ` `        ``if` `(n > ``1``) ` `            ``tri[``1``][``1``] = tri[``1``][``1``] + tri[``0``][``0``]; ` `            ``tri[``1``][``0``] = tri[``1``][``0``] + tri[``0``][``0``]; ` `     `  `        ``// Traverse remaining rows ` `        ``for``(``int` `i = ``2``; i < n; i++) { ` `            ``tri[i][``0``] = tri[i][``0``] + tri[i-``1``][``0``]; ` `            ``tri[i][i] = tri[i][i] + tri[i-``1``][i-``1``]; ` `     `  `            ``//Loop to traverse columns ` `            ``for` `(``int` `j = ``1``; j < i; j++){ ` `     `  `                ``// Checking the two conditions,  ` `                ``// directly below and below right. ` `                ``// Considering the greater one ` `                 `  `                ``// tri[i] would store the possible  ` `                ``// combinations of sum of the paths ` `                ``if` `(tri[i][j] + tri[i-``1``][j-``1``] >=  ` `                           ``tri[i][j] + tri[i-``1``][j]) ` `                     `  `                    ``tri[i][j] = tri[i][j]  ` `                                  ``+ tri[i-``1``][j-``1``]; ` `                     `  `                ``else` `                    ``tri[i][j] = tri[i][j] ` `                                    ``+ tri[i-``1``][j]; ` `            ``} ` `        ``} ` `         `  `        ``// array at n-1 index (tri[i]) stores  ` `        ``// all possible adding combination,  ` `        ``// finding the maximum one out of them ` `        ``int` `max = tri[n-``1``][``0``]; ` `         `  `        ``for``(``int` `i = ``1``; i < n; i++) ` `        ``{ ` `            ``if``(max < tri[n-``1``][i]) ` `                ``max = tri[n-``1``][i]; ` `        ``} ` `         `  `        ``return` `max; ` `    ``}  ` `         `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `tri[][] = {{``1``}, {``2``,``1``}, {``3``,``3``,``2``}}; ` `         `  `        ``System.out.println(maxSum(tri, ``3``)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python

 `# Python program to print maximum sum ` `# in a right triangle of numbers. ` ` `  `# tri[][] is a 2D array that stores the ` `# triangle, n is number of lines or rows. ` `def` `maxSum(tri, n): ` ` `  `    ``# Adding the element of row 1 to both the ` `    ``# elements of row 2 to reduce a step from ` `    ``# the loop ` `    ``if` `n > ``1``: ` `        ``tri[``1``][``1``] ``=` `tri[``1``][``1``]``+``tri[``0``][``0``] ` `        ``tri[``1``][``0``] ``=` `tri[``1``][``0``]``+``tri[``0``][``0``] ` ` `  `    ``# Traverse remaining rows ` `    ``for` `i ``in` `range``(``2``, n): ` `        ``tri[i][``0``] ``=` `tri[i][``0``] ``+` `tri[i``-``1``][``0``] ` `        ``tri[i][i] ``=` `tri[i][i] ``+` `tri[i``-``1``][i``-``1``] ` ` `  `        ``# Loop to traverse columns ` `        ``for` `j ``in` `range``(``1``, i): ` ` `  `            ``# Checking the two conditions, directly below ` `            ``# and below right. Considering the greater one ` ` `  `            ``# tri[i] would store the possible combinations ` `            ``# of sum of the paths ` `            ``if` `tri[i][j]``+``tri[i``-``1``][j``-``1``] >``=` `tri[i][j]``+``tri[i``-``1``][j]: ` `                ``tri[i][j] ``=` `tri[i][j] ``+` `tri[i``-``1``][j``-``1``] ` `            ``else``: ` `                ``tri[i][j] ``=` `tri[i][j]``+``tri[i``-``1``][j] ` ` `  `    ``# array at n-1 index (tri[i]) stores all possible ` `    ``# adding combination, finding the maximum one ` `    ``# out of them ` `    ``print` `max``(tri[n``-``1``]) ` ` `  `# driver program ` `tri ``=` `[[``1``], [``2``,``1``], [``3``,``3``,``2``]] ` `maxSum(tri, ``3``) `

## C#

 `// C# program to print  ` `// maximum sum in a right ` `// triangle of numbers ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// function to find  ` `    ``// maximum sum path ` `    ``static` `int` `maxSum(``int` `[,]tri,  ` `                      ``int` `n) ` `    ``{ ` `         `  `        ``// Adding the element of row 1  ` `        ``// to both the elements of row 2  ` `        ``// to reduce a step from the loop ` `        ``if` `(n > 1) ` `            ``tri[1, 1] = tri[1, 1] +  ` `                        ``tri[0, 0]; ` `            ``tri[1, 0] = tri[1, 0] +  ` `                        ``tri[0, 0]; ` `     `  `        ``// Traverse remaining rows ` `        ``for``(``int` `i = 2; i < n; i++)  ` `        ``{ ` `            ``tri[i, 0] = tri[i, 0] +  ` `                        ``tri[i - 1, 0]; ` `            ``tri[i, i] = tri[i, i] +  ` `                        ``tri[i - 1, i - 1]; ` `      `  `            ``//Loop to traverse columns ` `            ``for` `(``int` `j = 1; j < i; j++) ` `            ``{ ` `     `  `                ``// Checking the two conditions,  ` `                ``// directly below and below right. ` `                ``// Considering the greater one ` `                 `  `                ``// tri[i] would store the possible  ` `                ``// combinations of sum of the paths ` `                ``if` `(tri[i, j] + tri[i - 1, j - 1] >=  ` `                    ``tri[i, j] + tri[i - 1, j]) ` `                 `  `                    ``tri[i, j] = tri[i, j] +  ` `                                ``tri[i - 1, j - 1]; ` `                     `  `                ``else` `                    ``tri[i, j] = tri[i, j] +  ` `                                ``tri[i - 1, j]; ` `            ``} ` `        ``} ` `         `  `        ``// array at n-1 index (tri[i])  ` `        ``// stores all possible adding  ` `        ``// combination, finding the  ` `        ``// maximum one out of them ` `        ``int` `max = tri[n - 1, 0]; ` `         `  `        ``for``(``int` `i = 1; i < n; i++) ` `        ``{ ` `            ``if``(max < tri[n - 1, i]) ` `                ``max = tri[n - 1, i]; ` `        ``} ` `         `  `        ``return` `max; ` `    ``}  ` `         `  `// Driver Code ` `public` `static` `void` `Main () ` `{ ` `     `  `        ``int` `[,]tri = {{1,0,0},  ` `                      ``{2,1,0}, ` `                      ``{3,3,2}}; ` `         `  `        ``Console.Write(maxSum(tri, 3)); ` `} ` `} ` ` `  `// This code is contributed by ajit. `

## PHP

 ` 1) ` `        ``\$tri`` = ``\$tri`` + ``\$tri``; ` `        ``\$tri`` = ``\$tri`` + ``\$tri``; ` ` `  `    ``// Traverse remaining rows ` `    ``for``(``\$i` `= 2; ``\$i` `< ``\$n``; ``\$i``++) ` `    ``{ ` `        ``\$tri``[``\$i``] = ``\$tri``[``\$i``] +  ` `                      ``\$tri``[``\$i` `- 1]; ` `        ``\$tri``[``\$i``][``\$i``] = ``\$tri``[``\$i``][``\$i``] +  ` `                       ``\$tri``[``\$i` `- 1][``\$i` `- 1]; ` ` `  `        ``//Loop to traverse columns ` `        ``for` `(``\$j` `= 1; ``\$j` `< ``\$i``; ``\$j``++) ` `        ``{ ` ` `  `            ``// Checking the two conditions,  ` `            ``// directly below and below right. ` `            ``// Considering the greater one ` `             `  `            ``// tri[i] would store the possible  ` `            ``// combinations of sum of the paths ` `            ``if` `(``\$tri``[``\$i``][``\$j``] + ``\$tri``[``\$i` `- 1][``\$j` `- 1] >=  ` `                 ``\$tri``[``\$i``][``\$j``] + ``\$tri``[``\$i` `- 1][``\$j``]) ` `                 `  `                ``\$tri``[``\$i``][``\$j``] = ``\$tri``[``\$i``][``\$j``] +  ` `                               ``\$tri``[``\$i` `- 1][``\$j` `- 1]; ` `            ``else` `                ``\$tri``[``\$i``][``\$j``] = ``\$tri``[``\$i``][``\$j``] +  ` `                               ``\$tri``[``\$i` `- 1][``\$j``]; ` `        ``} ` `    ``} ` `     `  `    ``// array at n-1 index (tri[i])  ` `    ``// stores all possible adding  ` `    ``// combination, finding the  ` `    ``// maximum one out of them ` `     `  `    ``\$max` `= ``\$tri``[``\$n` `- 1]; ` `     `  `    ``for``(``\$i` `= 1; ``\$i` `< ``\$n``; ``\$i``++) ` `    ``{ ` `        ``if``(``\$max` `< ``\$tri``[``\$n` `- 1][``\$i``]) ` `            ``\$max` `= ``\$tri``[``\$n` `- 1][``\$i``]; ` `    ``} ` `     `  `    ``return` `\$max``; ` `} ` ` `  `// Driver Code ` `\$tri` `= ``array``(``array``(1), ` `             ``array``(2,1),  ` `             ``array``(3,3,2)); ` ` `  `echo` `maxSum(``\$tri``, 3); ` ` `  `// This code is contributed by ajit ` `?> `

Output :

```6
```

This article is contributed by Harshit Agrawal. 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.