We have given numbers in form of triangle, by starting at the top of the triangle and moving to adjacent numbers on the row below, find the maximum total from top to bottom.

Examples:

Input : 3 7 4 2 4 6 8 5 9 3 Output : 23 Explanation : 3 + 7 + 4 + 9 = 23 Input : 8 -4 4 2 2 6 1 1 1 1 Output : 19 Explanation : 8 + 4 + 6 + 1 = 19

We can go through the brute force by checking every possible path but that is much time taking so we should try to solve this problem with the help of dynamic programming which reduces the time complexity.

If we should left shift every element and put 0 at each empty position to make it a regular matrix, then our problem looks like minimum cost path.

So, after converting our input triangle elements into a regular matrix we should apply the dynamic programmic concept to find the maximum path sum.

Applying, DP in bottom-up manner we should solve our problem as:

**Example:**

3 7 4 2 4 6 8 5 9 3 Step 1 : 3 0 0 0 7 4 0 0 2 4 6 0 8 5 9 3 Step 2 : 3 0 0 0 7 4 0 0 10 13 15 0 Step 3 : 3 0 0 0 20 19 0 0 Step 4: 23 0 0 0 output : 23

/* Dynamic Programming implementation of Max sum problem in a triangle */ #include<bits/stdc++.h> using namespace std; #define N 3 // Function for finding maximum sum int maxPathSum(int tri[][N], int m, int n) { // loop for bottom-up calculation for (int i=m-1; i>=0; i--) { for (int j=0; j<=i; j++) { // for each element, check both // elements just below the number // and below right to the number // add the maximum of them to it if (tri[i+1][j] > tri[i+1][j+1]) tri[i][j] += tri[i+1][j]; else tri[i][j] += tri[i+1][j+1]; } } // return the top element // which stores the maximum sum return tri[0][0]; } /* Driver program to test above functions */ int main() { int tri[N][N] = { {1, 0, 0}, {4, 8, 0}, {1, 5, 3} }; cout << maxPathSum(tri, 2, 2); return 0; }

Output:

14

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