Maximum path sum in a triangle.

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

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

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

C++

// C++ program for 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; 
} 

Java

// Java Program for Dynamic  
// Programming implementation of 
// Max sum problem in a triangle  
import java.io.*; 
  
class GFG { 
          
    static int N = 3; 
      
    // Function for finding maximum sum 
    static int maxPathSum(int tri[][], 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 */
    public static void main (String[] args) 
    { 
        int tri[][] = { {1, 0, 0}, 
                        {4, 8, 0}, 
                        {1, 5, 3} }; 
        System.out.println ( maxPathSum(tri, 2, 2)); 
    } 
} 
  
// This code is contributed by vt_m 

Python3

# Python program for 
# Dynamic Programming 
# implementation of Max 
# sum problem in a 
# triangle 
  
N = 3
  
# Function for finding maximum sum 
def maxPathSum(tri, m, n): 
  
    # loop for bottom-up calculation 
    for i in range(m-1, -1, -1): 
        for j in range(i+1): 
  
            # 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 function 
  
tri = [[1, 0, 0], 
       [4, 8, 0], 
       [1, 5, 3]] 
print(maxPathSum(tri, 2, 2)) 
  
# This code is contributed 
# by Soumen Ghosh. 

C#

// C# Program for Dynamic Programming 
// implementation of Max sum problem  
// in a triangle  
using System; 
  
class GFG { 
      
    // Function for finding maximum sum 
    static int maxPathSum(int [,]tri,  
                          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 */
    public static void Main () 
    { 
        int [,]tri = { {1, 0, 0}, 
                        {4, 8, 0}, 
                        {1, 5, 3} }; 
                          
        Console.Write (  
             maxPathSum(tri, 2, 2)); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// PHP program for Dynamic 
// Programming implementation of 
// Max sum problem in a triangle  
  
// Function for finding 
// maximum sum 
function maxPathSum($tri, $m, $n) 
{ 
    // loop for bottom-up 
    // calculation 
    for ( $i = $m - 1; $i >= 0; $i--) 
    { 
        for ($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 Code 
$tri= array(array(1, 0, 0), 
            array(4, 8, 0), 
            array(1, 5, 3)); 
echo maxPathSum($tri, 2, 2); 
  
// This code is contributed by ajit 
?> 

Output :

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