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Maximum path sum in a triangle.

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  • Difficulty Level : Medium
  • Last Updated : 19 Jan, 2023
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We have given numbers in form of a 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

Method 1: 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. 

Implementation of Recursive Approach:

C++




// C++ program for
// Recursive 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 i, int j, int row, int col){
     if(j == col ){
         return 0;
     }
   
     if(i == row-1 ){
         return tri[i][j] ;
     }
   
     return tri[i][j] + max(maxPathSum(tri, i+1, j, row, col),
                            maxPathSum(tri, i+1, j+1, row, col)) ;
}
 
/* Driver program to test above functions */
int main()
{
   int tri[N][N] = {  {1, 0, 0},
                      {4, 8, 0},
                      {1, 5, 3} };
   cout << maxPathSum(tri, 0, 0, 3, 3);
   return 0;
}

Java




// Java program for
// Recursive implementation of
// Max sum problem in a triangle
import java.io.*;
 
class GFG {
    static int N = 3;
 
    //  Function for finding maximum sum
    public static int maxPathSum(int tri[][], int i, int j,
                                 int row, int col)
    {
        if (j == col) {
            return 0;
        }
 
        if (i == row - 1) {
            return tri[i][j];
        }
 
        return tri[i][j]
            + Math.max(
                maxPathSum(tri, i + 1, j, row, col),
                maxPathSum(tri, i + 1, j + 1, row, col));
    }
 
    /* 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.print(maxPathSum(tri, 0, 0, 3, 3));
    }
}
 
// This code is contributed by Rohit Pradhan

Python3




# Python program for
# Recursive implementation of
# Max sum problem in a triangle
N = 3
 
#  Function for finding maximum sum
def maxPathSum(tri, i, j, row, col):
     if(j == col ):
         return 0
   
     if(i == row-1 ):
         return tri[i][j]
   
     return tri[i][j] + max(maxPathSum(tri, i+1, j, row, col),
                            maxPathSum(tri, i+1, j+1, row, col))
 
# Driver program to test above functions
tri = [  [1, 0, 0],[4, 8, 0],[1, 5, 3] ]
print(maxPathSum(tri, 0, 0, 3, 3))
 
# This code is contributed by shinjanpatra

C#




// C# program for Recursive implementation
// of Max sum problem in a triangle
using System;
 
public class GFG
{
  static int N = 3;
 
  public static int max(int a, int b){
    if(a > b) return a;
    return b;
  }
 
  //  Function for finding maximum sum
  public static int maxPathSum(int[,] tri, int i, int j, int row, int col)
  {
    if (j == col) {
      return 0;
    }
 
    if (i == row - 1) {
      return tri[i,j];
    }
 
    return tri[i,j]
      + max(
      maxPathSum(tri, i + 1, j, row, col),
      maxPathSum(tri, i + 1, j + 1, row, col)
    );
  }
 
  // Driver program to test above functions
  public static void Main(string[] args)
  {
    int[,] tri = {{ 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
    Console.WriteLine(maxPathSum(tri, 0, 0, 3, 3));
  }
}
 
// This code is contributed by ajaymakavana.

Javascript




<script>
 
// JavaScript program for
// Recursive implementation of
// Max sum problem in a triangle
 
const N = 3
 
//  Function for finding maximum sum
function maxPathSum(tri, i, j, row, col){
     if(j == col ){
         return 0;
     }
   
     if(i == row-1 ){
         return tri[i][j] ;
     }
   
     return tri[i][j] + Math.max(maxPathSum(tri, i+1, j, row, col),
                            maxPathSum(tri, i+1, j+1, row, col)) ;
}
 
/* Driver program to test above functions */
let tri = [  [1, 0, 0],[4, 8, 0],[1, 5, 3] ];
document.write(maxPathSum(tri, 0, 0, 3, 3));
 
// This code is contributed by shinjanpatra
 
</script>

Output

14

Complexity Analysis:

  • Time Complexity: O(2N*N)
  • Auxiliary Space:  O(N)

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 programming concept to find the maximum path sum. 

Method 2: DP Top-Down
Since there are overlapping subproblems, we can avoid the repeated work done in method 1 by storing the min-cost path calculated so far using top-down approach

C++




// C++ program for Dynamic
// Programming implementation (Top-Down) 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 i, int j, int row, int col, vector<vector<int>> &dp){
     if(j == col ){
         return 0;
     }
   
     if(i == row-1 ){
         return tri[i][j] ;
     }
   
     if(dp[i][j] != -1){
         return dp[i][j] ;
     }
   
     return dp[i][j] = tri[i][j] + max(maxPathSum(tri, i+1, j, row, col, dp),
                                   maxPathSum(tri, i+1, j+1, row, col, dp)) ;
}
 
/* Driver program to test above functions */
int main()
{
   int tri[N][N] = {  {1, 0, 0},
                      {4, 8, 0},
                      {1, 5, 3} };
   vector<vector<int>> dp(N, vector<int>(N, -1) ) ;
   cout << maxPathSum(tri, 0, 0, N, N, dp);
   return 0;
}

Java




// Java program for Dynamic
// Programming implementation (Top-Down) of
// Max sum problem in a triangle
 
import java.io.*;
import java.util.*;
 
class GFG {
 
  //  Function for finding maximum sum
  static int maxPathSum(int[][] tri, int i, int j,
                        int row, int col, int[][] dp)
  {
    if (j == col) {
      return 0;
    }
 
    if (i == row - 1) {
      return tri[i][j];
    }
 
    if (dp[i][j] != -1) {
      return dp[i][j];
    }
 
    return dp[i][j]
      = tri[i][j]
      + Math.max(
      maxPathSum(tri, i + 1, j, row, col, dp),
      maxPathSum(tri, i + 1, j + 1, row, col,
                 dp));
  }
 
  public static void main(String[] args)
  {
    int n = 3;
    int[][] tri
      = { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
    int[][] dp = new int[n][n];
    for (int[] row : dp) {
      Arrays.fill(row, -1);
    }
    System.out.print(maxPathSum(tri, 0, 0, n, n, dp));
  }
}
 
// This code is contributed by lokeshmvs21.

Python3




# C++ program for Dynamic
# Programming implementation (Top-Down) of
# Max sum problem in a triangle
N = 3
 
#  Function for finding maximum sum
def maxPathSum(tri, i, j, row, col, dp):
     if(j == col):
         return 0
   
     if(i == row-1):
         return tri[i][j]
   
     if(dp[i][j] != -1):
         return -1
   
     dp[i][j] = tri[i][j] + max(maxPathSum(tri, i+1, j, row, col, dp),
                                   maxPathSum(tri, i+1, j+1, row, col, dp))
     return dp[i][j]
 
# Driver program to test above functions
tri = [ [1, 0, 0],
        [4, 8, 0],
        [1, 5, 3] ]
dp = [[-1 for i in range(N)]for j in range(N)]
print(maxPathSum(tri, 0, 0, N, N, dp))
 
 
# This code is contributed by shinjanpatra

C#




// C# program for Dynamic
// Programming implementation (Top-Down) of
// Max sum problem in a triangle
using System;
class GFG
{
 
  //  Function for finding maximum sum
  static int maxPathSum(int[, ] tri, int i, int j,
                        int row, int col, int[, ] dp)
  {
    if (j == col) {
      return 0;
    }
 
    if (i == row - 1) {
      return tri[i, j];
    }
 
    if (dp[i, j] != -1) {
      return dp[i, j];
    }
 
    return dp[i, j]
      = tri[i, j]
      + Math.Max(
      maxPathSum(tri, i + 1, j, row, col, dp),
      maxPathSum(tri, i + 1, j + 1, row, col,
                 dp));
  }
  static void Main()
  {
    int[, ] tri
      = { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
    int N = 3;
    int[, ] dp = new int[N, N];
    for (int i = 0; i < N; i++)
      for (int j = 0; j < N; j++)
        dp[i, j] = -1;
    Console.Write(maxPathSum(tri, 0, 0, N, N, dp));
  }
}
 
// This code is contributed by garg28harsh.

Javascript




<script>
 
b// JavaScript program for Dynamic
// Programming implementation (Top-Down) of
// Max sum problem in a triangle
const N = 3
 
//  Function for finding maximum sum
function maxPathSum(tri, i, j, row, col, dp){
     if(j == col)
         return 0
   
     if(i == row-1)
         return tri[i][j]
   
     if(dp[i][j] != -1)
         return -1
   
     dp[i][j] = tri[i][j] + Math.max(maxPathSum(tri, i+1, j, row, col, dp),
                                   maxPathSum(tri, i+1, j+1, row, col, dp))
     return dp[i][j]
}
 
// Driver program to test above functions
let tri = [ [1, 0, 0],
        [4, 8, 0],
        [1, 5, 3] ]
let dp = new Array(N).fill(-1).map(()=>new Array(N).fill(-1))
document.write(maxPathSum(tri, 0, 0, N, N, dp),"</br>")
 
// This code is contributed by shinjanpatra
 
</script>

Output

14

Complexity Analysis:

  • Time Complexity:  O(m*n) where m = no of rows and n = no of columns
  • Auxiliary Space:  O(n2)

Method 3: DP(Bottom – UP)
Since there are overlapping subproblems, we can avoid the repeated work done in method 1 by storing the min-cost path calculated so far using the bottom-up approach thus reducing stack space

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 n, vector<vector<int>> &dp)
{
     // loop for bottom-up calculation
     for(int j = 0; j < n; j++ ){
            dp[n-1][j] = tri[n-1][j] ;
        }
         
        for(int i = n-2; i >= 0; i--){
            for(int j = i; j >= 0; j-- ){
                dp[i][j] = tri[i][j] + max(dp[i+1][j] , dp[i+1][j+1]) ;
            }
        }
         
        return dp[0][0] ;
      
}
 
/* Driver program to test above functions */
int main()
{
   int tri[N][N] = {  {1, 0, 0},
                      {4, 8, 0},
                      {1, 5, 3} };
   vector<vector<int>> dp(N, vector<int>(N, -1) ) ;
   cout << maxPathSum(tri, N, dp);
   return 0;
}

Java




// Java program for Dynamic
// Programming implementation of
// Max sum problem in a triangle
public class Main
{
 
  //  Function for finding maximum sum
  static int maxPathSum(int[][] tri, int n, int[][] dp)
  {
 
    // loop for bottom-up calculation
    for (int j = 0; j < n; j++) {
      dp[n - 1][j] = tri[n - 1][j];
    }
 
    for (int i = n - 2; i >= 0; i--) {
      for (int j = i; j >= 0; j--) {
        dp[i][j] = tri[i][j]
          + Math.max(dp[i + 1][j],
                     dp[i + 1][j + 1]);
      }
    }
 
    return dp[0][0];
  }
 
  public static void main(String[] args)
  {
    int N = 3;
    int[][] tri
      = { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
    int[][] dp = new int[N][N];
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        dp[i][j] = -1;
      }
    }
 
    System.out.println(maxPathSum(tri, N, dp));
  }
}
 
// This code is contributed by garg28harsh.

Python3




# Python program for Dynamic
# Programming implementation of
# Max sum problem in a triangle
 
# Number of rows in the triangle
N = 3
 
# Function for finding maximum sum
 
def maxPathSum(tri, n, dp):
    # Loop for bottom-up calculation
    for j in range(n):
        dp[n-1][j] = tri[n-1][j]
 
    for i in range(n-2, -1, -1):
        for j in range(i, -1, -1):
            dp[i][j] = tri[i][j] + max(dp[i+1][j], dp[i+1][j+1])
 
    # Return the maximum sum
    return dp[0][0]
 
 
# Driver program to test above functions
if __name__ == '__main__':
    # Triangle of numbers
    tri = [[1, 0, 0],
           [4, 8, 0],
           [1, 5, 3]]
 
    # Initialize dp array with -1
    dp = [[-1 for j in range(N)] for i in range(N)]
 
    # Print the maximum sum
    print(maxPathSum(tri, N, dp))
# This code is contributed by divyansh2212

C#




// C# program for Dynamic
// Programming implementation of
// Max sum problem in a triangle
 
using System;
 
public class GFG {
 
    //  Function for finding maximum sum
    static int maxPathSum(int[, ] tri, int n, int[, ] dp)
    {
 
        // loop for bottom-up calculation
        for (int j = 0; j < n; j++) {
            dp[n - 1, j] = tri[n - 1, j];
        }
 
        for (int i = n - 2; i >= 0; i--) {
            for (int j = i; j >= 0; j--) {
                dp[i, j] = tri[i, j]
                           + Math.Max(dp[i + 1, j],
                                      dp[i + 1, j + 1]);
            }
        }
 
        return dp[0, 0];
    }
 
    public static void Main(string[] args)
    {
        int N = 3;
        int[, ] tri = new int[, ] { { 1, 0, 0 },
                                    { 4, 8, 0 },
                                    { 1, 5, 3 } };
        int[, ] dp = new int[N, N];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                dp[i, j] = -1;
            }
        }
 
        Console.WriteLine(maxPathSum(tri, N, dp));
    }
}
// This code is contributed by karandeep1234

Javascript




// Javascript program for Dynamic
// Programming implementation of
// Max sum problem in a triangle
let N = 3
 
//  Function for finding maximum sum
function maxPathSum(tri, n, dp)
{
     // loop for bottom-up calculation
     for(let j = 0; j < n; j++ ){
            dp[n-1][j] = tri[n-1][j] ;
        }
         
        for(let i = n-2; i >= 0; i--){
            for(let j = i; j >= 0; j-- ){
                dp[i][j] = tri[i][j] + Math.max(dp[i+1][j] , dp[i+1][j+1]) ;
            }
        }
         
        return dp[0][0] ;
      
}
 
/* Driver program to test above functions */
let tri = [  [1, 0, 0],
                  [4, 8, 0],
                  [1, 5, 3] ];
let dp = new Array(N);
for(let i = 0; i < N; i++)
    dp[i] = new Array(N).fill(-1);
console.log(maxPathSum(tri, N, dp));
  
 // This code is contributed by poojaagarwal2.

Output

14

Complexity Analysis:

  • Time Complexity: O(m*n) where m = no of rows and n = no of columns
  • Auxiliary Space: O(n2)

Method 4: Space Optimization (Without Changning input matrix)
We do not need a 2d matrix we only need a 1d array that stores the minimum of the immediate next column  and thus we can reduce space

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 n, vector<vector<int>> &dp)
{
        vector<int> front(n, -1) , curr(n, -1) ;
        
        for(int j = 0; j < n; j++ ){
            front[j] = tri[n-1][j] ;
        }
         
        for(int i = n-2; i >= 0; i--){
            for(int j = i; j >= 0; j-- ){
                curr[j] = tri[i][j] + max(front[j] , front[j+1]) ;
            }
            front = curr ;
        }
         
        return front[0] ;
      
}
 
/* Driver program to test above functions */
int main()
{
   int tri[N][N] = {  {1, 0, 0},
                      {4, 8, 0},
                      {1, 5, 3} };
   vector<vector<int>> dp(N, vector<int>(N, -1) ) ;
   cout << maxPathSum(tri, N, dp);
   return 0;
}

Java




// Java program for Dynamic
// Programming implementation of
// Max sum problem in a triangle
public class GFG
{
   
  //  Function for finding maximum sum
  static int maxPathSum(int[][] tri, int n, int[][] dp)
  {
 
    int front[] = new int[n];
    int curr[] = new int[n];
    for (int j = 0; j < n; j++) {
      front[j] = tri[n - 1][j];
      curr[j] = -1;
    }
 
    for (int i = n - 2; i >= 0; i--) {
      for (int j = i; j >= 0; j--) {
        curr[j]
          = tri[i][j]
          + Math.max(front[j], front[j + 1]);
      }
      front = curr;
    }
 
    return front[0];
  }
  public static void main(String[] args)
  {
    int N = 3;
    int[][] tri
      = { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
    int[][] dp = new int[N][N];
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        dp[i][j] = -1;
      }
    }
 
    System.out.println(maxPathSum(tri, N, dp));
  }
}
 
// This code is contributed by garg28harsh.

Python3




# Python program for Dynamic
# Programming implementation of
# Max sum problem in a triangle
 
N=3
 
# Function for finding maximum sum
def maxPathSum(tri,n,dp):
    front=[-1]*n
    curr=[-1]*n
    for j in range(n):
        front[j] = tri[n-1][j]
     
    for i in range(n-2,-1,-1):
        for j in range(i,-1,-1):
            curr[j] = tri[i][j] + max(front[j] ,front[j+1])
        front=curr
     
    return front[0]
     
# Driver program to test above functions
tri=[[1, 0, 0],[4,8,0],[1,5,3]]
dp= [[-1 for i in range(N)] for j in range(N)]
print(maxPathSum(tri,N,dp))
 
# This code is contributed by Pushpesh Raj.

C#




// C# program for Dynamic
// Programming implementation of
// Max sum problem in a triangle
using System;
public class GFG {
 
  // Function for finding maximum sum
  static int maxPathSum(int[, ] tri, int n, int[, ] dp)
  {
 
    int[] front = new int[n];
    int[] curr = new int[n];
    for (int j = 0; j < n; j++) {
      front[j] = tri[n - 1, j];
      curr[j] = -1;
    }
 
    for (int i = n - 2; i >= 0; i--) {
      for (int j = i; j >= 0; j--) {
        curr[j]
          = tri[i, j]
          + Math.Max(front[j], front[j + 1]);
      }
      front = curr;
    }
 
    return front[0];
  }
  public static void Main(string[] args)
  {
    int N = 3;
    int[, ] tri
      = { { 1, 0, 0 }, { 4, 8, 0 }, { 1, 5, 3 } };
    int[, ] dp = new int[N, N];
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        dp[i, j] = -1;
      }
    }
 
    Console.WriteLine(maxPathSum(tri, N, dp));
  }
}
 
// This code is contributed by karandeep1234

Javascript




// Javascript program for Dynamic
// Programming implementation of
// Max sum problem in a triangle
let N = 3
 
//  Function for finding maximum sum
function maxPathSum(tri,  n, dp)
{
        let front= new Array(n).fill(-1);
        let curr=new Array(n).fill(-1);
        
        for(let j = 0; j < n; j++ ){
            front[j] = tri[n-1][j] ;
        }
         
        for(let i = n-2; i >= 0; i--){
            for(let j = i; j >= 0; j-- ){
                curr[j] = tri[i][j] + Math.max(front[j] , front[j+1]) ;
            }
            front = curr ;
        }
         
        return front[0] ;
      
}
 
/* Driver program to test above functions */
let tri = [  [1, 0, 0],[4, 8, 0],[1, 5, 3]];
dp = new Array(N);
for(let i = 0; i < N; i++)
    dp[i] = new Array(N).fill(-1);
console.log(maxPathSum(tri, N, dp));
  
 // This code is contributed by ritaagarwal.

Output

14

Complexity Analysis:

  • Time Complexity: O(m*n) where m = no of rows and n = no of columns
  • Auxiliary Space:  O(n)

Method 5: Space Optimization (Changing input matrix)
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
?>

Javascript




<script>
// Javascript Program for Dynamic
// Programming implementation of
// Max sum problem in a triangle
    let N = 3;
       
    // Function for finding maximum sum
    function maxPathSum(tri, m, n)
    {
     
        // loop for bottom-up calculation
        for (let i = m - 1; i >= 0; i--)
        {
            for (let 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   
    let tri = [[1, 0, 0],
                  [4, 8, 0],
                  [1, 5, 3]];
       document.write( maxPathSum(tri, 2, 2));
               
// This code is contributed by susmitakundugoaldanga.           
</script>

Output

14

Complexity Analysis:

  • Time Complexity: O(m*n) where m = no of rows and n = no of columns
  • Auxiliary Space:  O(1)

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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