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Maximum sum of a path in a Right Number Triangle

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  • Difficulty Level : Easy
  • Last Updated : 24 May, 2022
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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

 

Method 1: Recursion

The idea is to find the largest sum ending at every cell of the last row and return a maximum of these sums. We can recursively compute these sums by recursively considering the above two cells. 

C++




// C++ program for
// Recursion implementation of
// Min Sum Path in a Triangle
#include <bits/stdc++.h>
using namespace std;
 
// Util function to find minimum sum for a path
int helper(vector<vector<int>>& tri, int i, int j){
    // Base Case 
    if(i == tri.size() ){
      return 0 ;
    }
   
    int mx ;
    // Add current to the maximum  of the next paths
    mx = tri[i][j] + max(helper(tri, i+1,j), helper(tri,i+1, j+1)) ;
    //return maximum
    return mx ;
}
 
 
int maxSumPath(vector<vector<int>>& tri) {
    return helper(tri, 0, 0) ;
}
 
/* Driver program to test above functions */
int main()
{
    vector<vector<int> > tri{ { 1 },
                            { 2, 1 },
                            { 3, 3, 2 } };
    cout << maxSumPath(tri);
    return 0;
}

Java




// Java program for
// Recursion implementation of
// Min Sum Path in a Triangle
import java.io.*;
 
class GFG
{
 
  // Util function to find minimum sum for a path
  public static int helper(int tri[][], int i, int j)
  {
 
    // Base Case
    if (i == tri.length) {
      return 0;
    }
 
    int mx;
     
    // Add current to the maximum  of the next paths
    mx = tri[i][j]
      + Math.max(helper(tri, i + 1, j),
                 helper(tri, i + 1, j + 1));
     
    // return maximum
    return mx;
  }
 
  public static int maxSumPath(int tri[][])
  {
    return helper(tri, 0, 0);
  }
 
  /* Driver program to test above functions */
  public static void main(String[] args)
  {
    int tri[][] = { { 1 }, { 2, 1 }, { 3, 3, 2 } };
    System.out.print(maxSumPath(tri));
  }
}
 
// This code is contributed by Rohit Pradhan

Output

6

Complexity Analysis:

  • Time Complexity: O(2N*N) where N = number of rows and M = number of columns
  • Auxiliary Space: O(N)

Method 2: Dynamic Programming – Top-Down Approach

Since there are overlapping subproblems, we use dynamic programming to find the maximum sum ending at a particular cell of the last row.
Below is the implementation of the above idea.

C++




// C++ program for Dynamic
// Programming implementation of
// Min Sum Path in a Triangle
#include <bits/stdc++.h>
using namespace std;
 
// Util function to find minimum sum for a path
int helper(vector<vector<int>>& tri, int i, int j, vector<vector<int>>& dp){
    // Base Case 
    if(i == tri.size() ){
      return 0 ;
    }
   
   // To avoid solving overlapping subproblem
   if(dp[i][j] != -1){
     return dp[i][j] ;
   }
   
 
    // Add current to the minimum  of the next paths
    // and store it in dp matrix
    return dp[i][j] = tri[i][j] + max(helper(tri, i+1,j, dp), helper(tri,i+1, j+1, dp)) ;
     
     
}
 
 
int maxSumPath(vector<vector<int>>& tri) {
    int n = tri.size() ;
    // Initializating of dp matrix
    vector<vector<int>> dp(n, vector<int>(n, -1) ) ;
    // calling helper function
    return helper(tri, 0, 0, dp) ;
}
 
/* Driver program to test above functions */
int main()
{
    vector<vector<int> > tri{ { 1 },
                            { 2, 1 },
                            { 3, 3, 2 } };
    cout << maxSumPath(tri);
    return 0;
}

Output

6

Complexity Analysis:

  • Time Complexity: O(N*M) where N = number of rows and M = number of columns
  • Auxiliary Space: O(N2)

Method 3: Dynamic Programming: Bottom-Up Approach

Since there are overlapping subproblems, we use dynamic programming to find the maximum sum ending at a particular cell of the last row. Below is the implementation of the above idea.

C++




// C++ program for Dynamic
// Programming implementation of
// Min Sum Path in a Triangle
#include <bits/stdc++.h>
using namespace std;
 
// Util function to find minimum sum for a path
int helper(vector<vector<int>>& tri, int i, int j, vector<vector<int>>& dp){
    int n = tri.size() ;
      // 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] ;
     
     
}
 
 
int maxSumPath(vector<vector<int>>& tri) {
    int n = tri.size() ;
    // Initializating of dp matrix
    vector<vector<int>> dp(n, vector<int>(n, -1) ) ;
    // calling helper function
    return helper(tri, 0, 0, dp) ;
}
 
/* Driver program to test above functions */
int main()
{
    vector<vector<int> > tri{ { 1 },
                            { 2, 1 },
                            { 3, 3, 2 } };
    cout << maxSumPath(tri);
    return 0;
}

Output

6

Complexity Analysis:

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

Method 4: Space Optimization(Without changing the input matrix)

We don’t need a matrix of m*n size. It will store all the results.

 We just need the triangle row minimum of the immediate bottom row

So using this approach we can optimize the space from O(m*n) to O(n).

Approach remains the same as that of Method 3.

C++




// C++ program for Dynamic
// Programming implementation of
// Min Sum Path in a Triangle
#include <bits/stdc++.h>
using namespace std;
 
// Util function to find minimum sum for a path
int helper(vector<vector<int>>& tri, int i, int j, vector<vector<int>>& dp){
    int n = tri.size() ;
    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] ;
     
}
 
 
int maxSumPath(vector<vector<int>>& tri) {
    int n = tri.size() ;
    // Initializating of dp matrix
    vector<vector<int>> dp(n, vector<int>(n, -1) ) ;
    // calling helper function
    return helper(tri, 0, 0, dp) ;
}
 
/* Driver program to test above functions */
int main()
{
    vector<vector<int> > tri{ { 1 },
                            { 2, 1 },
                            { 3, 3, 2 } };
    cout << maxSumPath(tri);
    return 0;
}

Output

6

Complexity Analysis:

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

Method 5: Space Optimization (Changing the input matrix)
 

C++




// C++ program to print maximum sum
// in a right triangle of numbers
#include<bits/stdc++.h>
using namespace std;
 
// function to find maximum sum path
int maxSum(int tri[][3], 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 program
int main(){
     
    int tri[3][3] = {{1}, {2,1}, {3,3,2}};
 
    cout<<maxSum(tri, 3);
     
    return 0;
}

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.

Python3




# 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




<?php
// PHP program to print maximum sum
// in a right triangle of numbers
 
// function to find maximum sum path
function 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 = 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 ($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][0];
     
    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
?>

Javascript




<script>
 
    // Javascript program to print maximum sum
    // in a right triangle of numbers
     
    // function to find maximum sum path
    function 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(let 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 (let 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
        let max = tri[n-1][0];
           
        for(let i = 1; i < n; i++)
        {
            if(max < tri[n-1][i])
                max = tri[n-1][i];
        }
           
        return max;
    }
     
    let tri = [[1], [2,1], [3,3,2]];
           
      document.write(maxSum(tri, 3));
         
</script>

Output

6

Complexity Analysis:

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

This article is contributed by Harshit Agrawal. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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