Find the largest contiguous pair sum in given Array

Given an integer array arr[] of size N, the task is to find contiguous pair {a, b} such that sum of both elements in the pair is maximum. If there are more than one such pairs with maximum sum then print any of such pair. In the case of multiple pairs with the largest sum, print any one of them.
Examples: 

Input: arr[] = {1, 2, 3, 4} 
Output: 3 4 
Explanation: 
Here, the contiguous pairs in the array are: 
{1, 2} -> Sum 3 
{2, 3} -> Sum 5 
{3, 4} -> Sum 7 
Maxiumum sum is 7 for the pair is (3, 4) so this is answer.

Input: arr[] = {11, -5, 9, -3, 2} 
Output: 11 -5 
The contiguous pairs with their respective sums are : 
{11, -5} -> Sum 6 
{-5, 9} -> Sum 4 
{9, -3} -> Sum 6 
{-3, 2} -> Sum -1 
The maximum sum obtained is 6 from the pairs (11, -5) and (9, -3). 
 

Approach: 
Follow the steps below to solve the problem: 

  1. Generate all the continuous pairs one by one and calculate there sum.
  2. Compare the sum of every pair with the maximum sum and update the pair corresponding to the maximum sum accordingly.
  3. Return the pair representing the maximum sum.

Below is the implementation of the above approach :



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// C++ program to find the 
// a contiguous pair from the
// which has the largest sum
#include <bits/stdc++.h>
using namespace std;
  
// Function to find and return
// the largest sum contiguous pair
vector<int> largestSumpair(int arr[], int n)
{
      
    // Stores the contiguous pair 
    vector<int> pair;
      
    // Intialize maximum sum
    int max_sum = INT_MIN, i;
      
    for(i = 1; i < n; i++) 
    {
          
        // Compare sum of pair with max_sum
        if (max_sum < (arr[i] + arr[i - 1]))
        {
            max_sum = arr[i] + arr[i - 1];
            if (pair.empty()) 
            
                  
                // Insert the pair
                pair.push_back(arr[i - 1]);
                pair.push_back(arr[i]);
            }
            else
            {
                pair[0] = arr[i - 1];
                pair[1] = arr[i];
            }
        }
        return pair;
    }
}
      
// Driver Code 
int main() 
    int arr[] = {11, -5, 9, -3, 2};
    int N = sizeof(arr) / sizeof(arr[0]);
      
    vector<int> pair = largestSumpair(arr, N);
    for(auto it = pair.begin(); it != pair.end(); ++it)
    {
        cout << *it << ' '
    }
      
    return 0; 
  
// This code is contributed by shubhamsingh10
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// Java program to find the 
// a contiguous pair from the 
// which has the largest sum 
import java.util.*;
  
class GFG{
      
// Function to find and return 
// the largest sum contiguous pair 
public static Vector<Integer> largestSumpair(int[] arr,
                                             int n) 
      
    // Stores the contiguous pair 
    Vector<Integer> pair = new Vector<Integer>(); 
      
    // Intialize maximum sum 
    int max_sum = Integer.MIN_VALUE, i; 
      
    for(i = 1; i < n; i++) 
    
          
        // Compare sum of pair with max_sum 
        if (max_sum < (arr[i] + arr[i - 1])) 
        
            max_sum = arr[i] + arr[i - 1]; 
              
            if (pair.isEmpty()) 
            
                  
                // Insert the pair 
                pair.add(arr[i - 1]); 
                pair.add(arr[i]); 
            
            else
            
                pair.set(0, arr[i - 1]);
                pair.set(1, arr[i]);
            
        
    
    return pair;
  
// Driver Code    
public static void main(String[] args) 
{
    int arr[] = { 11, -5, 9, -3, 2 }; 
    int N = arr.length; 
      
    Vector<Integer> pair = new Vector<Integer>(); 
    pair = largestSumpair(arr, N);
    for (Integer it : pair)
    {         
        System.out.print(it + " ");
    }
}
}
  
// This code is contributed by divyeshrabadiya07
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# Python3 program to find the 
# a contiguous pair from the
# which has the largest sum
  
# importing sys
import sys
  
# Function to find and return
# the largest sum contiguous pair
def largestSumpair(arr, n):
# Stores the contiguous pair 
    pair = []
  
# Intialize maximum sum
    max_sum = -sys.maxsize-1
  
    for i in range(1, n):
      
        # Compare sum of pair with max_sum
        if max_sum < ( arr[i] + arr[i-1] ):
            max_sum = arr[i] + arr[i-1]
          
            if pair == []:
                # Insert the pair
                pair.append(arr[i-1])
                pair.append(arr[i])
            else:
                pair[0] = arr[i-1]
                pair[1] = arr[i]
  
    return pair
      
      
# Driver Code 
arr = [11, -5, 9, -3, 2
N = len(arr) 
pair = largestSumpair(arr, N)
print(pair[0], end =" ")
print(pair[1], end =" ")
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// C# program to find the 
// a contiguous pair from the 
// which has the largest sum 
using System;
using System.Collections; 
using System.Collections.Generic;
  
class GFG{
  
// Function to find and return 
// the largest sum contiguous pair 
public static ArrayList largestSumpair(int[] arr,
                                       int n) 
      
    // Stores the contiguous pair 
    ArrayList pair = new ArrayList(); 
      
    // Intialize maximum sum 
    int max_sum = int.MinValue, i; 
      
    for(i = 1; i < n; i++) 
    
          
        // Compare sum of pair with max_sum 
        if (max_sum < (arr[i] + arr[i - 1])) 
        
            max_sum = arr[i] + arr[i - 1]; 
              
            if (pair.Count == 0) 
            
                  
                // Insert the pair 
                pair.Add(arr[i - 1]); 
                pair.Add(arr[i]); 
            
            else
            
                pair[0] = arr[i - 1];
                pair[1] = arr[i];
            
        
    
    return pair;
  
// Driver code
public static void Main(string[] args)
{
    int []arr = { 11, -5, 9, -3, 2 }; 
    int N = arr.Length; 
      
    ArrayList pair = new ArrayList(); 
    pair = largestSumpair(arr, N);
      
    foreach(int it in pair)
    {         
        Console.Write(it + " ");
    }
}
}
  
// This code is contributed by rutvik_56
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Output: 
11 -5

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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