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Find the largest contiguous pair sum in given Array

  • Last Updated : 13 Jul, 2021

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 :
 

C++




// 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;
     
    // Initialize 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

Java




// 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>();
     
    // Initialize 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

Python3




# 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 = []
 
# Initialize 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 =" ")

C#




// 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();
     
    // Initialize 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
Output: 



11 -5

 

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

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