Find the largest contiguous pair sum in given Array
Last Updated :
24 Nov, 2022
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
Maximum 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:
- Generate all the continuous pairs one by one and calculate their sum.
- Compare the sum of every pair with the maximum sum and update the pair corresponding to the maximum sum accordingly.
- Return the pair representing the maximum sum.
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>
using namespace std;
vector< int > largestSumpair( int arr[], int n)
{
vector< int > pair;
int max_sum = INT_MIN, i;
for (i = 1; i < n; i++) {
if (max_sum < (arr[i] + arr[i - 1])) {
max_sum = arr[i] + arr[i - 1];
if (pair.empty()) {
pair.push_back(arr[i - 1]);
pair.push_back(arr[i]);
}
else {
pair[0] = arr[i - 1];
pair[1] = arr[i];
}
}
}
return pair;
}
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;
}
|
Java
import java.util.*;
class GFG {
public static Vector<Integer> largestSumpair( int [] arr,
int n)
{
Vector<Integer> pair = new Vector<Integer>();
int max_sum = Integer.MIN_VALUE, i;
for (i = 1 ; i < n; i++) {
if (max_sum < (arr[i] + arr[i - 1 ])) {
max_sum = arr[i] + arr[i - 1 ];
if (pair.isEmpty()) {
pair.add(arr[i - 1 ]);
pair.add(arr[i]);
}
else {
pair.set( 0 , arr[i - 1 ]);
pair.set( 1 , arr[i]);
}
}
}
return pair;
}
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 + " " );
}
}
}
|
Python3
import sys
def largestSumpair(arr, n):
pair = []
max_sum = - sys.maxsize - 1
for i in range ( 1 , n):
if max_sum < (arr[i] + arr[i - 1 ]):
max_sum = arr[i] + arr[i - 1 ]
if pair = = []:
pair.append(arr[i - 1 ])
pair.append(arr[i])
else :
pair[ 0 ] = arr[i - 1 ]
pair[ 1 ] = arr[i]
return pair
arr = [ 11 , - 5 , 9 , - 3 , 2 ]
N = len (arr)
pair = largestSumpair(arr, N)
print (pair[ 0 ], end = " " )
print (pair[ 1 ], end = " " )
|
C#
using System;
using System.Collections;
using System.Collections.Generic;
class GFG {
public static ArrayList largestSumpair( int [] arr, int n)
{
ArrayList pair = new ArrayList();
int max_sum = int .MinValue, i;
for (i = 1; i < n; i++) {
if (max_sum < (arr[i] + arr[i - 1])) {
max_sum = arr[i] + arr[i - 1];
if (pair.Count == 0) {
pair.Add(arr[i - 1]);
pair.Add(arr[i]);
}
else {
pair[0] = arr[i - 1];
pair[1] = arr[i];
}
}
}
return pair;
}
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 + " " ); }
}
}
|
Javascript
<script>
function largestSumpair(arr, n)
{
var pair = [];
var max_sum = Number.MIN_VALUE, i;
for (i = 1; i < n; i++) {
if (max_sum < (arr[i] + arr[i - 1])) {
max_sum = arr[i] + arr[i - 1];
if (pair.length != 0) {
pair[0] = arr[i - 1];
pair[1]= arr[i];
}
else {
pair[0] = arr[i - 1];
pair[1] = arr[i];
}
}
}
return pair;
}
var arr = [ 11, -5, 9, -3, 2 ];
var N = arr.length;
pair = largestSumpair(arr, N);
for ( var it of pair) {
document.write(it + " " );
}
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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