Find two non-overlapping pairs having equal sum in an Array

Difficulty Level : Basic
Last Updated : 28 Oct, 2020

Given an unsorted array of integers. The task is to find any two non-overlapping pairs whose sum is equal.
Two pairs are said to be non-overlapping if all the elements of the pairs are at different indices. That is, pair (A_i, A_j) and pair (A_m, A_n) are said to be non-overlapping if i ≠ j ≠ m ≠ n.

Examples:

Input: arr[] = {8, 4, 7, 8, 4}
Output: Pair First(4, 8)
        Pair Second (8, 4)

Input: arr[] = {8, 4, 7, 4}
Output: No such non-overlapping pairs present.
Note: (8, 4) and (8, 4) forms overlapping pair
as index of 8 is same in both pairs.

The idea is to use a map of key-value pair to store all occurrences of a sum. The key in the map will store the sum and the corresponding value will be a list of pair of indices (i, j) with that sum.

The idea is to traverse the array and generate all possible pairs.
If a new sum is found, insert it directly to the map otherwise check if any of the previously encountered pairs with that sum does not overlap with current pair.
If not, print both of the pairs.

Below is implementation of the above approach:

C++

// C++ programs to find two non-overlapping
// pairs having equal sum in an Array
 
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
 
typedef pair<int, int> Pair;
 
// Function to find two non-overlapping
// with same sum in an array
void findPairs(int arr[], int n)
{
    // first create an empty map
    // key -> which is sum of a pair of
    // elements in the array
    // value -> vector storing index of
    // every pair having that sum
    unordered_map<int, vector<Pair> > map;
 
    // consider every pair (arr[i], arr[j])
    // and where (j > i)
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // calculate sum of current pair
            int sum = arr[i] + arr[j];
 
            // if sum is already present in the map
            if (map.find(sum) != map.end()) {
 
                // check every pair having equal sum
                for (auto pair : map.find(sum)->second) {
                    int m = pair.first, n = pair.second;
 
                    // if pairs don't overlap,
                    // print them and return
                    if ((m != i && m != j) && (n != i && n != j)) {
                        cout << "Pair First(" << arr[i] << ", "
                             << arr[j] << ")\nPair Second ("
                             << arr[m] << ", " << arr[n] << ")";
                        return;
                    }
                }
            }
 
            // Insert current pair into the map
            map[sum].push_back({ i, j });
        }
    }
 
    // If no such pair found
    cout << "No such non-overlapping pairs present";
}
 
// Driver Code
int main()
{
    int arr[] = { 8, 4, 7, 8, 4 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    findPairs(arr, size);
 
    return 0;
}

Java

// Java program to find two non-overlapping
// pairs having equal sum in an Array
 
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
 
// Declare a pair
class Pair {
    public int x, y;
 
    Pair(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}
 
class GFG {
    // Function to find two non-overlapping
    // pairs with same sum in an array
    public static void findPairs(int[] A)
    {
        // first create an empty map
        // key -> which is sum of a pair
        // of elements in the array
        // value -> list storing index of
        // every pair having that sum
        Map<Integer, List<Pair> > map = new HashMap<>();
 
        // consider every pair (A[i], A[j]) where (j > i)
        for (int i = 0; i < A.length - 1; i++) {
            for (int j = i + 1; j < A.length; j++) {
 
                // calculate sum of current pair
                int sum = A[i] + A[j];
 
                // if sum is already present in the map
                if (map.containsKey(sum)) {
 
                    // check every pair having desired sum
                    for (Pair pair : map.get(sum)) {
                        int x = pair.x;
                        int y = pair.y;
 
                        // if pairs don't overlap, print
                        // them and return
                        if ((x != i && x != j) && (y != i && y != j)) {
                            System.out.println("Pair First(" + A[i] + ", "
                                               + A[j] + ")");
 
                            System.out.println("Pair Second (" + A[x] + ", "
                                               + A[y] + ")");
 
                            return;
                        }
                    }
                }
 
                // Insert current pair into the map
                map.putIfAbsent(sum, new ArrayList<>());
                map.get(sum).add(new Pair(i, j));
            }
        }
 
        System.out.print("No such non-overlapping pairs present");
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] A = { 8, 4, 7, 8, 4 };
 
        findPairs(A);
    }
}

Python3

# Python3 programs to find two non-overlapping
# pairs having equal Sum in an Array
 
# Function to find two non-overlapping
# with same Sum in an array
def findPairs(arr, size):
 
    # first create an empty Map
    # key -> which is Sum of a pair of
    # elements in the array
    # value -> vector storing index of
    # every pair having that Sum
    Map = {}
 
    # consider every pair (arr[i], arr[j])
    # and where (j > i)
    for i in range(0, size - 1):
        for j in range(i + 1, size):
             
            # calculate Sum of current pair
            Sum = arr[i] + arr[j]
 
            # if Sum is already present in the Map
            if Sum in Map:
 
                # check every pair having equal Sum
                for pair in Map[Sum]:
                    m, n = pair
 
                    # if pairs don't overlap,
                    # print them and return
                    if ((m != i and m != j) and
                        (n != i and n != j)):
                         
                        print("Pair First ({}, {})".
                               format(arr[i], arr[j]))
                        print("Pair Second ({}, {})".
                               format(arr[m], arr[n]))
                        return
                     
            # Insert current pair into the Map
            if Sum not in Map:
                Map[Sum] = []
            Map[Sum].append((i, j))
     
    # If no such pair found
    print("No such non-overlapping pairs present")
 
# Driver Code
if __name__ == "__main__":
 
    arr = [8, 4, 7, 8, 4]
    size = len(arr)
 
    findPairs(arr, size)
     
# This code is contributed by Rituraj Jain

C#

// C# program to find two non-overlapping
// pairs having equal sum in an Array
using System;
using System.Collections.Generic;
 
// Declare a pair
class Pair
{
    public int x, y;
 
    public Pair(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}
 
class GFG{
     
// Function to find two non-overlapping
// pairs with same sum in an array
public static void findPairs(int[] A)
{
     
    // First create an empty map
    // key -> which is sum of a pair
    // of elements in the array
    // value -> list storing index of
    // every pair having that sum
    Dictionary<int,
               List<Pair>> map = new Dictionary<int,
                                                List<Pair>>();
 
    // Consider every pair (A[i], A[j])
    // where (j > i)
    for(int i = 0; i < A.Length - 1; i++)
    {
        for(int j = i + 1; j < A.Length; j++)
        {
             
            // Calculate sum of current pair
            int sum = A[i] + A[j];
 
            // If sum is already present in the map
            if (map.ContainsKey(sum))
            {
                 
                // Check every pair having desired sum
                foreach(Pair pair in map[sum])
                {
                    int x = pair.x;
                    int y = pair.y;
 
                    // If pairs don't overlap, print
                    // them and return
                    if ((x != i && x != j) &&
                        (y != i && y != j))
                    {
                        Console.WriteLine("Pair First(" + A[i] +
                                          ", " + A[j] + ")");
 
                        Console.WriteLine("Pair Second (" + A[x] +
                                          ", " + A[y] + ")");
                       
                        return;
                    }
                }
            }
 
            // Insert current pair into the map
            map[sum] = new List<Pair>();
            map[sum].Add(new Pair(i, j));
        }
    }
 
    Console.Write("No such non-overlapping pairs present");
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] A = { 8, 4, 7, 8, 4 };
 
    findPairs(A);
}
}
 
// This code is contributed by Princi Singh

Output:

Pair First(4, 8)
Pair Second (8, 4)

My Personal Notes

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