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

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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// CPP 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 

Output:

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

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

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