Find four elements a, b, c and d in an array such that a+b = c+d

Given an array of distinct integers, find if there are two pairs (a, b) and (c, d) such that a+b = c+d, and a, b, c and d are distinct elements. If there are multiple answers, then print any of them.

Example: 

Input:   {3, 4, 7, 1, 2, 9, 8}
Output:  (3, 8) and (4, 7)
Explanation: 3+8 = 4+7

Input:   {3, 4, 7, 1, 12, 9};
Output:  (4, 12) and (7, 9)
Explanation: 4+12 = 7+9

Input:  {65, 30, 7, 90, 1, 9, 8};
Output:  No pairs found

Expected Time Complexity: O(n²)

A Simple Solution is to run four loops to generate all possible quadruples of the array elements. For every quadruple (a, b, c, d), check if (a+b) = (c+d). The time complexity of this solution is O(n⁴).
An Efficient Solution can solve this problem in O(n²) time. The idea is to use hashing. We use sum as key and pair as the value in the hash table. 

Loop i = 0 to n-1 :
    Loop j = i + 1 to n-1 :
        calculate sum
        If in hash table any index already exist
            Then print (i, j) and previous pair 
            from hash table  
        Else update hash table
    EndLoop;
EndLoop;

Below are implementations of the above idea. In the below implementation, the map is used instead of a hash. The time complexity of map insert and search is actually O(Log n) instead of O(1). So below implementation is O(n² Log n).

C++

// Find four different elements a,b,c and d of array such that
// a+b = c+d
#include<bits/stdc++.h>
using namespace std;

bool findPairs(int arr[], int n)
{
    // Create an empty Hash to store mapping from sum to
    // pair indexes
    map<int, pair<int, int> > Hash;

    // Traverse through all possible pairs of arr[]
    for (int i = 0; i < n; ++i)
    {
        for (int j = i + 1; j < n; ++j)
        {
            // If sum of current pair is not in hash,
            // then store it and continue to next pair
            int sum = arr[i] + arr[j];
            if (Hash.find(sum) == Hash.end())
                Hash[sum] = make_pair(i, j);

            else // Else (Sum already present in hash)
            {
                // Find previous pair
                pair<int, int> pp = Hash[sum];// pp->previous pair

                // Since array elements are distinct, we don't
                // need to check if any element is common among pairs
                cout << "(" << arr[pp.first] << ", " << arr[pp.second]
                    << ") and (" << arr[i] << ", " << arr[j] << ")n";
                return true;
            }
        }
    }

    cout << "No pairs found";
    return false;
}

// Driver program
int main()
{
    int arr[] = {3, 4, 7, 1, 2, 9, 8};
    int n = sizeof arr / sizeof arr[0];
    findPairs(arr, n);
    return 0;
}

Java

// Java Program to find four different elements a,b,c and d of
// array such that a+b = c+d
import java.io.*;
import java.util.*;

class ArrayElements
{
    // Class to represent a pair
    class pair
    {
        int first, second;
        pair(int f,int s)
        {
            first = f; second = s;
        }
    };

    boolean findPairs(int arr[])
    {
        // Create an empty Hash to store mapping from sum to
        // pair indexes
        HashMap<Integer,pair> map = new HashMap<Integer,pair>();
        int n=arr.length;

        // Traverse through all possible pairs of arr[]
        for (int i=0; i<n; ++i)
        {
            for (int j=i+1; j<n; ++j)
            {
                // If sum of current pair is not in hash,
                // then store it and continue to next pair
                int sum = arr[i]+arr[j];
                if (!map.containsKey(sum))
                    map.put(sum,new pair(i,j));

                else // Else (Sum already present in hash)
                {
                    // Find previous pair
                    pair p = map.get(sum);

                    // Since array elements are distinct, we don't
                    // need to check if any element is common among pairs
                    System.out.println("("+arr[p.first]+", "+arr[p.second]+
                                      ") and ("+arr[i]+", "+arr[j]+")");
                    return true;
                }
            }
        }
        return false;
    }

    // Testing program
    public static void main(String args[])
    {
        int arr[] = {3, 4, 7, 1, 2, 9, 8};
        ArrayElements a = new ArrayElements();
        a.findPairs(arr);
    }
}
// This code is contributed by Aakash Hasija

Python3

# Python Program to find four different elements a,b,c and d of
# array such that a+b = c+d

# function to find a, b, c, d such that
# (a + b) = (c + d)


def find_pair_of_sum(arr: list, n: int):
    map = {}

    for i in range(n):
        for j in range(i+1, n):
            sum = arr[i] + arr[j]

            if sum in map:
                print(f"{map[sum]} and ({arr[i]}, {arr[j]})")
                return
            else:
                map[sum] = (arr[i], arr[j])


# Driver code
if __name__ == "__main__":
    arr = [3, 4, 7, 1, 2, 9, 8]
    n = len(arr)
    find_pair_of_sum(arr, n)

C#

using System;
using System.Collections.Generic;

// C# Program to find four different elements a,b,c and d of
// array such that a+b = c+d

public class ArrayElements
{
    // Class to represent a pair
    public class pair
    {
        private readonly ArrayElements outerInstance;

        public int first, second;
        public pair(ArrayElements outerInstance, int f, int s)
        {
            this.outerInstance = outerInstance;
            first = f;
            second = s;
        }
    }

    public virtual bool findPairs(int[] arr)
    {
        // Create an empty Hash to store mapping from sum to
        // pair indexes
        Dictionary<int, pair> map = new Dictionary<int, pair>();
        int n = arr.Length;

        // Traverse through all possible pairs of arr[]
        for (int i = 0; i < n; ++i)
        {
            for (int j = i + 1; j < n; ++j)
            {
                // If sum of current pair is not in hash,
                // then store it and continue to next pair
                int sum = arr[i] + arr[j];
                if (!map.ContainsKey(sum))
                {
                    map[sum] = new pair(this, i,j);
                }

                else // Else (Sum already present in hash)
                {
                    // Find previous pair
                    pair p = map[sum];

                    // Since array elements are distinct, we don't
                    // need to check if any element is common among pairs
                    Console.WriteLine("(" + arr[p.first] + ", " + arr[p.second] + ") and (" + arr[i] + ", " + arr[j] + ")");
                    return true;
                }
            }
        }
        return false;
    }

    // Testing program
    public static void Main(string[] args)
    {
        int[] arr = new int[] {3, 4, 7, 1, 2, 9, 8};
        ArrayElements a = new ArrayElements();
        a.findPairs(arr);
    }
}

// This code is contributed by Shrikant13

Javascript

<script>
// Find four different elements a,b,c and d of array such that
// a+b = c+d


function findPairs(arr, n) {
    // Create an empty Hash to store mapping from sum to
    // pair indexes
    let Hash = new Map();

    // Traverse through all possible pairs of arr[]
    for (let i = 0; i < n; ++i) {
        for (let j = i + 1; j < n; ++j) {
            // If sum of current pair is not in hash,
            // then store it and continue to next pair
            let sum = arr[i] + arr[j];
            if (!Hash.has(sum))
                Hash.set(sum, [i, j]);

            else // Else (Sum already present in hash)
            {
                // Find previous pair
                let pp = Hash.get(sum);// pp->previous pair

                // Since array elements are distinct, we don't
                // need to check if any element is common among pairs
                document.write("(" + arr[pp[0]] + ", " + arr[pp[1]] + ") and (" + arr[i] + ", " + arr[j] + ")");
                return true;
            }
        }
    }

    document.write("No pairs found");
    return false;
}

// Driver program

let arr = [3, 4, 7, 1, 2, 9, 8];
let n = arr.length
findPairs(arr, n);
</script>

Output:

(3, 8) and (4, 7)

Time Complexity : O(n² logn)

Auxiliary Space: O(n)

Thanks to Gaurav Ahirwar for suggesting above solutions.
Exercise:
1) Extend the above solution with duplicates allowed in array. 
2) Further extend the solution to print all quadruples in output instead of just one. And all quadruples should be printed  in lexicographical order (smaller values before greater ones). Assume we have two solutions S1 and S2. 

S1 : a1 b1 c1 d1 ( these are values of indices in the array )  
S2 : a2 b2 c2 d2

S1 is lexicographically smaller than S2 if
  a1 < a2 OR
  a1 = a2 AND b1 < b2 OR
  a1 = a2 AND b1 = b2 AND c1 < c2 OR 
  a1 = a2 AND b1 = b2 AND c1 = c2 AND d1 < d2 

See this for solution of exercise.

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Find all pairs (a,b) and (c,d) in array which satisfy ab = cd
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