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

08 Aug, 2021
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++
Java
Python3
C#
Javascript
// 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 programint main(){    int arr[] = {3, 4, 7, 1, 2, 9, 8};    int n = sizeof arr / sizeof arr[0];    findPairs(arr, n);    return 0;}
// Java Program to find four different elements a,b,c and d of// array such that a+b = c+dimport 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
# 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 codeif __name__ == "__main__":    arr = [3, 4, 7, 1, 2, 9, 8]    n = len(arr)    find_pair_of_sum(arr, n)
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
<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.lengthfindPairs(arr, n);</script>
Output:  
(3, 8) and (4, 7)
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 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 int the array )  
S2 : a2 b2 c2 d2

S1 is lexicographically smaller than S2 iff
  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.Related Article : Find all pairs (a,b) and (c,d) in array which satisfy ab = cdPlease write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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