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 array element. For every quadruple (a, b, c, d), check if (a+b) = (c+d). 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 value in 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 above idea. In below implementation, map is used instead of hash. 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/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 |
Python
# 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 findPairs(arr, n): # Create an empty hashmap to store mapping # from sum to pair indexes Hash = {} # Traverse through all possible pairs of arr[] for i in range(n - 1): for j in range(i + 1, n): sum = arr[i] + arr[j] # Sum already present in hash if sum in Hash.keys(): # print previous pair and current prev = Hash.get(sum) print (str(prev) + " and (%d, %d)" %(arr[i], arr[j])) return True else: # sum is not in hash # store it and continue to next pair Hash[sum] = (arr[i], arr[j]) return False # driver program arr = [3, 4, 7, 1, 2, 9, 8] n = len(arr) findPairs(arr, n) # This code is contributed by Aditi Sharma |
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 |
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 = cd
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
leave a comment
0 Comments