Given an array arr[] of N distinct integers. The task is to find the sum of two array integers a[i] + a[j] which occurs maximum number of times. In the case of multiple answers, print all of them.
Examples:
Input: arr[] = {1, 8, 3, 11, 4, 9, 2, 7}
Output:
10
12
11
The sum 10, 12 and 11 occur 3 times
7 + 4 = 11, 8 + 3 = 11, 9 + 2 = 11
1 + 9 = 10, 8 + 2 = 10, 7 + 3 = 10
1 + 11 = 12, 8 + 4 = 12, 9 + 3 = 12
Input: arr[] = {3, 1, 7, 11, 9, 2, 12}
Output:
12
14
10
13
Approach: The following steps can be followed to solve the problem:
- Iterate over every pair of elements.
- Use a hash-table to count the number of times every sum pair occurs.
- At the end iterate over the hash-table and find the sum pair which occurs the maximum number of times.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to find the sum pairs// that occur the mostvoid findSumPairs(int a[], int n){ // Hash-table unordered_map<int, int> mpp; for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // Keep a count of sum pairs mpp[a[i] + a[j]]++; } } // Variables to store // maximum occurrence int occur = 0; // Iterate in the hash table for (auto it : mpp) { if (it.second > occur) { occur = it.second; } } // Print all sum pair which occur // maximum number of times for (auto it : mpp) { if (it.second == occur) cout << it.first << endl; }}// Driver codeint main(){ int a[] = { 1, 8, 3, 11, 4, 9, 2, 7 }; int n = sizeof(a) / sizeof(a[0]); findSumPairs(a, n); return 0;} |
Java
// Java implementation of above approachimport java.util.*;class GFG {// Function to find the sum pairs// that occur the moststatic void findSumPairs(int a[], int n){ // Hash-table Map<Integer,Integer> mpp = new HashMap<>(); for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // Keep a count of sum pairs mpp.put(a[i] + a[j],mpp.get(a[i] + a[j])==null?1:mpp.get(a[i] + a[j])+1); } } // Variables to store // maximum occurrence int occur = 0; // Iterate in the hash table for (Map.Entry<Integer,Integer> entry : mpp.entrySet()) { if (entry.getValue() > occur) { occur = entry.getValue(); } } // Print all sum pair which occur // maximum number of times for (Map.Entry<Integer,Integer> entry : mpp.entrySet()) { if (entry.getValue() == occur) System.out.println(entry.getKey()); }}// Driver codepublic static void main(String args[]) { int a[] = { 1, 8, 3, 11, 4, 9, 2, 7 }; int n = a.length; findSumPairs(a, n);}}/* This code is contributed by PrinciRaj1992 */ |
Python3
# Python 3 implementation of the approach# Function to find the sum pairs# that occur the mostdef findSumPairs(a, n): # Hash-table mpp = {i:0 for i in range(21)} for i in range(n - 1): for j in range(i + 1, n, 1): # Keep a count of sum pairs mpp[a[i] + a[j]] += 1 # Variables to store # maximum occurrence occur = 0 # Iterate in the hash table for key, value in mpp.items(): if (value > occur): occur = value # Print all sum pair which occur # maximum number of times for key, value in mpp.items(): if (value == occur): print(key)# Driver codeif __name__ == '__main__': a = [1, 8, 3, 11, 4, 9, 2, 7] n = len(a) findSumPairs(a, n)# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of above approachusing System;using System.Collections.Generic; class GFG {// Function to find the sum pairs// that occur the moststatic void findSumPairs(int []a, int n){ // Hash-table Dictionary<int,int> mpp = new Dictionary<int,int>(); for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // Keep a count of sum pairs if(mpp.ContainsKey(a[i] + a[j])) { var val = mpp[a[i] + a[j]]; mpp.Remove(a[i] + a[j]); mpp.Add(a[i] + a[j], val + 1); } else { mpp.Add(a[i] + a[j], 1); } } } // Variables to store // maximum occurrence int occur = 0; // Iterate in the hash table foreach(KeyValuePair<int, int> entry in mpp) { if (entry.Value > occur) { occur = entry.Value; } } // Print all sum pair which occur // maximum number of times foreach(KeyValuePair<int, int> entry in mpp) { if (entry.Value == occur) Console.WriteLine(entry.Key); }}// Driver codepublic static void Main(String []args) { int []a = { 1, 8, 3, 11, 4, 9, 2, 7 }; int n = a.Length; findSumPairs(a, n);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript implementation of the approach// Function to find the sum pairs// that occur the mostfunction findSumPairs( a, n){ // Hash-table let mpp = new Map(); for (let i = 0; i < n - 1; i++) { for (let j = i + 1; j < n; j++) { if(mpp[a[i]+a[j]]) mpp[a[i]+a[j]]++; else mpp[a[i]+a[j]] = 1; } } // Variables to store // maximum occurrence let occur = 0; // Iterate in the hash table for (var it in mpp) { if (mpp[it] > occur) { occur = mpp[it]; } } // Print all sum pair which occur // maximum number of times for (var it in mpp) { if (mpp[it] == occur) document.write( it ,'<br>'); }}// Driver codelet a = [ 1, 8, 3, 11, 4, 9, 2, 7 ];let len = a.length;findSumPairs(a, len);</script> |
Output:
10 12 11
Time Complexity: O(N*N), as we are using nested loops for traversing N*N times. Where N is the number of elements in the array.
Auxiliary Space: O(N*N), as we are using extra space for the map. Where N is the number of elements in the array.
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