Given an unsorted array arr, the task is to find the count of the distinct triplets in which the sum of any two elements is the third element.
Examples:
Input: arr[] = {1, 3, 4, 15, 19}
Output: 2
Explanation:
In the given array there are two triplets such that the sum of the two elements is equal to the third element: {{1, 3, 4}, {4, 15, 19}}Input: arr[] = {7, 2, 5, 4, 3, 6, 1, 9, 10, 12}
Output: 18
Approach:
- Sort the given array
- Create a Hash map for the array to check that a particular element is present or not.
- Iterate over the Array with two Loops to select any two elements at different positions and check that the sum of those two elements is present in the hash map in O(1) time.
- If the sum of the two elements is present in the hash map, then increment the count of the triplets.
Below is the implementation of the above approach:
C++
// C++ Implementation to find the// Count the triplets in which sum// of two elements is the third element#include <bits/stdc++.h>using namespace std;// Function to find the count// the triplets in which sum// of two elements is the third elementint triplets(vector<int>arr){ // Dictionary to check a element is // present or not in array map<int, int> k; map<pair<int, int>, int> mpp; // List to check for // duplicates of Triplets vector<vector<int >> ssd; // Set initial count to zero int count = 0; // Sort the array sort(arr.begin(),arr.end()); int i = 0; while (i < arr.size()) { // Add all the values as key // value pairs to the dictionary if (k.find(arr[i]) == k.end()) k[arr[i]] = 1; i += 1; } // Loop to choose two elements int j = 0; while (j < arr.size() - 1) { int q = j + 1; while (q < arr.size()) { // Check for the sum and duplicate if ( k.find(arr[j] + arr[q]) != k.end() and mpp[{arr[j], arr[q]}] != arr[j] + arr[q]) { count += 1; ssd.push_back({arr[j], arr[q], arr[j] + arr[q]}); mpp[{arr[j], arr[q]}] = arr[j] + arr[q]; } q += 1; } j += 1; } return count;}// Driver Codeint main(){ vector<int>arr = {7, 2, 5, 4, 3, 6, 1, 9, 10, 12}; int count = triplets(arr); printf("%d",count); return 0;}// This code is contributed by mohit kumar 29 |
Java
// Java implementation to find the // Count the triplets in which sum // of two elements is the third element import java.io.*;import java.util.ArrayList;import java.util.Collections;import java.util.HashSet;class GFG{ // Function to find the count the triplets// in which sum of two elements is the third elementpublic static int triplets(ArrayList<Integer> arr){ // Dictionary to check a element is // present or not in array HashSet<Integer> k = new HashSet<>(); // List to check for // duplicates of Triplets HashSet<ArrayList<Integer>> ssd = new HashSet<>(); // Set initial count to zero int count = 0; // Sort the array Collections.sort(arr); int i = 0; // Add all the values as key // value pairs to the dictionary while (i < arr.size()) { if (!k.contains(arr.get(i))) { k.add(arr.get(i)); } i += 1; } int j = 0; // Loop to choose two elements while (j < arr.size() - 1) { int q = j + 1; // Check for the sum and duplicate while (q < arr.size()) { ArrayList<Integer> trip = new ArrayList<>(); trip.add(arr.get(j)); trip.add(arr.get(q)); trip.add(arr.get(j) + arr.get(q)); if (k.contains(arr.get(j) + arr.get(q)) && !ssd.contains(trip)) { count += 1; ArrayList<Integer> nums = new ArrayList<>(); nums.add(arr.get(j)); nums.add(arr.get(q)); nums.add(arr.get(j) + arr.get(q)); ssd.add(nums); } q += 1; } j += 1; } return count;}// Driver Codepublic static void main(String[] args){ ArrayList<Integer> arr = new ArrayList<>(); arr.add(7); arr.add(2); arr.add(5); arr.add(4); arr.add(3); arr.add(6); arr.add(1); arr.add(9); arr.add(10); arr.add(12); int count = triplets(arr); System.out.println(count);}}// This code is contributed by aditya7409 |
Python3
# Python3 Implementation to find the # Count the triplets in which sum # of two elements is the third element# Function to find the count# the triplets in which sum # of two elements is the third elementdef triplets(arr): # Dictionary to check a element is # present or not in array k = set() # List to check for # duplicates of Triplets ssd = [] # Set initial count to zero count = 0 # Sort the array arr.sort() i = 0 while i < len(arr): # Add all the values as key # value pairs to the dictionary if arr[i] not in k: k.add(arr[i]) i += 1 # Loop to choose two elements j = 0 while j < len(arr) - 1: q = j + 1 while q < len(arr): # Check for the sum and duplicate if arr[j] + arr[q] in k and\ [arr[j], arr[q], arr[j] + arr[q]] not in ssd: count += 1 ssd.append([arr[j], arr[q], arr[j] + arr[q]]) q += 1 j += 1 return count# Driver Codeif __name__ == "__main__": arr = [7, 2, 5, 4, 3, 6, 1, 9, 10, 12] count = triplets(arr) print(count) |
C#
// C# Implementation to find the// Count the triplets in which sum// of two elements is the third elementusing System;using System.Collections.Generic;class GFG { // Function to find the count // the triplets in which sum // of two elements is the third element static int triplets(List<int> arr) { // Dictionary to check a element is // present or not in array Dictionary<int, int> k = new Dictionary<int, int>(); Dictionary<Tuple<int, int>, int> mpp = new Dictionary<Tuple<int, int>, int>(); // List to check for // duplicates of Triplets List<List<int>> ssd = new List<List<int>>(); // Set initial count to zero int count = 0; // Sort the array arr.Sort(); int i = 0; while (i < arr.Count) { // Add all the values as key // value pairs to the dictionary if (!k.ContainsKey(arr[i])) k[arr[i]] = 1; i += 1; } // Loop to choose two elements int j = 0; while (j < arr.Count - 1) { int q = j + 1; while (q < arr.Count) { // Check for the sum and duplicate if(!k.ContainsKey(arr[j] + arr[q])) { q += 1; continue; } if (mpp.ContainsKey(new Tuple<int,int>(arr[j], arr[q])) && mpp[new Tuple<int,int>(arr[j], arr[q])] != (arr[j] + arr[q])) { count += 1; ssd.Add(new List<int>(new int[]{arr[j], arr[q], arr[j] + arr[q]})); mpp[new Tuple<int,int>(arr[j], arr[q])] = arr[j] + arr[q]; } else{ count += 1; ssd.Add(new List<int>(new int[]{arr[j], arr[q], arr[j] + arr[q]})); mpp[new Tuple<int,int>(arr[j], arr[q])] = arr[j] + arr[q]; } q += 1; } j += 1; } return count; } // Driver code static void Main() { List<int> arr = new List<int>(new int[]{7, 2, 5, 4, 3, 6, 1, 9, 10, 12}); int count = triplets(arr); Console.Write(count); }}// This code is contributed by divyeshrabadiya07 |
Javascript
<script>// Javascript implementation to find the// Count the triplets in which sum// of two elements is the third element // Function to find the count the triplets// in which sum of two elements is the third elementfunction triplets(arr){ // Dictionary to check a element is // present or not in array let k = new Set(); // List to check for // duplicates of Triplets let ssd = new Set(); // Set initial count to zero let count = 0; // Sort the array arr.sort((a, b) => a - b); let i = 0; // Add all the values as key // value pairs to the dictionary while (i < arr.length) { if (!k.has(arr[i])) { k.add(arr[i]); } i += 1; } let j = 0; // Loop to choose two elements while (j < arr.length - 1) { let q = j + 1; // Check for the sum and duplicate while (q < arr.length) { let trip = new Array(); trip.push(arr[j]); trip.push(arr[q]); trip.push(arr[j] + arr[q]); if (k.has(arr[j] + arr[q]) && !ssd.has(trip)) { count += 1; let nums = new Array(); nums.push(arr[j]); nums.push(arr[q]); nums.push(arr[j] + arr[q]); ssd.add(nums); } q += 1; } j += 1; } return count;} // Driver Code let arr = new Array(); arr.push(7); arr.push(2); arr.push(5); arr.push(4); arr.push(3); arr.push(6); arr.push(1); arr.push(9); arr.push(10); arr.push(12); let count = triplets(arr); document.write(count);// This code is contributed by gfgking</script> |
Output:
18
Time Complexity: O(N2*logN)
Auxiliary Space: O(3*N2)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
