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 element int 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 Code int 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 element public 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 Code public 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 element def 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 Code if __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 element using 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 element function 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)