Given an array of distinct positive integers arr[] of length N, the task is to count all the triplets such that sum of two elements equals the third element.
Examples:
Input: arr[] = {1, 5, 3, 2}
Output: 2
Explanation:
In the given array, there are two such triplets such that sum of the two numbers is equal to the third number, those are –
(1, 2, 3), (3, 2, 5)Input: arr[] = {3, 2, 7}
Output: 0
Explanation:
In the given array there are no such triplets such that sum of two numbers is equal to the third number.
Approach: The idea is to create a frequency array of the numbers which are present in the array and then check for each pair of the element that sum of the pair elements is present in the array or not with the help of frequency array in O(1) time.
Algorithm:
- Declare a frequency array to store the frequency of the numbers.
- Iterate over the elements of the array and increment the count of that number in the frequency array.
- Run two loops to choose two different indexes of the matrix and check that if the sum of the elements at those indexes is having frequency more than 0 in the frequency array.
If frequency of the sum is greater than 0: Increment the count of the triplets.
Note: We have assumed in the program that the value of array elements lies in the range [1, 100].
Below is the implementation of the above approach:
// C++ implementation to count the // triplets such that the sum of the // two numbers is equal to third number #include <bits/stdc++.h> using namespace std;
// Function to find the count of the // triplets such that sum of two // numbers is equal to the third number int countTriplets( int arr[], int n){
int freq[100] = {0};
// Loop to count the frequency
for ( int i=0; i < n; i++){
freq[arr[i]]++;
}
int count = 0;
// Loop to count for triplets
for ( int i = 0;i < n; i++){
for ( int j = i+1; j < n; j++){
if (freq[arr[i] + arr[j]]){
count++;
}
}
}
return count;
} // Driver Code int main()
{ int n = 4;
int arr[] = {1, 5, 3, 2};
// Function Call
cout << countTriplets(arr, n);
return 0;
} |
// Java implementation to count the // triplets such that the sum of the // two numbers is equal to third number class GFG{
// Function to find the count of the // triplets such that sum of two // numbers is equal to the third number static int countTriplets( int arr[], int n){
int []freq = new int [ 100 ];
// Loop to count the frequency
for ( int i = 0 ; i < n; i++){
freq[arr[i]]++;
}
int count = 0 ;
// Loop to count for triplets
for ( int i = 0 ; i < n; i++){
for ( int j = i + 1 ; j < n; j++){
if (freq[arr[i] + arr[j]] > 0 ){
count++;
}
}
}
return count;
} // Driver Code public static void main(String[] args)
{ int n = 4 ;
int arr[] = { 1 , 5 , 3 , 2 };
// Function Call
System.out.print(countTriplets(arr, n));
} } // This code is contributed by Rajput-Ji |
# Python 3 implementation to count the # triplets such that the sum of the # two numbers is equal to third number # Function to find the count of the # triplets such that sum of two # numbers is equal to the third number def countTriplets(arr, n):
freq = [ 0 for i in range ( 100 )]
# Loop to count the frequency
for i in range (n):
freq[arr[i]] + = 1
count = 0
# Loop to count for triplets
for i in range (n):
for j in range (i + 1 , n, 1 ):
if (freq[arr[i] + arr[j]]):
count + = 1
return count
# Driver Code if __name__ = = '__main__' :
n = 4
arr = [ 1 , 5 , 3 , 2 ]
# Function Call
print (countTriplets(arr, n))
# This code is contributed by Surendra_Gangwar |
// C# implementation to count the // triplets such that the sum of the // two numbers is equal to third number using System;
class GFG{
// Function to find the count of the // triplets such that sum of two // numbers is equal to the third number static int countTriplets( int []arr, int n){
int []freq = new int [100];
// Loop to count the frequency
for ( int i = 0; i < n; i++){
freq[arr[i]]++;
}
int count = 0;
// Loop to count for triplets
for ( int i = 0; i < n; i++){
for ( int j = i + 1; j < n; j++){
if (freq[arr[i] + arr[j]] > 0){
count++;
}
}
}
return count;
} // Driver Code public static void Main( string [] args)
{ int n = 4;
int []arr = {1, 5, 3, 2};
// Function Call
Console.WriteLine(countTriplets(arr, n));
} } // This code is contributed by Yahs_R |
2
Performance Analysis:
- Time Complexity: O(N^{2}).
- Auxiliary Space: O(N).
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