Skip to content
Related Articles

Related Articles

Improve Article

Count Triplets such that one of the numbers can be written as sum of the other two

  • Difficulty Level : Hard
  • Last Updated : 30 Jun, 2021

Given an array A[] of N integers. The task is to find the number of triples (i, j, k) , where i, j, k are indices and (1 <= i < j < k <= N), such that in the set { A_i   A_j   A_k   } at least one of the numbers can be written as the sum of the other two.
Examples
 

Input : A[] = {1, 2, 3, 4, 5}
Output : 4
The valid triplets are:
(1, 2, 3), (1, 3, 4), (1, 4, 5), (2, 3, 5)

Input : A[] = {1, 1, 1, 2, 2}
Output : 6

 

This is a counting problem. Let’s say f(x) represents the frequency of number x   in our array.
There exist four cases: 
 

  1. All three numbers are equal to 0. The number of ways = f(0)C3 (where pCq is the number of ways of choosing q numbers from p numbers).
  2. One number is equal to 0, the other two are equal to some x > 0: f(0) * f(x)C2.
  3. Two numbers are equal to some x>0, the third is 2*x: f(x)C2 * f(2 * x).
  4. The three numbers are x, y and x + y, 0 < x, y: f(x) * f(y) * f(x + y).

Below is the implementation of the above approach: 
 

C++




// C++ program to count Triplets such that at
// least one of the numbers can be written
// as sum of the other two
#include<bits/stdc++.h>
using namespace std;
 
    // Function to count the number of ways
    // to choose the triples
    int countWays(int arr[], int n)
    {
        // compute the max value in the array
        // and create frequency array of size
        // max_val + 1.
        // We can also use HashMap to store
        // frequencies. We have used an array
        // to keep remaining code simple.
        int max_val = 0;
        for (int i = 0; i < n; i++)
            max_val = max(max_val, arr[i]);
        int freq[max_val + 1]={0};
        for (int i = 0; i < n; i++)
            freq[arr[i]]++;
 
        int ans = 0; // stores the number of ways
 
        // Case 1: 0, 0, 0
        ans += freq[0] * (freq[0] - 1) * (freq[0] - 2) / 6;
 
        // Case 2: 0, x, x
        for (int i = 1; i <= max_val; i++)
            ans += freq[0] * freq[i] * (freq[i] - 1) / 2;
 
        // Case 3: x, x, 2*x
        for (int i = 1; 2 * i <= max_val; i++)
            ans += freq[i] * (freq[i] - 1) / 2 * freq[2 * i];
 
        // Case 4: x, y, x + y
        // iterate through all pairs (x, y)
        for (int i = 1; i <= max_val; i++) {
            for (int j = i + 1; i + j <= max_val; j++)
                ans += freq[i] * freq[j] * freq[i + j];
        }
 
        return ans;
    }
 
    // Driver code
    int main()
    {
        int arr[]={ 1, 2, 3, 4, 5 };
        int n = sizeof(arr)/sizeof(int);
        cout<<(countWays(arr, n));
        return 0;
    }
 
//contributed by Arnab Kundu

Java




// Java program to count Triplets such that at
// least one of the numbers can be written
// as a sum of the other two
 
class GFG {
 
    // Function to count the number of ways
    // to choose the triples
    static int countWays(int[] arr, int n)
    {
        // compute the max value in the array
        // and create frequency array of size
        // max_val + 1.
        // We can also use HashMap to store
        // frequencies. We have used an array
        // to keep remaining code simple.
        int max_val = 0;
        for (int i = 0; i < n; i++)
            max_val = Math.max(max_val, arr[i]);
        int[] freq = new int[max_val + 1];
        for (int i = 0; i < n; i++)
            freq[arr[i]]++;
 
        int ans = 0; // stores the number of ways
 
        // Case 1: 0, 0, 0
        ans += freq[0] * (freq[0] - 1) * (freq[0] - 2) / 6;
 
        // Case 2: 0, x, x
        for (int i = 1; i <= max_val; i++)
            ans += freq[0] * freq[i] * (freq[i] - 1) / 2;
 
        // Case 3: x, x, 2*x
        for (int i = 1; 2 * i <= max_val; i++)
            ans += freq[i] * (freq[i] - 1) / 2 * freq[2 * i];
 
        // Case 4: x, y, x + y
        // iterate through all pairs (x, y)
        for (int i = 1; i <= max_val; i++) {
            for (int j = i + 1; i + j <= max_val; j++)
                ans += freq[i] * freq[j] * freq[i + j];
        }
 
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = new int[] { 1, 2, 3, 4, 5 };
        int n = arr.length;
        System.out.println(countWays(arr, n));
    }
}

Python3




# Python3 program to count Triplets such
# that at least one of the numbers can be
# written as sum of the other two
import math as mt
 
# Function to count the number of ways
# to choose the triples
def countWays(arr, n):
 
    # compute the max value in the array
    # and create frequency array of size
    # max_val + 1.
    # We can also use HashMap to store
    # frequencies. We have used an array
    # to keep remaining code simple.
    max_val = 0
    for i in range(n):
        max_val = max(max_val, arr[i])
 
    freq = [0 for i in range(max_val + 1)]
 
    for i in range(n):
        freq[arr[i]] += 1
 
    ans = 0 # stores the number of ways
 
    # Case 1: 0, 0, 0
    ans += (freq[0] * (freq[0] - 1) *
           (freq[0] - 2) // 6)
 
    # Case 2: 0, x, x
    for i in range(1, max_val + 1):
        ans += (freq[0] * freq[i] *
               (freq[i] - 1) // 2)
 
    # Case 3: x, x, 2*x
    for i in range(1, (max_val + 1) // 2):
        ans += (freq[i] *
               (freq[i] - 1) // 2 * freq[2 * i])
 
    # Case 4: x, y, x + y
    # iterate through all pairs (x, y)
    for i in range(1, max_val + 1):
        for j in range(i + 1, max_val - i + 1):
            ans += freq[i] * freq[j] * freq[i + j]
 
    return ans
 
# Driver code
arr = [ 1, 2, 3, 4, 5]
n = len(arr)
print(countWays(arr, n))
 
# This code is contributed by
# mohit kumar 29

C#




// C# program to count Triplets
// such that at least one of the
// numbers can be written as sum
// of the other two
using System;
 
class GFG
{
 
// Function to count the number
// of ways to choose the triples
static int countWays(int[] arr, int n)
{
    // compute the max value in the array
    // and create frequency array of size
    // max_val + 1.
    // We can also use HashMap to store
    // frequencies. We have used an array
    // to keep remaining code simple.
    int max_val = 0;
    for (int i = 0; i < n; i++)
        max_val = Math.Max(max_val, arr[i]);
         
    int[] freq = new int[max_val + 1];
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;
 
    int ans = 0; // stores the number of ways
 
    // Case 1: 0, 0, 0
    ans += freq[0] * (freq[0] - 1) *
                     (freq[0] - 2) / 6;
 
    // Case 2: 0, x, x
    for (int i = 1; i <= max_val; i++)
        ans += freq[0] * freq[i] *
                        (freq[i] - 1) / 2;
 
    // Case 3: x, x, 2*x
    for (int i = 1;
             2 * i <= max_val; i++)
        ans += freq[i] * (freq[i] - 1) /
                      2 * freq[2 * i];
 
    // Case 4: x, y, x + y
    // iterate through all pairs (x, y)
    for (int i = 1; i <= max_val; i++)
    {
        for (int j = i + 1;
                 i + j <= max_val; j++)
            ans += freq[i] * freq[j] *
                             freq[i + j];
    }
 
    return ans;
}
 
// Driver code
public static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
    Console.WriteLine(countWays(arr, n));
}
}
 
// This code is contributed by shs..

PHP




<?php
// PHP program to count Triplets such that at
// least one of the numbers can be written
// as sum of the other two
 
// Function to count the number of ways
// to choose the triples
function countWays($arr, $n)
{
    // compute the max value in the array
    // and create frequency array of size
    // max_val + 1.
    // We can also use HashMap to store
    // frequencies. We have used an array
    // to keep remaining code simple.
    $max_val = 0;
    for ($i = 0; $i < $n; $i++)
        $max_val = max($max_val, $arr[$i]);
    $freq = array_fill(0, $max_val + 1, 0);
    for ($i = 0; $i < $n; $i++)
        $freq[$arr[$i]]++;
 
    $ans = 0; // stores the number of ways
 
    // Case 1: 0, 0, 0
    $ans += (int)($freq[0] * ($freq[0] - 1) *
                             ($freq[0] - 2) / 6);
 
    // Case 2: 0, x, x
    for ($i = 1; $i <= $max_val; $i++)
        $ans += (int)($freq[0] * $freq[$i] *
                     ($freq[$i] - 1) / 2);
 
    // Case 3: x, x, 2*x
    for ($i = 1; 2 * $i <= $max_val; $i++)
        $ans += (int)($freq[$i] * ($freq[$i] - 1) / 2 *
                                   $freq[2 * $i]);
 
    // Case 4: x, y, x + y
    // iterate through all pairs (x, y)
    for ($i = 1; $i <= $max_val; $i++)
    {
        for ($j = $i + 1; $i + $j <= $max_val; $j++)
            $ans += $freq[$i] * $freq[$j] *
                                $freq[$i + $j];
    }
 
    return $ans;
}
 
// Driver code
$arr = array( 1, 2, 3, 4, 5 );
$n = count($arr);
echo countWays($arr, $n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// JavaScript program to count Triplets such that at
// least one of the numbers can be written
// as a sum of the other two   
     
    // Function to count the number of ways
    // to choose the triples
    function countWays(arr,n)
    {
        // compute the max value in the array
        // and create frequency array of size
        // max_val + 1.
        // We can also use HashMap to store
        // frequencies. We have used an array
        // to keep remaining code simple.
        let max_val = 0;
        for (let i = 0; i < n; i++)
            max_val = Math.max(max_val, arr[i]);
        let freq = new Array(max_val + 1);
        for(let i=0;i<freq.length;i++)
        {
            freq[i]=0;
        }
        for (let i = 0; i < n; i++)
            freq[arr[i]]++;
   
        let ans = 0; // stores the number of ways
   
        // Case 1: 0, 0, 0
        ans += freq[0] * (freq[0] - 1) * (freq[0] - 2) / 6;
   
        // Case 2: 0, x, x
        for (let i = 1; i <= max_val; i++)
            ans += freq[0] * freq[i] * (freq[i] - 1) / 2;
   
        // Case 3: x, x, 2*x
        for (let i = 1; 2 * i <= max_val; i++)
            ans += freq[i] * (freq[i] - 1) / 2 * freq[2 * i];
   
        // Case 4: x, y, x + y
        // iterate through all pairs (x, y)
        for (let i = 1; i <= max_val; i++) {
            for (let j = i + 1; i + j <= max_val; j++)
                ans += freq[i] * freq[j] * freq[i + j];
        }
   
        return ans;
    }
     
    // Driver code
    let arr=[1, 2, 3, 4, 5];
    let n = arr.length;
    document.write(countWays(arr, n));
 
     
 
// This code is contributed by patel2127
 
</script>
Output: 
4

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :