Given an array arr[], the task is to find the number of pairs (arr[i], arr[j]) in the array such that arr[i] + arr[j] = arr[i] * arr[j]
Examples:
Input: arr[] = {2, 2, 3, 4, 6}
Output: 1
(2, 2) is the only possible pair as (2 + 2) = (2 * 2) = 4.
Input: arr[] = {1, 2, 3, 4, 5}
Output: 0
Approach: The only possible pairs of integers that will satisfy the given conditions are (0, 0) and (2, 2). So the task now is to count the number of 0s and 2s in the array and store them in cnt0 and cnt2 respectively and then the required count will be (cnt0 * (cnt0 – 1)) / 2 + (cnt2 * (cnt2 – 1)) / 2.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count// of the required pairsint sumEqualProduct(int a[], int n){ int zero = 0, two = 0; // Find the count of 0s // and 2s in the array for (int i = 0; i < n; i++) { if (a[i] == 0) { zero++; } if (a[i] == 2) { two++; } } // Find the count of required pairs int cnt = (zero * (zero - 1)) / 2 + (two * (two - 1)) / 2; // Return the count return cnt;}// Driver codeint main(){ int a[] = { 2, 2, 3, 4, 2, 6 }; int n = sizeof(a) / sizeof(a[0]); cout << sumEqualProduct(a, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function to return the count // of the required pairs static int sumEqualProduct(int a[], int n) { int zero = 0, two = 0; // Find the count of 0s // and 2s in the array for (int i = 0; i < n; i++) { if (a[i] == 0) { zero++; } if (a[i] == 2) { two++; } } // Find the count of required pairs int cnt = (zero * (zero - 1)) / 2 + (two * (two - 1)) / 2; // Return the count return cnt; } // Driver code public static void main(String[] args) { int a[] = { 2, 2, 3, 4, 2, 6 }; int n = a.length; System.out.print(sumEqualProduct(a, n)); }}// This code is contributed by Rajput-Ji |
Python3
# Python 3 implementation of the approach # Function to return the count # of the required pairs def sumEqualProduct(a, n): zero = 0 two = 0 # Find the count of 0s # and 2s in the array for i in range(n): if a[i] == 0: zero += 1 if a[i] == 2: two += 1 # Find the count of required pairs cnt = (zero * (zero - 1)) // 2 + \ (two * (two - 1)) // 2 # Return the count return cnt # Driver code a = [ 2, 2, 3, 4, 2, 6 ] n = len(a) print(sumEqualProduct(a, n))# This code is contributed by Ankit kumar |
C#
// C# implementation of the approachusing System;class GFG{// Function to return the count// of the required pairsstatic int sumEqualProduct(int []a, int n){ int zero = 0, two = 0; // Find the count of 0s // and 2s in the array for (int i = 0; i < n; i++) { if (a[i] == 0) { zero++; } if (a[i] == 2) { two++; } } // Find the count of required pairs int cnt = (zero * (zero - 1)) / 2 + (two * (two - 1)) / 2; // Return the count return cnt;}// Driver codepublic static void Main(String[] args){ int []a = { 2, 2, 3, 4, 2, 6 }; int n = a.Length; Console.Write(sumEqualProduct(a, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the approach// Function to return the count// of the required pairsfunction sumEqualProduct(a, n){ var zero = 0, two = 0; // Find the count of 0s // and 2s in the array for (var i = 0; i < n; i++) { if (a[i] == 0) { zero++; } if (a[i] == 2) { two++; } } // Find the count of required pairs var cnt = (zero * (zero - 1)) / 2 + (two * (two - 1)) / 2; // Return the count return cnt;}// Driver codevar a = [2, 2, 3, 4, 2, 6];var n = a.length;document.write( sumEqualProduct(a, n));// This code is contributed by importantly.</script> |
Output:
3
Time Complexity : O(N)
Auxiliary Space : O(1)
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