Given an array arr[] of size N, the task is to count possible pairs of array elements (arr[i], arr[j]) such that (arr[i] + arr[j]) * (arr[i] – arr[j]) is 0.
Examples:
Input: arr[] = {2, -2, 1, 1}
Output : 2
Explanation:
(arr[0] + arr[1]) * (arr[0] – arr[1]) = 0
(arr[3] + arr[4]) * (arr[3] – arr[4]) = 0Input: arr[] = {5, 9, -9, -9}
Output : 3
Naive approach:
Generate all the pairs and for each pair check if the given condition holds true, for all such valid pair increment the count.
- Initialise a variable count for storing the answer.
- Generate all the pairs
- Check if the given condition holds true (i.e, arr[i] + arr[j]) * (arr[i] – arr[j]) is 0)
- Increment the count by 1.
- Check if the given condition holds true (i.e, arr[i] + arr[j]) * (arr[i] – arr[j]) is 0)
- Print the count.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to count required // number of pairs void countPairs(int arr[], int N) { int count = 0; // Generate all the possible pair for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // Check if the given condition holds true if ((arr[i] + arr[j]) * (arr[i] - arr[j]) == 0) { count++; } } } // Print the total count cout << count << endl; } // Driver Code int main() { // Given arr[] int arr[] = { 2, -2, 1, 1 }; // Size of the array int N = sizeof(arr) / sizeof(arr[0]); // Function Call countPairs(arr, N); return 0; } |
Java
import java.io.*; class GFG { // Function to count required // number of pairs public static void countPairs(int[] arr, int N) { int count = 0; // Generate all the possible pair for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // Check if the given condition holds true if ((arr[i] + arr[j]) * (arr[i] - arr[j]) == 0) { count++; } } } // Print the total count System.out.println(count); } // Driver Code public static void main(String[] args) { // Given arr[] int[] arr = { 2, -2, 1, 1 }; // Size of the array int N = arr.length; // Function Call countPairs(arr, N); } } |
Python3
def countPairs(arr, N): count = 0 for i in range(N): for j in range(i+1, N): if (arr[i] + arr[j]) * (arr[i] - arr[j]) == 0: count += 1 print(count) arr = [2, -2, 1, 1] N = len(arr) countPairs(arr, N) |
C#
using System; public class Gfg { static void Main(string[] args) { // Given arr[] int[] arr = { 2, -2, 1, 1 }; // Size of the array int N = arr.Length; // Function Call countPairs(arr, N); } // Function to count required // number of pairs static void countPairs(int[] arr, int N) { int count = 0; // Generate all the possible pair for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { // Check if the given condition holds true if ((arr[i] + arr[j]) * (arr[i] - arr[j]) == 0) { count++; } } } // Print the total count Console.WriteLine(count); } } |
Javascript
// Javascript program for the above approach // Function to count required // number of pairs function countPairs(arr, N) { let count = 0; // Generate all the possible pair for (let i = 0; i < N; i++) { for (let j = i + 1; j < N; j++) { // Check if the given condition holds true if ((arr[i] + arr[j]) * (arr[i] - arr[j]) == 0) { count++; } } } // Print the total count document.write(count); } // Driver Code // Given arr[] let arr = [2, -2, 1, 1 ]; // Size of the array let N = arr.length; // Function Call countPairs(arr, N); |
Output
2
Time Complexity: O(n*n)
Auxiliary Space: O(1)
Approach: It can be observed that the equation (arr[i] + arr[j]) * (arr[i] – arr[j]) = 0 can be reduced to arr[i]2 = arr[j]2. Therefore, the task reduces to counting pairs having absolute value equal. Follow the steps below to solve the problem:
- Initialize an array hash[] to store the frequency of the absolute value of each array element.
- Calculate the count of pairs by adding (hash[x] * (hash[x] – 1))/ 2 for every array distinct absolute values.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; #define MAXN 100005 // Function to count required // number of pairs int countPairs(int arr[], int N) { // Stores count of pairs int desiredPairs = 0; // Initialize hash with 0 int hash[MAXN] = { 0 }; // Count frequency of each element for (int i = 0; i < N; i++) { hash[abs(arr[i])]++; } // Calculate desired number of pairs for (int i = 0; i < MAXN; i++) { desiredPairs += ((hash[i]) * (hash[i] - 1)) / 2; } // Print desired pairs cout << desiredPairs; } // Driver Code int main() { // Given arr[] int arr[] = { 2, -2, 1, 1 }; // Size of the array int N = sizeof(arr) / sizeof(arr[0]); // Function Call countPairs(arr, N); return 0; } |
Java
// Java program for the above approach import java.io.*; import java.util.Arrays; class GFG{ static int MAXN = 100005; // Function to count required // number of pairs static void countPairs(int arr[], int N) { // Stores count of pairs int desiredPairs = 0; // Initialize hash with 0 int hash[] = new int[MAXN]; Arrays.fill(hash, 0); // Count frequency of each element for(int i = 0; i < N; i++) { hash[(Math.abs(arr[i]))]++; } // Calculate desired number of pairs for(int i = 0; i < MAXN; i++) { desiredPairs += ((hash[i]) * (hash[i] - 1)) / 2; } // Print desired pairs System.out.print(desiredPairs); } // Driver Code public static void main (String[] args) { // Given arr[] int arr[] = { 2, -2, 1, 1 }; // Size of the array int N = arr.length; // Function call countPairs(arr, N); } } // This code is contributed by code_hunt |
Python3
# Python3 program for # the above approach MAXN = 100005 # Function to count required # number of pairs def countPairs(arr, N): # Stores count of pairs desiredPairs = 0 # Initialize hash with 0 hash = [0] * MAXN # Count frequency of # each element for i in range(N): hash[abs(arr[i])] += 1 # Calculate desired number # of pairs for i in range(MAXN): desiredPairs += ((hash[i]) * (hash[i] - 1)) // 2 # Print desired pairs print (desiredPairs) # Driver Code if __name__ == "__main__": # Given arr[] arr = [2, -2, 1, 1] # Size of the array N = len(arr) # Function Call countPairs(arr, N) # This code is contributed by Chitranayal |
C#
// C# program for the above approach using System; class GFG{ static int MAXN = 100005; // Function to count required // number of pairs static void countPairs(int []arr, int N) { // Stores count of pairs int desiredPairs = 0; // Initialize hash with 0 int []hash = new int[MAXN]; // Count frequency of each element for(int i = 0; i < N; i++) { hash[(Math.Abs(arr[i]))]++; } // Calculate desired number of pairs for(int i = 0; i < MAXN; i++) { desiredPairs += ((hash[i]) * (hash[i] - 1)) / 2; } // Print desired pairs Console.Write(desiredPairs); } // Driver Code public static void Main(String[] args) { // Given []arr int []arr = { 2, -2, 1, 1 }; // Size of the array int N = arr.Length; // Function call countPairs(arr, N); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program for the above approach // Function to count required // number of pairs function countPairs(arr, N) { // Stores count of pairs let desiredPairs = 0; // Initialize hash with 0 let hash = new Uint8Array(9100005); // Count frequency of each element for (let i = 0; i < N; i++) { hash[(Math.abs(arr[i]))]++; } // Calculate desired number of pairs for (let i = 0; i < 9100005; i++) { desiredPairs += ((hash[i]) * (hash[i] - 1)) / 2; } // Print desired pairs document.write(desiredPairs); } // Driver Code // Given arr[] let arr = [2, -2, 1, 1]; // Size of the array let N = arr.length; // Function Call countPairs(arr, N); // This code is contributed by Mayank Tyagi </script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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!
