Count pairs from an array having product of their sum and difference equal to 1
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 1.
Examples:
Input: arr[] = {3, 1, 1, 0}
Output: 2
Explanation:
The two possible pairs are:
- (arr[1] + arr[3]) * (arr[1] – arr[3]) = 1
- (arr[2] + arr[3]) * (arr[2] – arr[3]) = 1
Input: arr[] = {12, 0, 1, 1, 14, 0, 9, 0}
Output: 4
Explanation:
The four possible pairs are as follows:
- (3, 6): (arr[3] + arr[6]) * (arr[3] – arr[6]) = 1
- (4, 6): (arr[4] + arr[6]) * (arr[4] – arr[6]) = 1
- (3, 8): (arr[3] + arr[8]) * (arr[3] – arr[8]) = 1
- (3, 6): (arr[4] + arr[8]) * (arr[4] – arr[8]) = 1
Naive Approach: The simplest approach is to traverse the array and generate all possible pairs from the given array and count those pairs whose product of sum and difference is 1. Finally, print the final count obtained after completing the above steps.
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: The given condition can be expressed for any pair of array elements (arr[i], arr[j]) as:
(arr[i] + arr[j]) * (arr[i] – arr[j]) = 1
=> (arr[i]2 – arr[j]2) = 1
Therefore, it can be concluded that the required conditions can be satisfied only by the pairs arr[i] = 1 and arr[j] = 0 and i < j. Follow the steps below to solve the problem:
- Initialize two variables, oneCount and desiredPairs to store the count of 1s and required pairs respectively.
- Traverse the given array and check the following:
- If arr[i] = 1: Increment oneCount by 1.
- If arr[i] = 0: Add the count of 1s obtained so far to desiredPairs.
- After completing the above steps, print the value of desiredPairs as the required answer.
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 the desired// number of pairsint countPairs(int arr[], int n){ // Initialize oneCount int oneCount = 0; // Initialize the desiredPair int desiredPair = 0; // Traverse the given array for (int i = 0; i < n; i++) { // If 1 is encountered if (arr[i] == 1) { oneCount++; } // If 0 is encountered if (arr[i] == 0) { // Update count of pairs desiredPair += oneCount; } } // Return the final count return desiredPair;}// Driver Codeint main(){ // Given array arr[] int arr[] = { 3, 1, 1, 0 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call cout << countPairs(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{// Function to count the desired// number of pairsstatic int countPairs(int arr[], int n){ // Initialize oneCount int oneCount = 0; // Initialize the desiredPair int desiredPair = 0; // Traverse the given array for(int i = 0; i < n; i++) { // If 1 is encountered if (arr[i] == 1) { oneCount++; } // If 0 is encountered if (arr[i] == 0) { // Update count of pairs desiredPair += oneCount; } } // Return the final count return desiredPair;}// Driver Codepublic static void main(String[] args){ // Given array arr[] int arr[] = { 3, 1, 1, 0 }; int N = arr.length; // Function call System.out.println(countPairs(arr, N));}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approach# Function to count the desired# number of pairsdef countPairs(arr, n): # Initialize oneCount oneCount = 0 # Initialize the desiredPair desiredPair = 0 # Traverse the given array for i in range(n): # If 1 is encountered if (arr[i] == 1): oneCount += 1 # If 0 is encountered if (arr[i] == 0): # Update count of pairs desiredPair += oneCount # Return the final count return desiredPair# Driver Codeif __name__ == '__main__': # Given array arr[] arr = [ 3, 1, 1, 0 ] N = len(arr) # Function call print(countPairs(arr, N))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to count the desired// number of pairsstatic int countPairs(int []arr, int n){ // Initialize oneCount int oneCount = 0; // Initialize the desiredPair int desiredPair = 0; // Traverse the given array for(int i = 0; i < n; i++) { // If 1 is encountered if (arr[i] == 1) { oneCount++; } // If 0 is encountered if (arr[i] == 0) { // Update count of pairs desiredPair += oneCount; } } // Return the readonly count return desiredPair;}// Driver Codepublic static void Main(String[] args){ // Given array []arr int []arr = { 3, 1, 1, 0 }; int N = arr.Length; // Function call Console.WriteLine(countPairs(arr, N));}}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program for the above approach// Function to count the desired// number of pairsfunction countPairs(arr, n){ // Initialize oneCount let oneCount = 0; // Initialize the desiredPair let desiredPair = 0; // Traverse the given array for(let i = 0; i < n; i++) { // If 1 is encountered if (arr[i] == 1) { oneCount++; } // If 0 is encountered if (arr[i] == 0) { // Update count of pairs desiredPair += oneCount; } } // Return the final count return desiredPair;}// Driver Code// Given array arr[]let arr = [ 3, 1, 1, 0 ];let N = arr.length;// Function calldocument.write(countPairs(arr, N));// This code is contributed by target_2</script> |
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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