Given an array arr[] of N elements. The task is to count the total number of indices (i, j) such that arr[i] != arr[j] and i < j.
Examples:
Input: arr[] = {1, 1, 2}
Output: 2
(1, 2) and (1, 2) are the only valid pairs.
Input: arr[] = {1, 2, 3}
Output: 3
Input: arr[] = {1, 1, 1}
Output: 0
Approach: Initialise a count variable cnt = 0 and run two nested loops to check every possible pair whether the the current pair is valid or not. If it is valid, then increment the count variable. Finally, print the count of valid pairs.
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 valid pairsint countPairs(int arr[], int n){ // To store the required count int cnt = 0; // For each index pair (i, j) for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // If current pair is valid // then increment the count if (arr[i] != arr[j]) cnt++; } } return cnt;}// Driven codeint main(){ int arr[] = { 1, 1, 2 }; int n = sizeof(arr) / sizeof(int); cout << countPairs(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function to return the // count of valid pairs static int countPairs(int arr[], int n) { // To store the required count int cnt = 0; // For each index pair (i, j) for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // If current pair is valid // then increment the count if (arr[i] != arr[j]) cnt++; } } return cnt; } // Driven code public static void main (String[] args) { int arr[] = { 1, 1, 2 }; int n = arr.length; System.out.println(countPairs(arr, n)); }}// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to return the# count of valid pairsdef countPairs(arr, n): # To store the required count cnt = 0; # For each index pair (i, j) for i in range(n): for j in range(i + 1, n): # If current pair is valid # then increment the count if (arr[i] != arr[j]): cnt += 1; return cnt;# Driver codeif __name__ == '__main__': arr = [ 1, 1, 2 ]; n = len(arr); print(countPairs(arr, n)); # This code is contributed by 29AjayKumar |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the // count of valid pairs static int countPairs(int []arr, int n) { // To store the required count int cnt = 0; // For each index pair (i, j) for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // If current pair is valid // then increment the count if (arr[i] != arr[j]) cnt++; } } return cnt; } // Driven code public static void Main() { int []arr = { 1, 1, 2 }; int n = arr.Length; Console.WriteLine(countPairs(arr, n)); }}// This code is contributed by AnkitRai01 |
Javascript
<script>// Javascript implementation of the approach// Function to return the// count of valid pairsfunction countPairs(arr, n){ // To store the required count var cnt = 0; // For each index pair (i, j) for (var i = 0; i < n; i++) { for (var j = i + 1; j < n; j++) { // If current pair is valid // then increment the count if (arr[i] != arr[j]) cnt++; } } return cnt;}// Driven codevar arr = [ 1, 1, 2 ];var n = arr.length;document.write(countPairs(arr, n));</script> |
2
Time Complexity: O(N2)
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