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 current pair is valid or not. If it is valid, then increment the count variable. Finally, print the count of valid pairs.
Algorithm:
- Define a static method named “countPairs” that takes two parameters, an integer array “arr” and an integer “n”, and returns an integer.
- Declare an integer variable “cnt” and initialize it to 0. This variable will store the count of valid pairs.
- Use a nested for-loop to iterate through each index pair (i, j) of the input array “arr”.
- If the current pair of values at indices i and j are different, then increment the “cnt” variable by 1. This indicates that the current pair is a valid pair.
- Return the final value of “cnt”.
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 pairs 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 int main() { int arr[] = { 1, 1, 2 }; int n = sizeof(arr) / sizeof(int); cout << countPairs(arr, n); return 0; } |
Java
// Java implementation of the approach 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 (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 pairs def 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 code if __name__ == '__main__': arr = [ 1, 1, 2 ]; n = len(arr); print(countPairs(arr, n)); # This code is contributed by 29AjayKumar |
C#
// C# implementation of the approach using 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 pairs function 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 code var arr = [ 1, 1, 2 ]; var n = arr.length; document.write(countPairs(arr, n)); </script> |
Output
2
Time Complexity: O(N2)
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Approach 2: Count unequal element pairs from the given Array Using Recursion.
Algorithm:
1. Create function “countPairsRec” which takes array “arr[ ]” and indices “i” and “j” .
2. If the value of “i” is larger than or equal to the value of “j”, the function returns 0 because there are no more pairs to compare.
3. Otherwise, determine whether the pair (arr[i], arr[j]) is a valid pair. If it is valid, increase the count by one and call the function again with the next pair.
4. If the pair (arr[i], arr[j]) is invalid, go to the next pair without increasing the count.
5. At the end of the function, return the count.
6. Create a new function called “countPairs” that accepts an integer array “arr” and an integer “n” as input parameters.
7. In the function, call the “countPairsRec” function with the array’s starting and ending indices.
Here’s the implementation:
C++
#include <bits/stdc++.h> using namespace std; // Function to recursively count the number of valid pairs int countPairsRec(int arr[], int i, int j) { // Base case: if we've compared all pairs, return 0 if (i >= j) { return 0; } // Recursive case: // If the current pair is valid, add 1 and move on to the next pair if (arr[i] != arr[j]) { return 1 + countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } // If the current pair is invalid, move on to the next pair without counting it else { return countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } } // Wrapper function to call the recursive function int countPairs(int arr[], int n) { return countPairsRec(arr, 0, n-1); } // Driver code int main() { int arr[] = {1, 2, 3}; int n = sizeof(arr) / sizeof(int); cout << countPairs(arr, n) << endl; return 0; } // This code is contributed by Vaibhav Saroj |
C
#include <stdio.h> // Function to recursively count the number of valid pairs int countPairsRec(int arr[], int i, int j) { // Base case: if we've compared all pairs, return 0 if (i >= j) { return 0; } // Recursive case: // If the current pair is valid, add 1 and move on to the next pair if (arr[i] != arr[j]) { return 1 + countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } // If the current pair is invalid, move on to the next pair without counting it else { return countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } } // Wrapper function to call the recursive function int countPairs(int arr[], int n) { return countPairsRec(arr, 0, n-1); } // Driver code int main() { int arr[] = {1, 2, 3}; int n = sizeof(arr) / sizeof(int); printf("%d\n", countPairs(arr, n)); return 0; } // This code is contributed by Vaibhav Saroj |
Java
/*package whatever //do not write package name here */import java.util.*; class Main { // Function to recursively count the number of valid pairs static int countPairsRec(int arr[], int i, int j) { // Base case: if we've compared all pairs, return 0 if (i >= j) { return 0; } // Recursive case: // If the current pair is valid, add 1 and move on to the next pair if (arr[i] != arr[j]) { return 1 + countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } // If the current pair is invalid, move on to the next pair without counting it else { return countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } } // Wrapper function to call the recursive function static int countPairs(int arr[], int n) { return countPairsRec(arr, 0, n-1); } // Driver code public static void main(String[] args) { int arr[] = {1, 2, 3}; int n = arr.length; System.out.println(countPairs(arr, n)); } } // This code is contributed by Vaibhav Saroj |
Python3
def countPairsRec(arr, i, j): # Base case: if we've compared all pairs, return 0 if i >= j: return 0 # Recursive case: # If the current pair is valid, add 1 and move on to the next pair if arr[i] != arr[j]: return 1 + countPairsRec(arr, i + 1, j) + countPairsRec(arr, i, j - 1) # If the current pair is invalid, move on to the next pair without counting it else: return countPairsRec(arr, i + 1, j) + countPairsRec(arr, i, j - 1) # Wrapper function to call the recursive function def countPairs(arr): return countPairsRec(arr, 0, len(arr) - 1) # Driver code arr = [1, 2, 3] print(countPairs(arr)) |
C#
using System; class Program { // Function to recursively count the number of valid pairs static int countPairsRec(int[] arr, int i, int j) { // Base case: if we've compared all pairs, return 0 if (i >= j) { return 0; } // Recursive case: // If the current pair is valid, add 1 and move on to the next pair if (arr[i] != arr[j]) { return 1 + countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } // If the current pair is invalid, move on to the next pair without counting it else { return countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } } // Wrapper function to call the recursive function static int countPairs(int[] arr, int n) { return countPairsRec(arr, 0, n-1); } // Driver code static void Main(string[] args) { int[] arr = {1, 2, 3}; int n = arr.Length; Console.WriteLine(countPairs(arr, n)); } } // This code is contributed by Vaibhav Saroj |
Javascript
function countPairsRec(arr, i, j) { // Base case: if we've compared all pairs, return 0 if (i >= j) { return 0; } // Recursive case: // If the current pair is valid, add 1 and move on to the next pair if (arr[i] !== arr[j]) { return 1 + countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } // If the current pair is invalid, move on to the next pair without counting it else { return countPairsRec(arr, i+1, j) + countPairsRec(arr, i, j-1); } } // Wrapper function to call the recursive function function countPairs(arr) { return countPairsRec(arr, 0, arr.length-1); } // Driver code const arr = [1, 2, 3]; console.log(countPairs(arr)); // This code is contributed by Vaibhav Saroj |
Output
3
The Recursive approach is contributed by Vaibhav Saroj .
Time Complexity: O(n^2)
Auxiliary Space: O(1)
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