Count number of Inversion in a Binary Array
Last Updated :
13 Jan, 2023
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Given a Binary Array arr[], the task is to count the number of inversions in it. The number of inversions in an array is the number of pairs of indices i, j such that i < j and a[i] > a[j].
Examples:
Input: arr[] = {1, 0, 1, 0, 0, 1, 0}
Output: 8
Explanation: Pairs of the index (i, j) are (0, 1), (0, 3), (0, 4), (0, 6), (2, 3), (2, 4), (2, 6), (5, 6).Input: arr[] = {0, 1}
Output: 0
Approach: Follow the below steps to solve the problem:
- Initialize the variable count = 0, for storing the count of zeroes occur so far.
- Ans, res = 0, for storing the number of inversions in an array.
- Now, run a loop i from the last element of the array to the front.
- And check, if arr[i] = 0, then, Increment count by 1.
- Else, Add count in res.
- Return res after executing the loop.
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to count the number of// inversions in the binary arrayint solve(bool* arr, int N){ // Initialize the variables int count = 0, res = 0; // Run a loop from back for (int i = N - 1; i >= 0; i--) { // arr[i] is 1 if (arr[i]) { res += count; } // arr[i] is 0 else { count++; } } // Return the resultant answer return res;}// Driver Codeint main(){ bool arr[] = { 1, 0, 1, 0, 0, 1, 0 }; int N = sizeof(arr) / sizeof(arr[0]); cout << solve(arr, N) << endl; bool arr2[] = { 1, 0 }; int M = sizeof(arr2) / sizeof(arr2[0]); cout << solve(arr2, M) << endl; return 0;} |
Java
// Java code to implement the above approachimport java.io.*;import java.util.*;class GFG { // Function to count the number of // inversions in the binary array public static int solve(boolean[] arr, int N) { // Initialize the variables int count = 0, res = 0; // Run a loop from back for (int i = N - 1; i >= 0; i--) { // arr[i] is 1 if (arr[i]) { res += count; } // arr[i] is 0 else { count++; } } // Return the resultant answer return res; } public static void main(String[] args) { boolean[] arr = { true, false, true, false, false, true, false }; int N = arr.length; System.out.println(solve(arr, N)); boolean[] arr2 = { true, false }; int M = arr2.length; System.out.println(solve(arr2, M)); }}// This code is contributed by lokesh. |
Python3
# Python code for the above approach# Function to count the number of# inversions in the binary arraydef solve(arr, N): # Initialize the variables count = 0 res = 0 # Run a loop from back for i in range(N-1, -1, -1): # arr[i] is 1 if arr[i]: res += count # arr[i] is 0 else: count += 1 # Return the resultant answer return res# Driver Codearr1 = [1, 0, 1, 0, 0, 1, 0]N = len(arr1)print(solve(arr1, N))arr2 = [1, 0]M = len(arr2)print(solve(arr2, M))# This code is contributed by Potta Lokesh |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;public class Gfg { static int solve(int[] arr, int N) { // Initialize the variables int count = 0, res = 0; // Run a loop from back for (int i = N - 1; i >= 0; i--) { // arr[i] is 1 if (arr[i]==1) { res += count; } // arr[i] is 0 else { count++; } } // Return the resultant answer return res; } // Driver Code public static void Main(string[] args) { int[] arr = { 1, 0, 1, 0, 0, 1, 0 }; int N = arr.Length; Console.WriteLine(solve(arr, N)); //Console.WriteLine("\n"); int[] arr2 = { 1, 0 }; int M = arr2.Length; Console.WriteLine(solve(arr2, M)); }}// This code is contributed by poojaagarwal2. |
Javascript
// Function to count the number of// inversions in the binary arrayfunction solve(arr, N){ // Initialize the variables let count = 0; let res = 0; // Run a loop from back for (let i = N - 1; i >= 0; i--) { // arr[i] is 1 if (arr[i]) { res += count; } // arr[i] is 0 else { count++; } } // Return the resultant answer return res;}// Test casesconsole.log(solve([1, 0, 1, 0, 0, 1, 0], 7)); // Output: 4console.log(solve([1, 0], 2)); // Output: 1// This code is contributed by ksam24000 |
Output
8 1
Time Complexity: O(N)
Auxiliary Space: O(1)
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