Given two non-negative integers N and K, the task is to invert the Kth most significant bit of N and print the number obtained after inverting the bit.
Examples:
Input: N = 10, K = 1
Output: 2
The binary representation of 10 is 1010.
After inverting the first bit it becomes 0010
whose decimal equivalent is 2.
Input: N = 56, K = 2
Output: 40
Approach: Find the number of bits in N, if the number of bits is less than K then N itself is the required answer else flip the Kth most significant bit of N and print the number obtained after flipping it.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to convert decimal number n// to its binary representation// stored as an array arr[]void decBinary(int arr[], int n){ int k = log2(n); while (n > 0) { arr[k--] = n % 2; n /= 2; }}// Function to convert the number// represented as a binary array// arr[] into its decimal equivalentint binaryDec(int arr[], int n){ int ans = 0; for (int i = 0; i < n; i++) ans += arr[i] << (n - i - 1); return ans;}// Function to return the updated integer// after flipping the kth bitint getNum(int n, int k){ // Number of bits in n int l = log2(n) + 1; // Find the binary // representation of n int a[l] = { 0 }; decBinary(a, n); // The number of bits in n // are less than k if (k > l) return n; // Flip the kth bit a[k - 1] = (a[k - 1] == 0) ? 1 : 0; // Return the decimal equivalent // of the number return binaryDec(a, l);}// Driver codeint main(){ int n = 56, k = 2; cout << getNum(n, k); return 0;} |
Java
// Java implementation of the approach class GFG { // Function to convert decimal number n // to its binary representation // stored as an array arr[] static void decBinary(int arr[], int n) { int k = (int)(Math.log(n) / Math.log(2)); while (n > 0) { arr[k--] = n % 2; n /= 2; } } // Function to convert the number // represented as a binary array // arr[] into its decimal equivalent static int binaryDec(int arr[], int n) { int ans = 0; for (int i = 0; i < n; i++) ans += arr[i] << (n - i - 1); return ans; } // Function to return the updated integer // after flipping the kth bit static int getNum(int n, int k) { // Number of bits in n int l = (int)(Math.log(n) / Math.log(2)) + 1; // Find the binary // representation of n int a[] = new int[l]; decBinary(a, n); // The number of bits in n // are less than k if (k > l) return n; // Flip the kth bit a[k - 1] = (a[k - 1] == 0) ? 1 : 0; // Return the decimal equivalent // of the number return binaryDec(a, l); } // Driver code public static void main (String[] args) { int n = 56; int k = 2; System.out.println(getNum(n, k)); } }// This code is contributed by AnkitRai01 |
Python
# Python implementation of the approach import math# Function to convert decimal number n # to its binary representation # stored as an array arr[] def decBinary(arr, n): k = int(math.log2(n)) while (n > 0): arr[k] = n % 2 k = k - 1 n = n//2# Function to convert the number # represented as a binary array # arr[] its decimal equivalent def binaryDec(arr, n): ans = 0 for i in range(0, n): ans = ans + (arr[i] << (n - i - 1)) return ans# Function to concatenate the binary # numbers and return the decimal result def getNum(n, k): # Number of bits in both the numbers l = int(math.log2(n)) + 1 # Convert the bits in both the gers # to the arrays a[] and b[] a = [0 for i in range(0, l)] decBinary(a, n) # The number of bits in n # are less than k if(k > l): return n # Flip the kth bit if(a[k - 1] == 0): a[k - 1] = 1 else: a[k - 1] = 0 # Return the decimal equivalent # of the number return binaryDec(a, l)# Driver code n = 56k = 2print(getNum(n, k))# This code is contributed by Sanjit_Prasad |
C#
// C# implementation of the approach using System;class GFG { // Function to convert decimal number n // to its binary representation // stored as an array []arr static void decBinary(int []arr, int n) { int k = (int)(Math.Log(n) / Math.Log(2)); while (n > 0) { arr[k--] = n % 2; n /= 2; } } // Function to convert the number // represented as a binary array // []arr into its decimal equivalent static int binaryDec(int []arr, int n) { int ans = 0; for (int i = 0; i < n; i++) ans += arr[i] << (n - i - 1); return ans; } // Function to return the updated integer // after flipping the kth bit static int getNum(int n, int k) { // Number of bits in n int l = (int)(Math.Log(n) / Math.Log(2)) + 1; // Find the binary // representation of n int []a = new int[l]; decBinary(a, n); // The number of bits in n // are less than k if (k > l) return n; // Flip the kth bit a[k - 1] = (a[k - 1] == 0) ? 1 : 0; // Return the decimal equivalent // of the number return binaryDec(a, l); } // Driver code public static void Main(String[] args) { int n = 56; int k = 2; Console.WriteLine(getNum(n, k)); } }// This code is contributed by PrinciRaj1992 |
Javascript
<script> // javascript implementation of the approach // Function to convert decimal number n // to its binary representation // stored as an array arr[] function decBinary(arr, n) { let k = parseInt(Math.log2(n), 10); while (n > 0) { arr[k--] = n % 2; n = parseInt(n/2,10); } } // Function to convert the number // represented as a binary array // arr[] into its decimal equivalent function binaryDec(arr, n) { let ans = 0; for (let i = 0; i < n; i++) ans += arr[i] << (n - i - 1); return ans; } // Function to return the updated integer // after flipping the kth bit function getNum(n,k) { // Number of bits in n let l = parseInt(Math.log2(n),10) + 1; // Find the binary // representation of n let a = new Array(l); a.fill(0); decBinary(a, n); // The number of bits in n // are less than k if (k > l) return n; // Flip the kth bit a[k - 1] = (a[k - 1] == 0) ? 1 : 0; // Return the decimal equivalent // of the number return binaryDec(a, l); } let n = 56, k = 2; document.write(getNum(n, k)); // This code is contributed by vaibhavrabadiya117.</script> |
Output:
40
Time Complexity: O(log(m) + log(n))
Auxiliary Space: O(log(m) + log(n))
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