Given an integer N, the task is to find the greatest number that can be obtained by flipping at most K bits in the binary representation of N.
Examples:
Input: N = 4, K = 1
Output: 6
The binary equivalent of 4 is 100.
On flipping the 1st 0, we get 110
which is equivalent to 6.
Input: N = 5, K = 2
Output: 7
Approach:
- If the number of 0s in the binary representation of N is less than K then flip all the 0s.
- If the number of 0s is greater than or equal to K then flip the most significant K 0s.
- Finally, print the maximized integer.
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 maximized// number by flipping atmost k bitsint maxNum(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); // To count the number of 0s flipped int cn = 0; for (int i = 0; i < l; i++) { if (a[i] == 0 && cn < k) { a[i] = 1; cn++; } } // Return the decimal equivalent // of the maximized number return binaryDec(a, l);}// Driver codeint main(){ int n = 4, k = 1; cout << maxNum(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 maximized // number by flipping atmost k bits static int maxNum(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); // To count the number of 0s flipped int cn = 0; for (int i = 0; i < l; i++) { if (a[i] == 0 && cn < k) { a[i] = 1; cn++; } } // Return the decimal equivalent // of the maximized number return binaryDec(a, l); } // Driver code public static void main (String[] args) { int n = 4, k = 1; System.out.println(maxNum(n, k)); } }// This code is contributed by AnkitRai01 |
Python3
# 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[] into 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 return the maximized # number by flipping atmost k bits def maxNum(n, k): # Number of bits in n l = int(math.log2(n)) + 1 # Find the binary representation of n a = [0 for i in range(0, l)] decBinary(a, n) # To count the number of 0s flipped cn = 0 for i in range(0, l): if (a[i] == 0 and cn < k): a[i] = 1 cn = cn + 1 # Return the decimal equivalent # of the maximized number return binaryDec(a, l)# Driver code n = 4k = 1print(maxNum(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 maximized // number by flipping atmost k bits static int maxNum(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); // To count the number of 0s flipped int cn = 0; for (int i = 0; i < l; i++) { if (a[i] == 0 && cn < k) { a[i] = 1; cn++; } } // Return the decimal equivalent // of the maximized number return binaryDec(a, l); } // Driver code public static void Main(String[] args) { int n = 4, k = 1; Console.WriteLine(maxNum(n, k)); } }// This code is contributed by Rajput-Ji |
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 = Math.log2(n); while (n > 0) { arr[k--] = n % 2; n = Math.floor(n / 2); }}// Function to convert the number// represented as a binary array// arr[] into its decimal equivalentfunction 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 maximized// number by flipping atmost k bitsfunction maxNum(n, k) { // Number of bits in n let l = Math.log2(n) + 1; // Find the binary representation of n let a = new Array(l).fill(0); decBinary(a, n); // To count the number of 0s flipped let cn = 0; for (let i = 0; i < l; i++) { if (a[i] == 0 && cn < k) { a[i] = 1; cn++; } } // Return the decimal equivalent // of the maximized number return binaryDec(a, l);}// Driver codelet n = 4, k = 1;document.write(maxNum(n, k));// This code is contributed by _saurabh_jaiswal.</script> |
Output:
6
Time Complexity: O(log2n)
Auxiliary Space: O(log2n)
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