Given two numbers n and k, we need to find the minimum number of flips required to maximize given number by flipping its bits such that the resulting number has exactly k set bits.
Note : K must be less than number of bits in n.
Examples :
Input : n = 14, k = 2 Output : Min Flips = 1 Explanation : Binary representation of 14 = 1110 Largest 4-digit Binary number with 2 set bit = 1100 Conversion from 1110 to 1100 requires 1 flipping Input : n = 145, k = 4 Output : Min Flips = 3 Explanation : Binary representation of 145 = 10010001 Largest 8-digit Binary number with 4 set bit = 11110000 Conversion from 10010001 to 11110000 requires 3 flipping
For the given number n and k find the largest number possible with k-set bits and having exactly same number of bits as n has as :
- size = log2(n) + 1 gives the number of bits of n.
- max = pow(2, k) – 1 gives largest possible number with k bits.
- max = max << (size – k) gives the largest number possible with k-set bits and having exactly same number of bits as n has
- Number of set bit in (n XOR max ) is our required number of flipping.
Illustration of above approach :
let n = 145 (10010001), k = 4
size = log2(n) + 1 = log2(145) + 1
= 7 + 1 = 8
max = pow(2, k) -1 = pow(2, 4) - 1
= 16 - 1 = 15 (1111)
max = max << (size - k) = 15 << (8 - 4)
= 240 (11110000)
number of set bit in = no. of set bit in
(n XOR max ) (145 ^ 240 )
= 3
C++
// CPP for finding min flip// for maximizing given n#include <bits/stdc++.h>using namespace std;// function for finding set bitint setBit(int xorValue){ int count = 0; while (xorValue) { if (xorValue % 2) count++; xorValue /= 2; } // return count of set bit return count;}// function for finding min flipint minFlip(int n, int k){ // number of bits in n int size = log2(n) + 1; // Find the largest number of // same size with k set bits int max = pow(2, k) - 1; max = max << (size - k); // Count bit differences to find // required flipping. int xorValue = (n ^ max); return (setBit(xorValue));}// driver programint main(){ int n = 27, k = 3; cout << "Min Flips = " << minFlip(n, k); return 0;} |
Java
// JAVA Code to find Minimum flips required// to maximize a number with k set bitsimport java.util.*;class GFG { // function for finding set bit static int setBit(int xorValue) { int count = 0; while (xorValue >= 1) { if (xorValue % 2 == 1) count++; xorValue /= 2; } // return count of set bit return count; } // function for finding min flip static int minFlip(int n, int k) { // number of bits in n int size = (int)(Math.log(n) / Math.log(2)) + 1; // Find the largest number of // same size with k set bits int max = (int)Math.pow(2, k) - 1; max = max << (size - k); // Count bit differences to find // required flipping. int xorValue = (n ^ max); return (setBit(xorValue)); } /* Driver program to test above function */ public static void main(String[] args) { int n = 27, k = 3; System.out.println("Min Flips = "+ minFlip(n, k)); }}// This code is contributed by Arnav Kr. Mandal. |
Python 3
# Python3 for finding min flip# for maximizing given nimport math# function for finding set bitdef setBit(xorValue): count = 0 while (xorValue): if (xorValue % 2): count += 1 xorValue = int(xorValue / 2) # return count # of set bit return count# function for # finding min flipdef minFlip(n, k): # number of bits in n size = int(math.log(n) / math.log(2) + 1) # Find the largest number of # same size with k set bits max = pow(2, k) - 1 max = max << (size - k) # Count bit differences to # find required flipping. xorValue = (n ^ max) return (setBit(xorValue))# Driver Coden = 27k = 3print("Min Flips = " , minFlip(n, k))# This code is contributed# by Smitha |
C#
// C# Code to find Minimum flips required// to maximize a number with k set bitsusing System;class GFG { // function for finding set bit static int setBit(int xorValue) { int count = 0; while (xorValue >= 1) { if (xorValue % 2 == 1) count++; xorValue /= 2; } // return count of set bit return count; } // function for finding min flip static int minFlip(int n, int k) { // number of bits in n int size = (int)(Math.Log(n) / Math.Log(2)) + 1; // Find the largest number of // same size with k set bits int max = (int)Math.Pow(2, k) - 1; max = max << (size - k); // Count bit differences to find // required flipping. int xorValue = (n ^ max); return (setBit(xorValue)); } // Driver Code public static void Main() { int n = 27, k = 3; Console.Write("Min Flips = "+ minFlip(n, k)); }}// This code is contributed by Nitin Mittal. |
PHP
<?php// PHP for finding min flip// for maximizing given n// function for finding set bitfunction setBit($xorValue){ $count = 0; while ($xorValue) { if ($xorValue % 2) $count++; $xorValue /= 2; } // return count of set bit return $count;}// function for finding min flipfunction minFlip($n, $k){ // number of bits in n $size = log($n) + 1; // Find the largest number of // same size with k set bits $max = pow(2, $k) - 1; $max = $max << ($size - $k); // Count bit differences to find // required flipping. $xorValue = ($n ^ $max); return (setBit($xorValue));}// Driver Code$n = 27; $k = 3;echo "Min Flips = " , minFlip($n, $k);// This code is contributed by vt_m.?> |
Javascript
<script>// Javascript for finding min flip// for maximizing given n// function for finding set bitfunction setBit( xorValue){ var count = 0; while (xorValue) { if (xorValue % 2) count++; xorValue = parseInt(xorValue / 2); } // return count of set bit return count;}// function for finding min flipfunction minFlip(n, k){ // number of bits in n var size = Math.log2(n) + 1; // Find the largest number of // same size with k set bits var max = Math.pow(2, k) - 1; max = (max << (size - k)); // Count bit differences to find // required flipping. var xorValue = (n ^ max); return (setBit(xorValue));}// driver programvar n = 27, k = 3;document.write( "Min Flips = " + minFlip(n, k));// This code is contributed by itsok.</script> |
Output :
Min Flips = 3
Time Complexity: O(log2n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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