给定一个整数N ,任务是找到通过翻转N的二进制表示中最多K位可以获得的最大数。
例子:
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
方法:
- 如果N的二进制表示中0的数量小于K则翻转所有0 。
- 如果0的数量大于或等于K,则翻转最重要的K 0s 。
- 最后,打印最大化的整数。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
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 equivalent
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
int 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 code
int 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 = 4
k = 1
print(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
输出:
6
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