最小化 Array 元素的替换以使按位与大于 K
给定一个包含N个整数的数组A [] 和一个整数K ，任务是找到数组元素的最小替换次数，使得所有数组元素的按位与严格大于K 。
例子：
Input: N = 4, K = 2, A[] = {3, 1, 2, 7}
Output: 2
Explanation: Change A[1] = 3 and A[2] = 11.
After performing two operations: Modified array: {3, 3, 11, 7}
Now, Bitwise AND of all the elements is 3 & 3 & 11 & 7 = 3
Input: N = 3, K = 1, A[] = {2, 2, 2}
Output: 0
方法：这个问题可以通过位操作来解决，思路如下：
Find the bit representation of the value K. For each bit assume that is set and calculate how many replacements are required to achieve that value as bitwise AND of array elements. The minimum among these values is the required answer.
按照以下步骤实施上述观察：
- 创建一个 32 位的掩码变量并将其初始化为零。
- 从第 31个最高有效位开始，继续检查 K 中的第 i个位是否已设置。
- 如果设置了第 i 个位，则还要在掩码中设置该位。
- 如果未设置该位，则创建一个临时变量（例如temp ），其值为 mask 但设置第i个位，以便数组的按位与可以严格大于K 。
- 对于数组中的每个元素，检查元素和temp的按位与是否等于temp 。
- 如果它是temp则无需更改数组的此元素。
- 通过从 N （数组的大小）中减去不需要更改的元素的数量来找到替换的数量（比如X ）。
- 从第31位到第0 位的所有迭代中X的最小值就是答案。
下面是上述方法的实现。
C++
// C++ code to implement the approach
#include
using namespace std;
// Function to find the
// minimum number of replacements
int count(int N, vector A, int K)
{
// Initially assuming we need to change
// all elements of the array
int ans = N;
// Initialize mask as 0
int mask = 0;
// Starting from 31st bit check all
// the bits of X if they are set or not
for (int i = 31; i >= 0; i--) {
// If bit is set,
// set that bit in mask also
if ((K >> i) & 1)
mask |= (1 << i);
// If bit is not set then
// count the elements of array
// requiring no change
// and change answer to
// minimum replacements.
else {
int good = 0;
int temp = mask | (1 << i);
for (auto element : A)
if ((element & temp) == temp)
good++;
ans = min(ans, N - good);
}
}
return ans;
}
// Driver code
int main()
{
int N = 4, K = 2;
vector A = { 3, 1, 2, 7 };
// Function call
int ans = count(N, A, K);
cout << ans << endl;
return 0;
} Java
// Java code to implement the approach
import java.io.*;
class GFG {
// Function to find the
// minimum number of replacements
static int count(int N, int[] A, int K)
{
// Initially assuming we need to change
// all elements of the array
int ans = N;
// Initialize mask as 0
int mask = 0;
// Starting from 31st bit check all
// the bits of X if they are set or not
for (int i = 31; i >= 0; i--) {
// If bit is set,
// set that bit in mask also
if (((K >> i) & 1) != 0)
mask |= (1 << i);
// If bit is not set then
// count the elements of array
// requiring no change
// and change answer to
// minimum replacements.
else {
int good = 0;
int temp = mask | (1 << i);
for( int element : A)
if ((element & temp) == temp) good++;
ans = Math.min(ans, N - good);
}
}
return ans;
}
// Driver code
public static void main (String[] args) {
int N = 4, K = 2;
int A[] = { 3, 1, 2, 7 };
// Function call
int ans = count(N, A, K);
System.out.println(ans);
}
}
// This code is contributed by hrithikgarg03188.Python
# Python code to implement the approach
# Function to find the
# minimum number of replacements
def count(N, A, K):
# Initially assuming we need to change
# all elements of the array
ans = N
# Initialize mask as 0
mask = 0
# Starting from 31st bit check all
# the bits of X if they are set or not
i = 31
while(i >= 0):
# If bit is set,
# set that bit in mask also
if ((K >> i) & 1):
mask |= (1 << i)
# If bit is not set then
# count the elements of array
# requiring no change
# and change answer to
# minimum replacements.
else:
good = 0
temp = mask | (1 << i)
for x in range(0, len(A)):
if ((A[x] & temp) == temp):
good += 1
ans = min(ans, N - good)
i -= 1
return ans
# Driver code
N = 4
K = 2
A = [3, 1, 2, 7]
# Function call
ans = count(N, A, K)
print(ans)
# This code is contributed by Samim Hossain Mondal.C#
// C# code to implement the approach
using System;
class GFG {
// Function to find the
// minimum number of replacements
static int count(int N, int[] A, int K)
{
// Initially assuming we need to change
// all elements of the array
int ans = N;
// Initialize mask as 0
int mask = 0;
// Starting from 31st bit check all
// the bits of X if they are set or not
for (int i = 31; i >= 0; i--) {
// If bit is set,
// set that bit in mask also
if (((K >> i) & 1) != 0)
mask |= (1 << i);
// If bit is not set then
// count the elements of array
// requiring no change
// and change answer to
// minimum replacements.
else {
int good = 0;
int temp = mask | (1 << i);
foreach(
int element in A) if ((element & temp)
== temp) good++;
ans = Math.Min(ans, N - good);
}
}
return ans;
}
// Driver code
public static void Main()
{
int N = 4, K = 2;
int[] A = { 3, 1, 2, 7 };
// Function call
int ans = count(N, A, K);
Console.WriteLine(ans);
}
}
// This code is contributed by Samim Hossain Mondal.Javascript
输出：
2时间复杂度： O(N * LogM) 其中 M 是数组的最大元素。
辅助空间： O(1)