最接近K的子数组的按位与

给定一个整数数组ARR的大小N和整数K []，任务是找到子阵列ARR [我…·J其中i≤j和计算所有子数组元素的按位说X然后打印| K – X |的最小值在X的所有可能值中。

例子：

Input: arr[] = {1, 6}, K = 3
Output: 2

Sub-array	Bitwise AND	|K – X|
{1}	1	2
{6}	6	3
{1, 6}	1	2

Input: arr[] = {4, 7, 10}, K = 2
Output: 0

方法1：
找到所有可能的子数组的按位与，并跟踪| K – X |的最小可能值。 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum possible value
// of |K - X| where X is the bitwise AND of
// the elements of some sub-array
int closetAND(int arr[], int n, int k)
{
    int ans = INT_MAX;
  
    // Check all possible sub-arrays
    for (int i = 0; i < n; i++) {
  
        int X = arr[i];
        for (int j = i; j < n; j++) {
            X &= arr[j];
  
            // Find the overall minimum
            ans = min(ans, abs(k - X));
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 4, 7, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
    cout << closetAND(arr, n, k);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.io.*;
  
class GFG {
  
    // Function to return the minimum possible value
    // of |K - X| where X is the bitwise AND of
    // the elements of some sub-array
    static int closetAND(int arr[], int n, int k)
    {
        int ans = Integer.MAX_VALUE;
  
        // Check all possible sub-arrays
        for (int i = 0; i < n; i++) {
  
            int X = arr[i];
            for (int j = i; j < n; j++) {
                X &= arr[j];
  
                // Find the overall minimum
                ans = Math.min(ans, Math.abs(k - X));
            }
        }
        return ans;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 4, 7, 10 };
        int n = arr.length;
        int k = 2;
        System.out.println(closetAND(arr, n, k));
    }
}
  
// This code is contributed by jit_t

Python3

# Python implementation of the approach
  
# Function to return the minimum possible value
# of |K - X| where X is the bitwise AND of
# the elements of some sub-array
def closetAND(arr, n, k):
  
    ans = 10**9
  
    # Check all possible sub-arrays
    for i in range(n):
  
        X = arr[i]
  
        for j in range(i,n):
            X &= arr[j]
  
            # Find the overall minimum
            ans = min(ans, abs(k - X))
          
    return ans
  
# Driver code
arr = [4, 7, 10]
n = len(arr)
k = 2;
print(closetAND(arr, n, k))
  
# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach 
using System;
  
class GFG
{ 
  
    // Function to return the minimum possible value 
    // of |K - X| where X is the bitwise AND of 
    // the elements of some sub-array 
    static int closetAND(int []arr, int n, int k) 
    { 
        int ans = int.MaxValue; 
  
        // Check all possible sub-arrays 
        for (int i = 0; i < n; i++)
        { 
  
            int X = arr[i]; 
            for (int j = i; j < n; j++)
            { 
                X &= arr[j]; 
  
                // Find the overall minimum 
                ans = Math.Min(ans, Math.Abs(k - X)); 
            } 
        } 
        return ans; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int []arr = { 4, 7, 10 }; 
        int n = arr.Length; 
        int k = 2; 
          
        Console.WriteLine(closetAND(arr, n, k)); 
    } 
} 
  
// This code is contributed by AnkitRai01

输出：

0

时间复杂度： O(n ² )

方法2：
可以观察到，在子阵列中执行“与”运算时， X的值可以保持恒定或减小，但永远不会增大。
因此，我们将从子数组的第一个元素开始，将按位与并比较| K – X |。直到X≤K为止的电流最小差，因为在那之后| K – X |将开始增加。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum possible value
// of |K - X| where X is the bitwise AND of
// the elements of some sub-array
int closetAND(int arr[], int n, int k)
{
    int ans = INT_MAX;
  
    // Check all possible sub-arrays
    for (int i = 0; i < n; i++) {
  
        int X = arr[i];
        for (int j = i; j < n; j++) {
            X &= arr[j];
  
            // Find the overall minimum
            ans = min(ans, abs(k - X));
  
            // No need to perform more AND operations
            // as |k - X| will increase
            if (X <= k)
                break;
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 4, 7, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 2;
    cout << closetAND(arr, n, k);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Function to return the minimum possible value
// of |K - X| where X is the bitwise AND of
// the elements of some sub-array
static int closetAND(int arr[], int n, int k)
{
    int ans = Integer.MAX_VALUE;
  
    // Check all possible sub-arrays
    for (int i = 0; i < n; i++) 
    {
  
        int X = arr[i];
        for (int j = i; j < n; j++) 
        {
            X &= arr[j];
  
            // Find the overall minimum
            ans = Math.min(ans, Math.abs(k - X));
  
            // No need to perform more AND operations
            // as |k - X| will increase
            if (X <= k)
                break;
        }
    }
    return ans;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = { 4, 7, 10 };
    int n = arr.length;
    int k = 2;
    System.out.println(closetAND(arr, n, k));
}
}
  
// This code is contributed by Princi Singh

Python3

# Python implementation of the approach
import sys
  
# Function to return the minimum possible value
# of |K - X| where X is the bitwise AND of
# the elements of some sub-array
def closetAND(arr, n, k):
    ans = sys.maxsize;
  
    # Check all possible sub-arrays
    for i in range(n):
  
        X = arr[i];
        for j in range(i,n):
            X &= arr[j];
  
            # Find the overall minimum
            ans = min(ans, abs(k - X));
  
            # No need to perform more AND operations
            # as |k - X| will increase
            if (X <= k):
                break;
    return ans;
  
# Driver code
arr = [4, 7, 10 ];
n = len(arr);
k = 2;
print(closetAND(arr, n, k));
  
# This code is contributed by PrinciRaj1992

C＃

// C# implementation of the approach
using System;     
      
class GFG 
{
  
// Function to return the minimum possible value
// of |K - X| where X is the bitwise AND of
// the elements of some sub-array
static int closetAND(int []arr, int n, int k)
{
    int ans = int.MaxValue;
  
    // Check all possible sub-arrays
    for (int i = 0; i < n; i++) 
    {
  
        int X = arr[i];
        for (int j = i; j < n; j++) 
        {
            X &= arr[j];
  
            // Find the overall minimum
            ans = Math.Min(ans, Math.Abs(k - X));
  
            // No need to perform more AND operations
            // as |k - X| will increase
            if (X <= k)
                break;
        }
    }
    return ans;
}
  
// Driver code
public static void Main(String[] args) 
{
    int []arr = { 4, 7, 10 };
    int n = arr.Length;
    int k = 2;
    Console.WriteLine(closetAND(arr, n, k));
}
}
  
// This code has been contributed by 29AjayKumar

输出：

0

时间复杂度： O(n ² )