Bitwise AND of sub-array closest to K

Given an integer array arr[] of size N and an integer K, the task is to find the sub-array arr[i….j] where i ≤ j and compute the bitwise AND of all sub-array elements say X then print the minimum value of |K – X| among all possible values of X.

Example:

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

Method 1:
Find the bitwise AND of all possible sub-arrays and keep track of the minimum possible value of |K – X|.

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP 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
function closetAND(&$arr, $n, $k)
{
    $ans = PHP_INT_MAX;
  
    // Check all possible sub-arrays
    for ($i = 0; $i < $n; $i++) 
    {
  
        $X = $arr[$i];
        for ($j = $i; $j < $n; $j++) 
        {
            $X &= $arr[$j];
  
            // Find the overall minimum
            $ans = min($ans, abs($k - $X));
        }
    }
    return $ans;
}
  
    // Driver code
    $arr = array( 4, 7, 10 );
    $n = sizeof($arr) / sizeof($arr[0]);
    $k = 2;
    echo closetAND($arr, $n, $k);
  
    return 0;
      
    // This code is contributed by ChitraNayal
?>
chevron_right

Output:
0

Time complexity: O(n2)

Method 2:
It can be observed that while performing AND operation in the sub-array, the value of X can remain constant or decrease but will never increase.
Hence, we will start from the first element of a sub-array and will be doing bitwise AND and comparing the |K – X| with the current minimum difference until X ≤ K because after that |K – X| will start increasing.

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

Output:
0

Time complexity: O(n2)




Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :