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Bitwise AND of sub-array closest to K
• Last Updated : 12 May, 2021

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:

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

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:

## 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);``    ``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`

## PHP

 ``

## Javascript

 ``
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:

## 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);``    ``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`

## Javascript

 ``
Output:
`0`

Time complexity: O(n2)

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