# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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 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 `

 `# 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# 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 `

 ` `

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++ 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 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 `

 `# 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# 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 `

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 :