# Maximum AND value of a pair in an array

• Difficulty Level : Hard
• Last Updated : 09 May, 2022

We are given an array of n positive elements. we need to find the maximum AND value generated by any pair of element from the array. AND is bitwise & operator.

Examples:

```Input : arr[] = {4, 8, 12, 16}
Output : Maximum AND value = 8

Input : arr[] = {4, 8, 16, 2}
Output : Maximum AND value = 0```

Naive Approach : Basic Approach is same as maximum xor value. We iterate over all the possible pair and calculate the AND value of those all. Pick the largest value among them. Time complexity of this solution is O(n^2).

## C++

 `// CPP Program to find maximum XOR value of a pair``#include``using` `namespace` `std;`  `// Function for finding maximum and value pair``int` `maxAND(``int` `arr[], ``int` `n)``{``    ``int` `res = 0;``    ``for` `(``int` `i=0; i

## Java

 `// Java Program to find maximum``// XOR value of a pair``import` `java.util.*;``import` `java.lang.*;``  ` `public` `class` `GfG{``     ` `// Function for finding maximum``// and value pair``static` `int` `maxAND(``int` `arr[], ``int` `n)``{``    ``int` `res = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``for` `(``int` `j = i + ``1``; j < n; j++)``        ``res = res > ( arr[i] & arr[j]) ?``              ``res : ( arr[i] & arr[j]);`` ` `    ``return` `res;``}`` ` `// driver function``public` `static` `void` `main(String argc[])``{``    ``int` `arr[] = {``4``, ``8``, ``6``, ``2``};``    ``int` `n = arr.length;``    ``System.out.println(``"Maximum AND Value = "` `+``                       ``maxAND(arr,n));``}``}` `// This code is contributed by Prerna Saini`

## Python3

 `# Python3 Program to find maximum XOR``# value of a pair` `# Function for finding maximum and value pair``def` `maxAND(arr, n) :``    ``res ``=` `0``    ` `    ``for` `i ``in` `range``(``0``, n) :``        ``for` `j ``in` `range``(i ``+` `1``, n) :``            ``res ``=` `max``(res, arr[i] & arr[j])``    ` `    ``return` `res` `# Driver function``arr ``=` `[``4``, ``8``, ``6``, ``2``]``n ``=` `len``(arr)``print``(``"Maximum AND Value = "``, maxAND(arr,n))``  ` `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# Program to find maximum``// XOR value of a pair``using` `System;` `public` `class` `GfG``{``    ` `// Function for finding maximum``// and value pair``static` `int` `maxAND(``int` `[]arr, ``int` `n)``{``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``for` `(``int` `j = i + 1; j < n; j++)``        ``res = res > ( arr[i] & arr[j]) ?``              ``res : ( arr[i] & arr[j]);` `    ``return` `res;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]arr = {4, 8, 6, 2};``    ``int` `n = arr.Length;``    ``Console.WriteLine(``"Maximum AND Value = "` `+``                       ``maxAND(arr, n));``}``}` `// This code is contributed by vt_m.`

## PHP

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## Javascript

 ``

Output:

`Maximum AND Value = 4`

Better Approach : Idea is based on properties of AND operator. AND operation of any two bits result in 1 if both bits are 1. We start from the MSB and check whether we have minimum of two elements of array having set value. If yes then that MSB will be part of our solution and be added to result otherwise we will discard that bit. Similarly, iterating from MSB to LSB (32 to 1) for bit position we can easily check which bit will be part of our solution and will keep adding all such bits to our solution.

Explanation : Lets consider 1st example of {4, 8, 12, 16} :

step 1: Write Bit-representation of each element :
4 = 100, 8 = 1000, 12 = 1100, 16 = 10000
step 2: Check for 1st MSB , pattern = 0 + 16 = 16. Now 5th bit in 16 is set but no other element has 5-bit as set bit so this will not add up to our RES, still RES = 0 and pattern = 0
step 3: Check 4th bit, pattern = 0 + 8 = 8. Now 8 and 12 both have set bit on 4th bit position so that will add up in our solution, RES = 8 and pattern = 8
step 4: Check 3rd bit, pattern = 8 + 4 = 12. Now only 12 has both bits set bit (same as pattern) so we will discard 3rd bit, RES = 8 and pattern = 8
step 5: Check 2nd bit, pattern = 8 + 2 = 10. No element has set bit same as pattern so we will discard 2nd bit, RES = 8 and pattern = 8
step 4: Check 1st bit, pattern = 8 + 1 = 9. No element has set bit same as pattern so we will discard 1st bit, RES = 8 and pattern = 8

## C++

 `// CPP Program to find maximum XOR value of a pair``#include``using` `namespace` `std;` `// Utility function to check number of elements``// having set msb as of pattern``int` `checkBit(``int` `pattern, ``int` `arr[], ``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `((pattern & arr[i]) == pattern)``            ``count++;``    ``return` `count;``}` `// Function for finding maximum and value pair``int` `maxAND (``int` `arr[], ``int` `n)``{``    ``int` `res = 0, count;` `    ``// iterate over total of 32bits from msb to lsb``    ``for` `(``int` `bit = 31; bit >= 0; bit--)``    ``{``        ``// find the count of element having set  msb``        ``count = checkBit(res | (1 << bit),arr,n);` `        ``// if count >= 2 set particular bit in result``        ``if` `( count >= 2 )       ``            ``res |= (1 << bit);       ``    ``}` `    ``return` `res;``}` `// Driver function``int` `main()``{``    ``int` `arr[] = {4, 8, 6, 2};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr);``    ``cout << ``"Maximum AND Value = "` `<< maxAND(arr,n);``    ``return` `0;``}`

## Java

 `// Java Program to find maximum``// XOR value of a pair``import` `java.util.*;``import` `java.lang.*;``  ` `public` `class` `GfG{` `// Utility function to check number of elements``// having set msb as of pattern``static` `int` `checkBit(``int` `pattern, ``int` `arr[], ``int` `n)``{``    ``int` `count = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``if` `((pattern & arr[i]) == pattern)``            ``count++;``    ``return` `count;``}`` ` `// Function for finding maximum and value pair``static` `int` `maxAND (``int` `arr[], ``int` `n)``{``    ``int` `res = ``0``, count;`` ` `    ``// iterate over total of 32bits``    ``// from msb to lsb``    ``for` `(``int` `bit = ``31``; bit >= ``0``; bit--)``    ``{``        ``// find the count of element``        ``// having set msb``        ``count = checkBit(res | (``1` `<< bit), arr, n);`` ` `        ``// if count >= 2 set particular``        ``// bit in result``        ``if` `( count >= ``2` `)    ``            ``res |= (``1` `<< bit);    ``    ``}`` ` `    ``return` `res;``}``  ` `// driver function``public` `static` `void` `main(String argc[])``{``    ``int` `arr[] = {``4``, ``8``, ``6``, ``2``};``    ``int` `n = arr.length;``    ``System.out.println(``"Maximum AND Value = "` `+``                       ``maxAND(arr, n));``}``}` `// This code is contributed by Prerna Saini`

## Python3

 `# Python3 Program to find maximum XOR``# value of a pair` `# Utility function to check number of``# elements having set msb as of pattern``def` `checkBit(pattern,arr,  n) :``    ``count ``=` `0``    ` `    ``for` `i ``in` `range``(``0``, n) :``        ``if` `((pattern & arr[i]) ``=``=` `pattern) :``            ``count ``=` `count ``+` `1``    ``return` `count` `# Function for finding maximum and``# value pair``def` `maxAND (arr,  n) :``    ``res ``=` `0``    ` `    ``# iterate over total of 32bits``    ``# from msb to lsb``    ``for` `bit ``in` `range``(``31``,``-``1``,``-``1``) :``      ` `        ``# find the count of element``        ``# having set  msb``        ``count ``=` `checkBit(res | (``1` `<< bit), arr, n)`` ` `        ``# if count >= 2 set particular``        ``# bit in result``        ``if` `( count >``=` `2` `) :``            ``res ``=``res | (``1` `<< bit)``            ` `    ``return` `res``    ` ` ` `# Driver function``arr ``=` `[``4``, ``8``, ``6``, ``2``]``n ``=` `len``(arr)``print``(``"Maximum AND Value = "``, maxAND(arr, n))` `# This code is contributed by Nikita Tiwari`

## C#

 `// C# Program to find maximum``// XOR value of a pair``using` `System;` `public` `class` `GfG``{` `// Utility function to check``// number of elements having``// set msb as of pattern``static` `int` `checkBit(``int` `pattern,``                    ``int` `[]arr,``                    ``int` `n)``{``    ``int` `count = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `((pattern & arr[i]) == pattern)``            ``count++;``    ``return` `count;``}` `// Function for finding maximum``// and value pair``static` `int` `maxAND (``int` `[]arr, ``int` `n)``{``    ``int` `res = 0, count;` `    ``// iterate over total of 32bits``    ``// from msb to lsb``    ``for` `(``int` `bit = 31; bit >= 0; bit--)``    ``{``        ` `        ``// find the count of element``        ``// having set msb``        ``count = checkBit(res | (1 << bit), arr, n);` `        ``// if count >= 2 set particular``        ``// bit in result``        ``if` `(count >= 2)    ``            ``res |= (1 << bit);    ``    ``}` `    ``return` `res;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]arr = {4, 8, 6, 2};``    ``int` `n = arr.Length;``    ``Console.WriteLine(``"Maximum AND Value = "` `+``                       ``maxAND(arr, n));``}``}` `// This code is contributed by vt_m.`

## PHP

 `= 0; ``\$bit``--)``    ``{``        ` `        ``// find the count of element``        ``// having set msb``        ``\$count` `= checkBit(``\$res` `| (1 << ``\$bit``),``                                   ``\$arr``, ``\$n``);` `        ``// if count >= 2 set particular``        ``// bit in result``        ``if` `( ``\$count` `>= 2 )``            ``\$res` `|= (1 << ``\$bit``);``    ``}` `    ``return` `\$res``;``}` `    ``// Driver Code``    ``\$arr` `= ``array``(4, 8, 6, 2);``    ``\$n` `= ``count``(``\$arr``);``    ``echo` `"Maximum AND Value = "` `, maxAND(``\$arr``,``\$n``);``    ` `// This code is contributed by vt_m.``?>`

## Javascript

 ``

Output:

`Maximum AND Value = 4`

Time Complexity : O(n * log(m)) where m is the maximum element from the array and n is the size of the array.

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