# Find a number which give minimum sum when XOR with every number of array of integers

Given an array arr[] of non-negative integers, the task is to find an integer X such that (arr XOR X) + (arr XOR X) + … + arr[n – 1] XOR X is minimum possible.
Examples:

Input: arr[] = {3, 9, 6, 2, 4}
Output: X = 2, Sum = 22
Input: arr[] = {6, 56, 78, 34}
Output: X = 2, Sum = 170

Approach: We will check ‘i’th bit of every number of the array in binary representation and count those numbers containing that ‘i’th bit set to ‘1’ because these set bits will contribute to maximize the sum rather than minimize. So we have to make this set ‘i’th bit to ‘0’ if the count is greater than N/2 and if the count is less than N/2 then the numbers having ‘i’th bit set are less and so it will not affect the answer. According to XOR operation on two bits, we know that when A XOR B and both A and B are the same then it gives the result as ‘0’ so we will make that ‘i’th bit in our number (num) to ‘1’, so that (1 XOR 1) will give ‘0’ and minimize the sum.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `#include ` `using` `namespace` `std;`   `// Function to find an integer X such that` `// the sum of all the array elements after` `// getting XORed with X is minimum` `void` `findX(``int` `arr[], ``int` `n)` `{` `    ``// Finding Maximum element of array` `    ``int``* itr = max_element(arr, arr + n);`   `    ``// Find Maximum number of bits required` `    ``// in the binary representation` `    ``// of maximum number` `    ``// so log2 is calculated` `    ``int` `p = log2(*itr) + 1;`   `    ``// Running loop from p times which is` `    ``// the number of bits required to represent` `    ``// all the elements of the array` `    ``int` `X = 0;` `    ``for` `(``int` `i = 0; i < p; i++) {` `        ``int` `count = 0;` `        ``for` `(``int` `j = 0; j < n; j++) {`   `            ``// If the bits in same position are set` `            ``// then count` `            ``if` `(arr[j] & (1 << i)) {` `                ``count++;` `            ``}` `        ``}`   `        ``// If count becomes greater than half of` `        ``// size of array then we need to make` `        ``// that bit '0' by setting X bit to '1'` `        ``if` `(count > (n / 2)) {`   `            ``// Again using shift operation to calculate` `            ``// the required number` `            ``X += 1 << i;` `        ``}` `    ``}`   `    ``// Calculate minimized sum` `    ``long` `long` `int` `sum = 0;` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``sum += (X ^ arr[i]);`   `    ``// Print solution` `    ``cout << ``"X = "` `<< X << ``", Sum = "` `<< sum;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 2, 3, 4, 5, 6 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``findX(arr, n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of above approach` `import` `java.lang.Math;` `import` `java.util.*;`   `class` `GFG` `{` `    ``// Function to find an integer X such that` `    ``// the sum of all the array elements after` `    ``// getting XORed with X is minimum` `    ``public` `static` `void` `findX(``int``[] a, ``int` `n)` `    ``{` `        `  `        ``// Finding Maximum element of array` `        ``Collections.sort(Arrays.asList(a), ``null``);` `        ``int` `itr = a[n-``1``];` `        `  `        ``// Find Maximum number of bits required` `        ``// in the binary representation` `        ``// of maximum number` `        ``// so log2 is calculated` `        ``int` `p = (``int``)(Math.log(itr)/Math.log(``2``)) + ``1``;`   `        ``// Running loop from p times which is` `        ``// the number of bits required to represent` `        ``// all the elements of the array` `        ``int` `x = ``0``;` `        ``for` `(``int` `i = ``0``; i < p; i++)` `        ``{` `            ``int` `count = ``0``;` `            ``for` `(``int` `j = ``0``; j < n; j++)` `            ``{` `                `  `                ``// If the bits in same position are set` `                ``// then count` `                ``if` `((a[j] & (``1` `<< i)) != ``0``)` `                    ``count++;` `            ``}`   `            ``// If count becomes greater than half of` `            ``// size of array then we need to make` `            ``// that bit '0' by setting X bit to '1'` `            ``if` `(count > (n / ``2``))` `            ``{` `                `  `                ``// Again using shift operation to calculate` `                ``// the required number` `                ``x += ``1` `<< i;` `            ``}` `        ``}`   `        ``// Calculate minimized sum` `        ``long` `sum = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``sum += (x ^ a[i]);` `        `  `        ``// Print solution` `        ``System.out.println(``"X = "` `+ x + ``", Sum = "` `+ sum);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] a = {``2``, ``3``, ``4``, ``5``, ``6``};` `        ``int` `n = a.length;`   `        ``findX(a, n);` `    ``}`   `}`   `// This code is contributed by` `// sanjeev2552`

## Python3

 `# Python 3 implementation of the approach` `from` `math ``import` `log2`   `# Function to find an integer X such that` `# the sum of all the array elements after` `# getting XORed with X is minimum` `def` `findX(arr, n):` `    `  `    ``# Finding Maximum element of array` `    ``itr ``=` `arr[``0``]` `    ``for` `i ``in` `range``(``len``(arr)):` `        `  `        ``# Find Maximum number of bits required` `        ``# in the binary representation` `        ``# of maximum number` `        ``# so log2 is calculated` `        ``if``(arr[i] > itr):` `            ``itr ``=` `arr[i]`   `    ``p ``=` `int``(log2(itr)) ``+` `1`   `    ``# Running loop from p times which is` `    ``# the number of bits required to represent` `    ``# all the elements of the array` `    ``X ``=` `0` `    ``for` `i ``in` `range``(p):` `        ``count ``=` `0` `        ``for` `j ``in` `range``(n):` `            `  `            ``# If the bits in same position are set` `            ``# then increase count` `            ``if` `(arr[j] & (``1` `<< i)):` `                ``count ``+``=` `1`   `        ``# If count becomes greater than half of` `        ``# size of array then we need to make` `        ``# that bit '0' by setting X bit to '1'` `        ``if` `(count > ``int``(n ``/` `2``)):` `            `  `            ``# Again using shift operation to calculate` `            ``# the required number` `            ``X ``+``=` `1` `<< i`   `    ``# Calculate minimized sum` `    ``sum` `=` `0` `    ``for` `i ``in` `range``(n):` `        ``sum` `+``=` `(X ^ arr[i])`   `    ``# Print solution` `    ``print``(``"X ="``, X, ``", Sum ="``, ``sum``)`   `# Driver code` `if` `__name__``=``=``'__main__'``:` `    ``arr ``=` `[``2``, ``3``, ``4``, ``5``, ``6``]` `    ``n ``=` `len``(arr)` `    ``findX(arr, n)` `    `  `# This code is contributed by` `# Surendra_Gangwar`

## C#

 `// C# implementation of above approach ` `using` `System;` `using` `System.Linq;`   `class` `GFG ` `{ ` `    ``// Function to find an integer X such that ` `    ``// the sum of all the array elements after ` `    ``// getting XORed with X is minimum` `    ``public` `static` `void` `findX(``int``[] a, ``int` `n) ` `    ``{ ` `        `  `        ``// Finding Maximum element of array ` `        ``int` `itr = a.Max(); ` `        `  `        ``// Find Maximum number of bits required ` `        ``// in the binary representation ` `        ``// of maximum number ` `        ``// so log2 is calculated ` `        ``int` `p = (``int``) Math.Log(itr, 2) + 1; `   `        ``// Running loop from p times which is ` `        ``// the number of bits required to represent ` `        ``// all the elements of the array ` `        ``int` `x = 0; ` `        ``for` `(``int` `i = 0; i < p; i++) ` `        ``{ ` `            ``int` `count = 0; ` `            ``for` `(``int` `j = 0; j < n; j++) ` `            ``{ ` `                `  `                ``// If the bits in same position are set ` `                ``// then count ` `                ``if` `((a[j] & (1 << i)) != 0) ` `                    ``count++; ` `            ``} `   `            ``// If count becomes greater than half of ` `            ``// size of array then we need to make ` `            ``// that bit '0' by setting X bit to '1' ` `            ``if` `(count > (n / 2)) ` `            ``{ ` `                `  `                ``// Again using shift operation to calculate ` `                ``// the required number ` `                ``x += 1 << i; ` `            ``} ` `        ``} `   `        ``// Calculate minimized sum ` `        ``long` `sum = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``sum += (x ^ a[i]); ` `        `  `        ``// Print solution ` `        ``Console.Write(``"X = "` `+ x + ``", Sum = "` `+ sum); ` `    ``} `   `    ``// Driver Code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] a = {2, 3, 4, 5, 6}; ` `        ``int` `n = a.Length; `   `        ``findX(a, n); ` `    ``} `   `} `   `// This code is contributed by ravikishor`

## Javascript

 ``

Output:

`X = 6, Sum = 14`

Time Complexity: O(N * log(A))
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next