# Minimize K whose XOR with given array elements leaves array unchanged

Given an array of N elements, the task is to find the minimum value of K such that Bitwise XOR of K with all array elements produces the same set of elements. If it is impossible to find any value of K then print “-1”.

Examples:

Input: arr[] = { 1, 0, 2, 3}
Output: 1
Explanation:
For K = 1,
1 xor 1 = 1
1 xor 0 = 0
1 xor 2 = 2
1 xor 3 = 3
Thus, K = 1 is the least possible positive value which leaves the array unaltered.

Input: arr[] = { 7, 1, 2, 3, 8}
Output: -1

Naive Approach: The naive approach is to iterate for all the possible value of K in the range [1, 1024] and check if Bitwise XOR of K with all the elements in the array gives the same array elements or not. If for any minimum value of K the Bitwise XOR produces the same array then print that value of K else print “-1”.

Time Complexity: O(K*N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using additional space. Below are the steps:

1. Insert all the elements into the set.
2. Iterate for all possible value of K in the range [0, 1024].
3. For every element in set, find its Bitwise XOR with K.
4. The first value of K for which all the elements generated after Bitwise XOR with the element in set is same as that of the given set, then print the value of K.
5. If no such K is obtained, print “-1”.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum ` `// value of K in given range ` `int` `min_value(``int` `arr[], ``int` `N) ` `{ ` `    ``int` `x, X, K; ` ` `  `    ``// Declare a set ` `    ``set<``int``> S; ` ` `  `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``S.insert(arr[i]); ` `    ``} ` ` `  `    ``// Initialize count variable ` `    ``int` `count = 0; ` ` `  `    ``// Iterate in range [1, 1024] ` `    ``for` `(``int` `i = 1; i <= 1024; i++) { ` ` `  `        ``// counter set as 0 ` `        ``count = 0; ` ` `  `        ``// Iterating through the Set ` `        ``for` `(``auto` `it = S.begin(); ` `             ``it != S.end(); it++) ` ` `  `        ``// Check if the XOR ` `        ``// calculated is present ` `        ``// in the Set ` `        ``{ ` ` `  `            ``X = ((i | *it) - (i & *it)); ` ` `  `            ``// If the value of Bitwise XOR ` `            ``// inside the given set then ` `            ``// increment count ` `            ``if` `(S.find(X) != S.end()) { ` `                ``count++; ` `            ``} ` `        ``} ` ` `  `        ``// Check if the value of count is ` `        ``// equal to the size of set ` `        ``if` `(count == S.size()) { ` `            ``K = i; ` ` `  `            ``// Return minimum value of K ` `            ``return` `K; ` `        ``} ` `    ``} ` ` `  `    ``// If no such K is found ` `    ``return` `-1; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array ` `    ``int` `arr[] = { 1, 0, 3, 3, 0, 2 }; ` ` `  `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``cout << min_value(arr, N); ` `    ``return` `0; ` `} `

Output:

```1
```

Time Complexity: O(K*N*log2N)
Auxiliary Space: O(1)

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