LeetCode 3287: Find the Maximum Sequence Value of Array

Explanation:

To solve this problem, we need to find the maximum value of the XOR operation within subsequences of size 2 * k in the given array. We can approach this problem by iterating through all possible subsequences of size 2 * k and calculating the XOR value for each subsequence. The maximum XOR value among all subsequences will be our answer.

Here is the algorithmic idea:

1. Iterate through all possible subsequences of size 2 * k in the array.
2. Calculate the XOR value for each subsequence.
3. Keep track of the maximum XOR value found.

Time Complexity: The time complexity of this approach is O(n * 2^(2k)), where n is the length of the input array and k is the given parameter.

Space Complexity: The space complexity is O(1) as we are not using any extra space other than a few variables.

class Solution {
    public int findMaximumXORValue(int[] nums, int k) {
        int n = nums.length;
        int max = 0;
        for (int i = 0; i < (1 << n); i++) {
            if (Integer.bitCount(i) == 2 * k) {
                int seq = 0;
                for (int j = 0; j < n; j++) {
                    if ((i & (1 << j)) != 0) {
                        seq ^= nums[j];
                    }
                }
                max = Math.max(max, seq);
            }
        }
        return max;
    }
}