3394. Minimum Array End¶

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3394. Minimum Array End

Medium

You are given two integers n and x. You have to construct an array of positive integers nums of size n where for every 0 <= i < n - 1, nums[i + 1] is greater than nums[i], and the result of the bitwise AND operation between all elements of nums is x.

Return the minimum possible value of nums[n - 1].

Example 1:

Example 2:

Constraints:

• 1 <= n, x <= 108

Solution¶

class Solution {
    public long minEnd(int n, int x) {
        long current = x;
        while ((n - 1) > 0) {
            n--;
            current = (current + 1) | x;
        }
        return current;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here