Approach

# 要找到從右邊算過來，最左邊的連續1把它減掉，這樣子OR出來就會是答案

Time Complexity

O(n)

Space Complexity

O(n)

Code

class Solution {
public:
    vector<int> minBitwiseArray(vector<int>& nums) {
        vector<int> ans;

        for (int n : nums) {

            if (n % 2 == 0) {
                ans.push_back(-1);
            }    
            else {
                int tmp = n, bitCount = 0;

                while ((tmp & 1) == 1) {
                    tmp >>= 1;
                    bitCount++;
                }

                ans.push_back(n - (1 << (bitCount - 1)));
            }
        
        }

        return ans;
    }
};