Leetcode 1558: Minimum Numbers of Function Calls to Make Target Array

 https://leetcode.com/problems/minimum-numbers-of-function-calls-to-make-target-array/description/

Notes:

This question can be solved greedily: 

1): if there is one element is odd, we need to reduce 1 to make it even; 

2): when all even, we can divide it by 2 

Since 2) is always faster than 1, so we need to do operation 2) whenever it is possible.

See the code below:

class Solution {
public:
    int minOperations(vector<int>& nums) {
        int res = 0;
        bool largerThan0 = true, allEven = true;
        while(largerThan0) {
            largerThan0 = false;
            allEven = true;
            for(auto &a : nums) {
                if(a>0) largerThan0 = true;
                if(a&1) {
                    --a;
                    ++res;
                    allEven = false;
                }
            }
            if(largerThan0 && allEven) {
                for(auto &a : nums) a /= 2;
                ++res;
                largerThan0 = true;
            }
        }
        return res;
    }
};