Welcome back to the leetcode daily problem, 24th of December.

Link Problem

Solution:

• We can see that there are two cases for alternating: strings with one or zero appear at the even indices, and vice versa. For example : 101010, 010101, etc.

• We can count the one or zero at their respective indexes, and the answer will be the minimum of two result variables.

Code:

class Solution {
public:
    int minOperations(string s) {
        int one = 0, zero = 0;
        const int n = (int) s.size();

        for(int i = 0; i < n; ++i) {
            if(i % 2 == s[i] - '0')
                ++zero;
            else
                ++one;
        }

        return min(zero, one);
    }
};