# 1888. Minimum Number of Flips to Make the Binary String Alternating
###### tags: `Leetcode` `Medium` `Sliding Window`
Link: https://leetcode.com/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/description/
## 思路
思路[参考](https://leetcode.com/problems/minimum-number-of-flips-to-make-the-binary-string-alternating/solutions/1253874/c-solution-sliding-window-o-n-time-o-1-space/)
因为operation1(可以把前面的字符搬到后面去) 所以我们construct a new ```s=s+s```
然后我们构建两个string 分别是01循环 ```s1```和10循环 ```s2```
然后我们用sliding window compare ```s1``` and ```s2``` with ```s```
找出最小需要flip多少个数字
## Code

class Solution {
    public int minFlips(String s) {
        s += s;
        int n = s.length();
        StringBuilder s1 = new StringBuilder();
        StringBuilder s2 = new StringBuilder();
        for(int i=0; i<n; i++){
            s1.append(i%2==0?'0':'1');
            s2.append(i%2==0?'1':'0');
        }
        int ans1 = 0, ans2 = 0, ans = Integer.MAX_VALUE;
        for(int i=0; i<n; i++){
            if(s.charAt(i)!=s1.charAt(i)) ans1++;
            if(s.charAt(i)!=s2.charAt(i)) ans2++;
            if(i>=n/2){
                if(s.charAt(i-n/2)!=s1.charAt(i-n/2)) ans1--;
                if(s.charAt(i-n/2)!=s2.charAt(i-n/2)) ans2--;
            }
            if(i>=n/2-1) ans = Math.min(ans, Math.min(ans1, ans2));
        }
        return ans;
    }
}