# 2483. Minimum Penalty for a Shop
###### tags: `Leetcode` `Medium`
Link: https://leetcode.com/problems/minimum-penalty-for-a-shop/description/
## 思路
penalty = 关之前'N'的数目+关的时候和关之后'Y'的数目
只需要从前往后遍历就可以计算出关之前'N'的数目
如果知道总共有几个'Y'-关之前遇到的'Y' 那么我们就得到了 关的时候和关之后'Y'的数目

写完code发现其实根本不需要多花O(n)算一共有多少个'Y'
我们只要直接找到noCount-customerCount最小的index即可
## Code
优化前

class Solution {
    public int bestClosingTime(String customers) {
        int total = 0;
        int n = customers.length();
        for(int i=0; i<n; i++){
            if(customers.charAt(i)=='Y') total++;
        }
        int bestTime = 0;
        int minPenalty = Integer.MAX_VALUE;
        int noCount = 0;
        int customerCount = 0;
        for(int i=0; i<=n; i++){
            int currPenalty = noCount+total-customerCount;
            if(currPenalty<minPenalty){
                minPenalty = currPenalty;
                bestTime = i;
            }
            if(i==n) continue;
            if(customers.charAt(i)=='Y') customerCount++;
            else noCount++;
        }
        return bestTime;
    }
}

优化后

class Solution {
    public int bestClosingTime(String customers) {
        int n = customers.length();
        int bestTime = 0;
        int minPenalty = Integer.MAX_VALUE;
        int noCount = 0;
        int customerCount = 0;
        for(int i=0; i<=n; i++){
            int currPenalty = noCount-customerCount;
            if(currPenalty<minPenalty){
                minPenalty = currPenalty;
                bestTime = i;
            }
            if(i==n) continue;
            if(customers.charAt(i)=='Y') customerCount++;
            else noCount++;
        }
        return bestTime;
    }
}