926.Flip String to Monotone Increasing
===

###### tags: `Medium`,`String`,`DP`

[926. Flip String to Monotone Increasing](https://leetcode.com/problems/flip-string-to-monotone-increasing/)

### 題目描述

A binary string is monotone increasing if it consists of some number of `0`'s (possibly none), followed by some number of `1`'s (also possibly none).

You are given a binary string `s`. You can flip `s[i]` changing it from `0` to `1` or from `1` to `0`.

Return *the minimum number of flips to make* `s` *monotone increasing.*

### 範例

**Example 1:**

Input: s = "00110"
Output: 1
Explanation: We flip the last digit to get 00111.

**Example 2:**

Input: s = "010110"
Output: 2
Explanation: We flip to get 011111, or alternatively 000111.

**Example 3:**

Input: s = "00011000"
Output: 2
Explanation: We flip to get 00000000.

**Constraints**:

* 1 <= `s.length` <= 10^5^
* `s[i]` is either `'0'` or `'1'`.

### 解答
#### Python

class Solution:
    def minFlipsMonoIncr(self, s: str) -> int:
        # f0: Answer to 0's monotone string steps
        # f1: Answer to followed by some number of 1's monotone string steps
        f0, f1 = 0, 0
        
        for ch in s:
            if ch == '0':
                f1 += 1
            else:
                f1 = min(f0, f1)
                f0 += 1
                
        return min(f0, f1)

> [name=Ron Chen][time=Jan 17, 2023]

#### C++
##### 思路:
1. 考慮一個合法的string, +"1"時合法不變, +"0"時結果有二, 一為把前面出現的1全變成0, 二為把當前0變成1

##### 做法:
1. 遍歷string, 紀錄1的數量, 出現0時更新結果, 看是改0還是1

class Solution {
public:
    int minFlipsMonoIncr(string s) {
        int cnt_1 = 0, res = 0;
        for(char c : s)
        {
            if(c == '1')
                cnt_1++;
            else
                res = min(res+1, cnt_1);
        }
        return res;
    }
};

> [name=XD] [time= Jan 17, 2023]

Time: $O(n)$
Extra Space: $O(1)$

### Reference

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