Description

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.

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'.

Solution

Dynamic programming, dp_0 is the minimum number of flips to end with 0, and dp_1 is the number of ending with 1, then:

$$\begin{gathered} d{p}_0=\{\begin{array}{ll}d{p}_0,& \text{if s[i] is 0}\\ d{p}_0+1,& \text{if s[i] is 1}\end{array} \\ d{p}_1=\{\begin{array}{ll}d{p}_0+1,& \text{if s[i] is 0}\\ \min (d{p}_1,d{p}_0),& \text{if s[i] is 1}\end{array} \end{gathered}$$

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:def minFlipsMonoIncr(self, s: str) -> int:dp = [0, 0]for c in s:if c == '1':dp = [dp[0] + 1, min(dp)]else:dp = [dp[0], dp[1] + 1]return min(dp)