# Flip String to Monotone Increasing

Published: Sep 12, 2022

Medium String Dynamic Programming

## Introduction

The string related problems often need much consideration to figure out how to solve. Once it becomes clear, the solution might be very simple.

## Problem 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.

Constraints:

• `1 <= s.length <= 10**5`
• `s[i]` is either ‘0’ or ‘1’

https://leetcode.com/problems/flip-string-to-monotone-increasing/

## Examples

``````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.
``````

## Analysis

Let’s start from a length 2. Valid patters are:

``````2: 00, 01, 11
3: 000, 001, 011, 111
4: 0000, 0001, 0011, 0111, 1111
...
``````

How many ones are there – this is a key to solve the problem. While going over the string one by one, count up ones if the character is ‘1’. Count up flips and count down ones if the character is ‘0’ and ones are not 0. That’s all to find the answer.

## Solution

``````class FlipStringToMonotoneIncreasing:
def minFlipsMonoIncr(self, s: str) -> int:
flips = 0
ones = 0
for c in s:
if c == '1':
ones += 1
elif (ones > 0):
flips += 1
ones -= 1
return flips
``````

## Complexities

• Time: `O(n)`
• Space: `O(1)`