0926 - Flip String to Monotone Increasing (Medium)

Problem Link

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

Problem Statement

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 makesmonotone 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'.

Approach 1: Dynamic Programming

Let dp[i] be the min flips to make [0, i) monotone increasing. Starting from i = 1, if s[i] = 1, we check the previous character s[i - 1]. If it is 1 (e.g. 11...), then it is monotone increasing already, sodp[i] = dp[i - 1]. However, if s[i - 1] = 0 (e.g. 10...), then we have two choices - we either flip this zero to make like 11... or we flip all the ones before this zero (e.g. 00...). Therefore, we can see the DP transition here.

• if s[i - 1] is 1, then dp[i] = dp[i - 1]
• else dp[i] = min(dp[i - 1] + 1, cnt1)

since dp[i] is always based on dp[i - 1], we can space-optimize it using two variables - cnt0 and cnt1 where cnt0 is dp[i] at index i and cnt1 is the number of 1s.

if (s[i] == 0) cnt0 = min(cnt0 + 1, cnt1);
else cnt1 += 1;

which is essentially same as

if (s[i] == 0) cnt0 += 1;
else cnt1 += 1;
cnt0 = min(cnt0, cnt1);

Alternatively, we can count the max of cnt0 and cnt1 and return s.size() - cnt1.

class Solution {
public:
    int minFlipsMonoIncr(string s) {
        // counters to count numbers of 0s and 1s
        int cnt0 = 0, cnt1 = 0;
        // for each character
        for(auto x : s) {
            // we count the number of zeros
            if (x == '0') cnt0++;
            // or the number of ones
            else cnt1++;
            // we can either flip all ones to zeros
            // e.g. [111]0000 -> [000]0000
            // or we can just flip the current zero
            // e.g. 001[0] -> 001[1]
            cnt0 = min(cnt0, cnt1);
        }
        return cnt0;
    }
};