README_EN.md

926. Flip String to Monotone Increasing

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Description

A string of '0's and '1's is monotone increasing if it consists of some number of '0's (possibly 0), followed by some number of '1's (also possibly 0.)

We are given a string S of '0's and '1's, and we may flip any '0' to a '1' or a '1' to a '0'.

Return the minimum number of flips to make S monotone increasing.

 

Example 1:

Input: "00110"

Output: 1

Explanation: We flip the last digit to get 00111.

Example 2:

Input: "010110"

Output: 2

Explanation: We flip to get 011111, or alternatively 000111.

Example 3:

Input: "00011000"

Output: 2

Explanation: We flip to get 00000000.

 

Note:

1. 1 <= S.length <= 20000
2. S only consists of '0' and '1' characters.

Solutions

Python3

class Solution:
    def minFlipsMonoIncr(self, s: str) -> int:
        n = len(s)
        left, right = [0] * (n + 1), [0] * (n + 1)
        ans = 0x3f3f3f3f
        for i in range(1, n + 1):
            left[i] = left[i - 1] + (1 if s[i - 1] == '1' else 0)
        for i in range(n - 1, -1, -1):
            right[i] = right[i + 1] + (1 if s[i] == '0' else 0)
        for i in range(0, n + 1):
            ans = min(ans, left[i] + right[i])
        return ans

Java

class Solution {
    public int minFlipsMonoIncr(String s) {
        int n = s.length();
        int[] left = new int[n + 1];
        int[] right = new int[n + 1];
        int ans = Integer.MAX_VALUE;
        for (int i = 1; i <= n; i++) {
            left[i] = left[i - 1] + (s.charAt(i - 1) == '1' ? 1 : 0);
        }
        for (int i = n - 1; i >= 0; i--) {
            right[i] = right[i + 1] + (s.charAt(i) == '0' ? 1 : 0);
        }
        for (int i = 0; i <= n; i++) {
            ans = Math.min(ans, left[i] + right[i]);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minFlipsMonoIncr(string s) {
        int n = s.size();
        vector<int> left(n + 1, 0), right(n + 1, 0);
        int ans = INT_MAX;
        for (int i = 1; i <= n; ++i) {
            left[i] = left[i - 1] + (s[i - 1] == '1');
        }
        for (int i = n - 1; i >= 0; --i) {
            right[i] = right[i + 1] + (s[i] == '0');
        }
        for (int i = 0; i <= n; i++) {
            ans = min(ans, left[i] + right[i]);
        }
        return ans;
    }
};

Go

func minFlipsMonoIncr(s string) int {
	n := len(s)
	left, right := make([]int, n+1), make([]int, n+1)
	ans := math.MaxInt32
	for i := 1; i <= n; i++ {
		left[i] = left[i-1]
		if s[i-1] == '1' {
			left[i]++
		}
	}
	for i := n - 1; i >= 0; i-- {
		right[i] = right[i+1]
		if s[i] == '0' {
			right[i]++
		}
	}
	for i := 0; i <= n; i++ {
		ans = min(ans, left[i]+right[i])
	}
	return ans
}

func min(x, y int) int {
	if x < y {
		return x
	}
	return y
}

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