README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2300-2399/2380.Time%20Needed%20to%20Rearrange%20a%20Binary%20String/README_EN.md	1481	Biweekly Contest 85 Q2	String Dynamic Programming Simulation

2380. Time Needed to Rearrange a Binary String

中文文档

Description

You are given a binary string s. In one second, all occurrences of "01" are simultaneously replaced with "10". This process repeats until no occurrences of "01" exist.

Return the number of seconds needed to complete this process.

 

Example 1:

Input: s = "0110101"
Output: 4
Explanation: 
After one second, s becomes "1011010".
After another second, s becomes "1101100".
After the third second, s becomes "1110100".
After the fourth second, s becomes "1111000".
No occurrence of "01" exists any longer, and the process needed 4 seconds to complete,
so we return 4.

Example 2:

Input: s = "11100"
Output: 0
Explanation:
No occurrence of "01" exists in s, and the processes needed 0 seconds to complete,
so we return 0.

 

Constraints:

• 1 <= s.length <= 1000
• s[i] is either '0' or '1'.

 

Follow up:

Can you solve this problem in O(n) time complexity?

Solutions

Solution 1

Python3

class Solution:
    def secondsToRemoveOccurrences(self, s: str) -> int:
        ans = 0
        while s.count('01'):
            s = s.replace('01', '10')
            ans += 1
        return ans

Java

class Solution {
    public int secondsToRemoveOccurrences(String s) {
        char[] cs = s.toCharArray();
        boolean find = true;
        int ans = 0;
        while (find) {
            find = false;
            for (int i = 0; i < cs.length - 1; ++i) {
                if (cs[i] == '0' && cs[i + 1] == '1') {
                    char t = cs[i];
                    cs[i] = cs[i + 1];
                    cs[i + 1] = t;
                    ++i;
                    find = true;
                }
            }
            if (find) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int secondsToRemoveOccurrences(string s) {
        bool find = true;
        int ans = 0;
        while (find) {
            find = false;
            for (int i = 0; i < s.size() - 1; ++i) {
                if (s[i] == '0' && s[i + 1] == '1') {
                    swap(s[i], s[i + 1]);
                    ++i;
                    find = true;
                }
            }
            if (find) {
                ++ans;
            }
        }
        return ans;
    }
};

Go

func secondsToRemoveOccurrences(s string) int {
	cs := []byte(s)
	ans := 0
	find := true
	for find {
		find = false
		for i := 0; i < len(cs)-1; i++ {
			if cs[i] == '0' && cs[i+1] == '1' {
				cs[i], cs[i+1] = cs[i+1], cs[i]
				i++
				find = true
			}
		}
		if find {
			ans++
		}
	}
	return ans
}

Solution 2

Python3

class Solution:
    def secondsToRemoveOccurrences(self, s: str) -> int:
        ans = cnt = 0
        for c in s:
            if c == '0':
                cnt += 1
            elif cnt:
                ans = max(ans + 1, cnt)
        return ans

Java

class Solution {
    public int secondsToRemoveOccurrences(String s) {
        int ans = 0, cnt = 0;
        for (char c : s.toCharArray()) {
            if (c == '0') {
                ++cnt;
            } else if (cnt > 0) {
                ans = Math.max(ans + 1, cnt);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int secondsToRemoveOccurrences(string s) {
        int ans = 0, cnt = 0;
        for (char c : s) {
            if (c == '0') {
                ++cnt;
            } else if (cnt) {
                ans = max(ans + 1, cnt);
            }
        }
        return ans;
    }
};

Go

func secondsToRemoveOccurrences(s string) int {
	ans, cnt := 0, 0
	for _, c := range s {
		if c == '0' {
			cnt++
		} else if cnt > 0 {
			ans = max(ans+1, cnt)
		}
	}
	return ans
}