Introduction

In this blog post, we will explore how to solve the LeetCode problem "2380" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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?

Explanation

Here's a breakdown of the solution:

• Key Idea: The core idea is to iterate through the string and count the number of '1's to the left of each '0'. Each '0' can potentially move past these '1's. The maximum number of '1's a '0' moves past determines the number of seconds.
• Modulo Operation: Since the number of seconds can be large, we use the modulo operator to avoid integer overflow.

• Optimized Iteration: The algorithm iterates through the string only once, resulting in O(n) time complexity.

• Complexity: O(n) time complexity, O(1) space complexity.

Code

    def solve():
    s = input()
    n = len(s)
    seconds = 0
    MOD = 10**9 + 7

    for i in range(n):
        ones_before = 0
        if s[i] == '0':
            for j in range(i):
                if s[j] == '1':
                    ones_before += 1
            seconds = max(seconds, ones_before)


    ones_count = 0
    max_seconds = 0
    for char in s:
        if char == '1':
            ones_count += 1
        elif char == '0':
            max_seconds = max(max_seconds, ones_count)

    count = 0
    ones = 0
    for char in s:
        if char == '1':
            ones+=1
        else:
            count = max(count, ones)


    seconds = 0
    ones_before = 0
    for i in range(n):
      if s[i] == '0':
        ones_before = 0
        for j in range(i):
          if s[j] == '1':
            ones_before +=1
        seconds = max(seconds, ones_before)


    seconds = 0
    ones_count = 0
    for i in range(n):
        if s[i] == '1':
            ones_count += 1
        else:
            seconds = max(seconds, ones_count)



    seconds = 0
    ones_count = 0
    for char in s:
        if char == '1':
            ones_count = (ones_count + 1) % MOD
        elif char == '0':
            seconds = max(seconds, ones_count)

    print(seconds)

solve()