Introduction

In this blog post, we will explore how to solve the LeetCode problem "3501" 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 of length n, where: '1' represents an active section. '0' represents an inactive section. You can perform at most one trade to maximize the number of active sections in s. In a trade, you: Convert a contiguous block of '1's that is surrounded by '0's to all '0's. Afterward, convert a contiguous block of '0's that is surrounded by '1's to all '1's. Additionally, you are given a 2D array queries, where queries[i] = [li, ri] represents a substring s[li...ri]. For each query, determine the maximum possible number of active sections in s after making the optimal trade on the substring s[li...ri]. Return an array answer, where answer[i] is the result for queries[i]. Note For each query, treat s[li...ri] as if it is augmented with a '1' at both ends, forming t = '1' + s[li...ri] + '1'. The augmented '1's do not contribute to the final count. The queries are independent of each other. Example 1: Input: s = "01", queries = [[0,1]] Output: [1] Explanation: Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 1. Example 2: Input: s = "0100", queries = [[0,3],[0,2],[1,3],[2,3]] Output: [4,3,1,1] Explanation: Query [0, 3] → Substring "0100" → Augmented to "101001" Choose "0100", convert "0100" → "0000" → "1111". The final string without augmentation is "1111". The maximum number of active sections is 4. Query [0, 2] → Substring "010" → Augmented to "10101" Choose "010", convert "010" → "000" → "111". The final string without augmentation is "1110". The maximum number of active sections is 3. Query [1, 3] → Substring "100" → Augmented to "11001" Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 1. Query [2, 3] → Substring "00" → Augmented to "1001" Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 1. Example 3: Input: s = "1000100", queries = [[1,5],[0,6],[0,4]] Output: [6,7,2] Explanation: Query [1, 5] → Substring "00010" → Augmented to "1000101" Choose "00010", convert "00010" → "00000" → "11111". The final string without augmentation is "1111110". The maximum number of active sections is 6. Query [0, 6] → Substring "1000100" → Augmented to "110001001" Choose "000100", convert "000100" → "000000" → "111111". The final string without augmentation is "1111111". The maximum number of active sections is 7. Query [0, 4] → Substring "10001" → Augmented to "1100011" Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 2. Example 4: Input: s = "01010", queries = [[0,3],[1,4],[1,3]] Output: [4,4,2] Explanation: Query [0, 3] → Substring "0101" → Augmented to "101011" Choose "010", convert "010" → "000" → "111". The final string without augmentation is "11110". The maximum number of active sections is 4. Query [1, 4] → Substring "1010" → Augmented to "110101" Choose "010", convert "010" → "000" → "111". The final string without augmentation is "01111". The maximum number of active sections is 4. Query [1, 3] → Substring "101" → Augmented to "11011" Because there is no block of '1's surrounded by '0's, no valid trade is possible. The maximum number of active sections is 2. Constraints: 1 <= n == s.length <= 105 1 <= queries.length <= 105 s[i] is either '0' or '1'. queries[i] = [li, ri] 0 <= li <= ri < n

Explanation

Here's the breakdown of the solution:

• Iterate through Queries: For each query [li, ri], extract the substring s[li:ri+1].
• Simulate Trade and Calculate Active Sections: Simulate the trade process: first converting a '1' block to '0's and then converting a '0' block to '1's. Keep track of the maximum number of active sections achieved after any valid trade. If no valid trade is possible, return the initial count of '1's in the substring.

• Augment String: Augment the substring with '1' at the beginning and the end for the trade simulation.

• Runtime Complexity: O(q * n), where q is the number of queries and n is the maximum length of the string.

• Storage Complexity: O(n) in the worst case, for creating the substring and potentially modified versions of it.

Code

    def solve():
    s = input()
    queries = []
    num_queries = int(input())
    for _ in range(num_queries):
        queries.append(list(map(int, input().split())))

    def count_ones(sub):
        count = 0
        for char in sub:
            if char == '1':
                count += 1
        return count

    def max_active_sections(sub):
        n = len(sub)
        max_ones = count_ones(sub)

        for i in range(n):
            for j in range(i, n):
                temp_sub = list(sub)  # Work with a copy to avoid modifying the original

                # Step 1: Find a '1' block surrounded by '0's and convert it to '0's.
                found_block = False
                for start in range(n):
                    end = start
                    while end < n and temp_sub[end] == '1':
                        end += 1

                    if start > 0 and end < n and temp_sub[start-1] == '0' and temp_sub[end] == '0' and start != end :
                        found_block = True
                        for k in range(start, end):
                            temp_sub[k] = '0'
                        break #Only apply the first trade

                if not found_block:
                    continue

                # Step 2: Find a '0' block surrounded by '1's and convert it to '1's.
                for start in range(n):
                    end = start
                    while end < n and temp_sub[end] == '0':
                        end += 1

                    if start > 0 and end < n and temp_sub[start-1] == '1' and temp_sub[end] == '1' and start != end:
                        for k in range(start, end):
                            temp_sub[k] = '1'

                        max_ones = max(max_ones, count_ones("".join(temp_sub)))
                        break #Only apply the first trade

        return max_ones

    results = []
    for li, ri in queries:
        sub = s[li:ri+1]
        augmented_sub = "1" + sub + "1"
        result = max_active_sections(augmented_sub)
        results.append(result)

    print(results)

solve()