Introduction

In this blog post, we will explore how to solve the LeetCode problem "1963" 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 0-indexed string s of even length n. The string consists of exactly n / 2 opening brackets '[' and n / 2 closing brackets ']'. A string is called balanced if and only if: It is the empty string, or It can be written as AB, where both A and B are balanced strings, or It can be written as [C], where C is a balanced string. You may swap the brackets at any two indices any number of times. Return the minimum number of swaps to make s balanced. Example 1: Input: s = "][][" Output: 1 Explanation: You can make the string balanced by swapping index 0 with index 3. The resulting string is "[[]]". Example 2: Input: s = "]]][[[" Output: 2 Explanation: You can do the following to make the string balanced: - Swap index 0 with index 4. s = "[]][][". - Swap index 1 with index 5. s = "[[][]]". The resulting string is "[[][]]". Example 3: Input: s = "[]" Output: 0 Explanation: The string is already balanced. Constraints: n == s.length 2 <= n <= 106 n is even. s[i] is either '[' or ']'. The number of opening brackets '[' equals n / 2, and the number of closing brackets ']' equals n / 2.

Explanation

Here's the approach, complexity analysis, and the Python code for solving this problem:

• High-level approach:

  • Iterate through the string, keeping track of the balance (number of open brackets minus number of closed brackets encountered so far).
  • Whenever the balance becomes negative, it indicates an imbalance that needs correction. Increment the swap count and reset balance by 2 (effectively swapping a closing bracket with an opening bracket from later in the string). We can swap because the problem constraints guarantee that an equal number of opening and closing brackets exist.
  • By tracking the imbalance and incrementing the swap count, we minimize the number of swaps to balance the string.

• Complexity Analysis:

  • Runtime Complexity: O(n), where n is the length of the string.
  • Storage Complexity: O(1).

Code

    def minSwaps(s: str) -> int:
    """
    Calculates the minimum number of swaps to make a string balanced.

    Args:
        s: The input string consisting of '[' and ']'.

    Returns:
        The minimum number of swaps to balance the string.
    """

    balance = 0
    swaps = 0

    for char in s:
        if char == '[':
            balance += 1
        else:
            balance -= 1

        if balance < 0:
            swaps += 1
            balance += 2

    return swaps