Introduction

In this blog post, we will explore how to solve the LeetCode problem "1234" 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 string s of length n containing only four kinds of characters: 'Q', 'W', 'E', and 'R'. A string is said to be balanced if each of its characters appears n / 4 times where n is the length of the string. Return the minimum length of the substring that can be replaced with any other string of the same length to make s balanced. If s is already balanced, return 0. Example 1: Input: s = "QWER" Output: 0 Explanation: s is already balanced. Example 2: Input: s = "QQWE" Output: 1 Explanation: We need to replace a 'Q' to 'R', so that "RQWE" (or "QRWE") is balanced. Example 3: Input: s = "QQQW" Output: 2 Explanation: We can replace the first "QQ" to "ER". Constraints: n == s.length 4 <= n <= 105 n is a multiple of 4. s contains only 'Q', 'W', 'E', and 'R'.

Explanation

Here's the breakdown of the approach and the Python code:

• High-Level Approach:

  • Calculate the required count for each character (n / 4).
  • Use a sliding window to find the smallest substring that, when removed, leaves a balanced string. The key is to count the excess of each character and then look for a window that eliminates this excess.
  • Optimize the sliding window by expanding the right boundary until the window contains at least the needed excess for each character. Then, shrink the left boundary while maintaining that condition to find the minimum window size.

• Complexity:

  • Runtime: O(n) - Linear due to the sliding window approach.
  • Storage: O(1) - Constant space for character counts.

Code

    def balancedString(s: str) -> int:
    n = len(s)
    k = n // 4
    count = {'Q': 0, 'W': 0, 'E': 0, 'R': 0}
    for char in s:
        count[char] += 1

    excess = {'Q': max(0, count['Q'] - k),
              'W': max(0, count['W'] - k),
              'E': max(0, count['E'] - k),
              'R': max(0, count['R'] - k)}

    if all(val == 0 for val in excess.values()):
        return 0

    left = 0
    min_len = n
    window_count = {'Q': 0, 'W': 0, 'E': 0, 'R': 0}

    for right in range(n):
        window_count[s[right]] += 1

        while all(window_count[char] >= excess[char] for char in 'QWER'):
            min_len = min(min_len, right - left + 1)
            window_count[s[left]] -= 1
            left += 1

    return min_len