Introduction

In this blog post, we will explore how to solve the LeetCode problem "76" 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

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "". The testcases will be generated such that the answer is unique. Example 1: Input: s = "ADOBECODEBANC", t = "ABC" Output: "BANC" Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t. Example 2: Input: s = "a", t = "a" Output: "a" Explanation: The entire string s is the minimum window. Example 3: Input: s = "a", t = "aa" Output: "" Explanation: Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string. Constraints: m == s.length n == t.length 1 <= m, n <= 105 s and t consist of uppercase and lowercase English letters. Follow up: Could you find an algorithm that runs in O(m + n) time?

Explanation

Here's an efficient solution to find the minimum window substring containing all characters of another string, along with explanations and complexity analysis:

• Sliding Window: Use a sliding window approach. Maintain two pointers, left and right, representing the window's boundaries. Expand the right pointer to find a window containing all characters of t.
• Character Frequency Tracking: Use a dictionary (or array for ASCII characters) to store the frequency of characters in t. Also, maintain a separate dictionary (or array) to track the frequency of characters within the current window.

• Contraction: Once a valid window is found (i.e., it contains all characters of t), contract the window from the left until it's no longer valid. Keep track of the minimum window found so far.

• Time Complexity: O(m + n), where m is the length of string s and n is the length of string t. Space Complexity: O(1). The character frequency dictionaries/arrays will store at most the size of the alphabet, which is constant.

Code

    def minWindow(s: str, t: str) -> str:
    """
    Finds the minimum window substring of s such that every character in t
    (including duplicates) is included in the window.
    """

    if not s or not t:
        return ""

    if len(s) < len(t):
        return ""

    t_freq = {}
    for char in t:
        t_freq[char] = t_freq.get(char, 0) + 1

    window_freq = {}
    left = 0
    right = 0
    matched = 0  # Number of characters in the window that satisfy the target frequency.
    min_len = float('inf')
    start_index = 0

    while right < len(s):
        char = s[right]
        window_freq[char] = window_freq.get(char, 0) + 1

        if char in t_freq and window_freq[char] == t_freq[char]:
            matched += 1

        while matched == len(t_freq):
            window_len = right - left + 1
            if window_len < min_len:
                min_len = window_len
                start_index = left

            left_char = s[left]
            window_freq[left_char] -= 1

            if left_char in t_freq and window_freq[left_char] < t_freq[left_char]:
                matched -= 1

            left += 1

        right += 1

    if min_len == float('inf'):
        return ""
    else:
        return s[start_index: start_index + min_len]