How do you solve Minimum Window Substring on LeetCode?

Minimum Window Substring (LeetCode #76) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a sliding window allows us to dynamically adjust our search space, making it efficient.

What is the brute force solution for Minimum Window Substring?

The brute force approach for Minimum Window Substring is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible substring of 's' to see if it contains all characters from 't'. This is straightforward but inefficient, especially for long strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Window Substring?

Minimum Window Substring uses the following concepts: Hash Table, String, Sliding Window. The recommended patterns to study are: Hash Map, Sliding Window.

What are the interview tips for Minimum Window Substring?

Always clarify the problem requirements and constraints before jumping into coding. Think about edge cases and how your solution handles them. Explain your thought process clearly as you code; it shows your understanding and helps the interviewer follow along.

What are common mistakes when solving Minimum Window Substring?

Not handling edge cases where 't' is longer than 's'. Failing to correctly update the counts in the frequency map when moving the left pointer.

#76

Minimum Window Substring

Hard

Hash TableStringSliding WindowHash MapSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses the sliding window technique with two pointers to efficiently find the minimum window. This reduces the time complexity significantly by avoiding unnecessary checks.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency map of characters in 't'.
2Step 2: Use two pointers to create a window in 's'. Expand the right pointer to include characters until the window contains all characters from 't'.
3Step 3: Once all characters are included, move the left pointer to minimize the window while still containing all characters from 't'.
4Step 4: Keep track of the minimum length window found.

solution.py31 lines

1from collections import Counter, defaultdict
2
3def min_window(s, t):
4    if not t or not s:
5        return ""
6    dict_t = Counter(t)
7    required = len(dict_t)
8    l, r = 0, 0
9    formed = 0
10    window_counts = defaultdict(int)
11    min_len = float('inf')
12    min_window = (0, 0)
13
14    while r < len(s):
15        character = s[r]
16        window_counts[character] += 1
17        if character in dict_t and window_counts[character] == dict_t[character]:
18            formed += 1
19
20        while l <= r and formed == required:
21            character = s[l]
22            if r - l + 1 < min_len:
23                min_len = r - l + 1
24                min_window = (l, r)
25            window_counts[character] -= 1
26            if character in dict_t and window_counts[character] < dict_t[character]:
27                formed -= 1
28            l += 1
29        r += 1
30
31    return "" if min_len == float('inf') else s[min_window[0]:min_window[1] + 1]

ℹ

Complexity note: The time complexity is O(n) because we traverse the string 's' at most twice (once with the right pointer and once with the left pointer). The space complexity is O(n) due to the storage of character counts in the frequency maps.

1Using a sliding window allows us to dynamically adjust our search space, making it efficient.
2Character frequency maps help in quickly checking if a window contains all required characters.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.