Minimum Window Substring in DSA Python - Time & Space Complexity

Time Complexity: Minimum Window Substring

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to find the smallest substring containing all characters from another string changes as the input grows.

How does the running time increase when the input strings get longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

from collections import Counter

def min_window(s: str, t: str) -> str:
    need = Counter(t)
    missing = len(t)
    left = start = end = 0

    for right, char in enumerate(s):
        if need[char] > 0:
            missing -= 1
        need[char] -= 1

        while missing == 0:
            if end == 0 or right - left + 1 < end - start:
                start, end = left, right + 1
            need[s[left]] += 1
            if need[s[left]] > 0:
                missing += 1
            left += 1

    return s[start:end]
    

This code finds the smallest substring in s that contains all characters of t.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The main loop moves the right pointer through string s, and the inner while loop moves the left pointer to shrink the window.
• How many times: Each character in s is visited at most twice--once when expanding the right pointer and once when moving the left pointer.

How Execution Grows With Input

As the input string s grows, the number of operations grows roughly in proportion to its length because each character is processed a limited number of times.

Input Size (n)	Approx. Operations
10	About 20 (each character visited twice)
100	About 200
1000	About 2000

Pattern observation: The operations grow linearly with input size, doubling roughly when input size doubles.

Final Time Complexity

Time Complexity: O(n + m)

This means the time to find the minimum window grows linearly with the size of both input strings s and t.

Common Mistake

[X] Wrong: "The nested while loop makes this algorithm quadratic, so it must be O(n²)."

[OK] Correct: Although there is a nested loop, each character is visited at most twice in total, so the total work is still linear, not quadratic.

Interview Connect

Understanding this sliding window approach and its linear time complexity shows you can handle problems that require careful pointer movement and counting, a valuable skill in many coding challenges.

Self-Check

"What if the input string t contains many repeated characters? How would that affect the time complexity?"

Practice

1. What is the main goal of the Minimum Window Substring problem?

easy

A. Find the smallest substring in a string that contains all characters of another string

B. Find the longest substring without repeating characters

C. Count the number of substrings that contain a specific character

D. Sort characters of a string in ascending order

Solution

1. Step 1: Understand the problem statement

   The problem asks to find the smallest part of a string that has all characters from another string.

2. Step 2: Compare options with problem goal

   Only Find the smallest substring in a string that contains all characters of another string matches the goal of finding the smallest substring containing all characters of another string.

3. Final Answer:

   Find the smallest substring in a string that contains all characters of another string -> Option A

4. Quick Check:

   Minimum Window Substring = Find smallest substring with all chars [OK]

Hint: Look for smallest substring containing all target chars [OK]

Common Mistakes:

• Confusing with longest substring problems
• Thinking about sorting instead of substring search
• Ignoring the requirement to include all characters

2. Which of the following Python code snippets correctly initializes a dictionary to count characters in string t for the Minimum Window Substring problem?

easy

A. count = dict.fromkeys(t, 0)

B. count = {0: char for char in t}

C. count = {char: 1 for char in t}

D. count = {char: t.count(char) for char in t}

Solution

1. Step 1: Understand dictionary initialization for counting

   We want a dictionary with keys as characters from t and values initialized to the count of each character in t.

2. Step 2: Check each option

   count = {char: t.count(char) for char in t} correctly counts the occurrences of each character in t.

3. Final Answer:

   count = {char: t.count(char) for char in t} -> Option D

4. Quick Check:

   Use dict comprehension with t.count(char) to initialize counts [OK]

Hint: Use dict comprehension with t.count(char) to initialize counts [OK]

Common Mistakes:

• Setting counts to 0 instead of actual counts
• Swapping keys and values in dict comprehension
• Using invalid dictionary syntax

3. What will be the output of the following Python code snippet?

def min_window(s, t):
    from collections import Counter
    dict_t = Counter(t)
    required = len(dict_t)
    l, r = 0, 0
    formed = 0
    window_counts = {}
    ans = float('inf'), None, None
    while r < len(s):
        character = s[r]
        window_counts[character] = window_counts.get(character, 0) + 1
        if character in dict_t and window_counts[character] == dict_t[character]:
            formed += 1
        while l <= r and formed == required:
            character = s[l]
            if r - l + 1 < ans[0]:
                ans = (r - l + 1, l, r)
            window_counts[character] -= 1
            if character in dict_t and window_counts[character] < dict_t[character]:
                formed -= 1
            l += 1
        r += 1
    return '' if ans[0] == float('inf') else s[ans[1]:ans[2]+1]

print(min_window('ADOBECODEBANC', 'ABC'))

medium

A. NONE

B. ADOBEC

C. BANC

D. CODEBA

Solution

1. Step 1: Understand the sliding window logic

   The code finds the smallest substring in 'ADOBECODEBANC' containing all chars 'A', 'B', 'C'.

2. Step 2: Trace the window updates

   The smallest such substring is 'BANC' starting at index 9 and ending at 12.

3. Final Answer:

   BANC -> Option C

4. Quick Check:

   Minimum window substring = BANC [OK]

Hint: Trace window expansion and contraction carefully [OK]

Common Mistakes:

• Confusing substring boundaries
• Not updating formed count correctly
• Returning empty string incorrectly

4. Identify the bug in the following code snippet for Minimum Window Substring and select the correct fix:

def min_window(s, t):
    from collections import Counter
    dict_t = Counter(t)
    required = len(dict_t)
    l, r = 0, 0
    formed = 0
    window_counts = {}
    ans = float('inf'), None, None
    while r < len(s):
        character = s[r]
        window_counts[character] = window_counts.get(character, 0) + 1
        if character in dict_t and window_counts[character] == dict_t[character]:
            formed += 1
        while l <= r and formed == required:
            character = s[l]
            if r - l + 1 < ans[0]:
                ans = (r - l + 1, l, r)
            window_counts[character] -= 1
            if character in dict_t and window_counts[character] < dict_t[character]:
                formed -= 1
            l += 1
        r += 1
    return s[ans[1]:ans[2]+1]

medium

A. Initialize window_counts with dict_t keys

B. Return empty string if no valid window found

C. Use a list instead of dict for window_counts

D. Remove formed variable

Solution

1. Step 1: Analyze return statement

   The code returns substring using ans indices without checking if a valid window was found.

2. Step 2: Fix missing empty string return

   If no window found, ans[0] remains infinity, so return empty string to avoid error.

3. Final Answer:

   Return empty string if no valid window found -> Option B

4. Quick Check:

   Check ans[0] before return to avoid errors [OK]

Hint: Always check if a valid window was found before returning [OK]

Common Mistakes:

• Returning substring without validation
• Ignoring empty result cases
• Overcomplicating window_counts initialization

5. Given strings s = 'aabbcc' and t = 'abc', which of the following modifications to the Minimum Window Substring algorithm will correctly find the smallest substring containing all characters of t even if t has duplicate characters like 'aabc'?

hard

A. Use a frequency counter for t and compare counts exactly in the window

B. Only check if characters exist in the window, ignoring counts

C. Sort s and t and find common prefix

D. Use a set for t instead of a counter

Solution

1. Step 1: Understand the need for frequency counts

   When t has duplicates, we must count how many times each character appears and match exactly.

2. Step 2: Choose the correct approach

   Using a frequency counter for t and comparing counts in the window ensures all duplicates are included.

3. Final Answer:

   Use a frequency counter for t and compare counts exactly in the window -> Option A

4. Quick Check:

   Count duplicates exactly using frequency counters [OK]

Hint: Count character frequencies to handle duplicates correctly [OK]

Common Mistakes:

• Ignoring duplicate characters in t
• Using sets which lose count information
• Sorting strings which doesn't help substring search