Minimum Window Substring in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the goal of the Minimum Window Substring problem?

To find the smallest substring in a given string that contains all characters of another given string, including duplicates.

beginner

Which data structure is commonly used to keep track of character counts in the Minimum Window Substring problem?

A dictionary (hash map) is used to store the frequency of characters needed and the frequency of characters in the current window.

intermediate

Explain the sliding window technique in the context of Minimum Window Substring.

The sliding window technique uses two pointers to represent a window in the string. The window expands to include characters until it contains all needed characters, then it contracts to find the smallest valid window.

intermediate

Why do we need to track the number of 'formed' characters in the Minimum Window Substring problem?

Tracking 'formed' characters helps know when the current window contains all required characters with correct frequencies, so we can try to minimize the window size.

advanced

What is the time complexity of the optimal solution for the Minimum Window Substring problem?

The optimal solution runs in O(n + m) time, where n is the length of the main string and m is the length of the target string, because each character is visited at most twice by the sliding window.

What does the sliding window represent in the Minimum Window Substring problem?

A The entire main string

B A fixed-size substring of the main string

C Only the characters from the target string

D A substring of the main string that may contain all required characters

The sliding window is a substring that expands and contracts to find the smallest substring containing all required characters.

Which of these is NOT needed to solve the Minimum Window Substring problem?

A Frequency count of characters in the target string

B Sorting the main string

C Two pointers to manage the window

D Frequency count of characters in the current window

Sorting the main string is not required; the solution uses frequency counts and two pointers.

When do we try to shrink the sliding window in the Minimum Window Substring problem?

A When the window contains all required characters

B When the window does not contain all required characters

C At the start of the algorithm

D Only after processing the entire string

We shrink the window to find the smallest valid substring once all required characters are included.

What does the 'formed' variable track in the algorithm?

A Number of unique characters in the current window matching the target frequency

B Number of unique characters in the target string

C Length of the current window

D Total characters processed so far

'Formed' counts how many unique characters in the current window meet the required frequency.

What is the main advantage of using a sliding window over checking all substrings?

A It sorts the string automatically

B It uses less memory

C It reduces time complexity from exponential to linear

D It finds the longest substring

Sliding window avoids checking all substrings, reducing time complexity significantly.

Describe step-by-step how the sliding window technique works to find the minimum window substring.

Think about expanding and contracting the window while checking character counts.

Explain why tracking character frequencies is important in the Minimum Window Substring problem.

Consider what happens if you ignore duplicates or frequencies.

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