Minimum Window Substring in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Minimum Window Substring Function

What is the output of the following Python code that finds the minimum window substring containing all characters of t in s?

DSA Python

def min_window(s, t):
    from collections import Counter
    need = Counter(t)
    missing = len(t)
    left = start = end = 0
    for right, char in enumerate(s, 1):
        if need[char] > 0:
            missing -= 1
        need[char] -= 1
        if missing == 0:
            while left < right and need[s[left]] < 0:
                need[s[left]] += 1
                left += 1
            if end == 0 or right - left < end - start:
                start, end = left, right
            need[s[left]] += 1
            missing += 1
            left += 1
    return s[start:end]

print(min_window("ADOBECODEBANC", "ABC"))

A "BANC"

B "ADOBEC"

C "CODEBA"

D "DOBECODEBA"

💡 Hint

Focus on the smallest substring that contains all characters A, B, and C.

The function finds the smallest substring in s that contains all characters of t. The substring "BANC" is the shortest that contains A, B, and C.

❓ Predict Output

intermediate

Minimum Window Substring with Repeated Characters

What is the output of this code when s contains repeated characters and t requires multiple occurrences?

DSA Python

def min_window(s, t):
    from collections import Counter
    need = Counter(t)
    missing = len(t)
    left = start = end = 0
    for right, char in enumerate(s, 1):
        if need[char] > 0:
            missing -= 1
        need[char] -= 1
        if missing == 0:
            while left < right and need[s[left]] < 0:
                need[s[left]] += 1
                left += 1
            if end == 0 or right - left < end - start:
                start, end = left, right
            need[s[left]] += 1
            missing += 1
            left += 1
    return s[start:end]

print(min_window("AAABBC", "AABC"))

A "AABB"

B "AAABBC"

C "AABBC"

D "ABBC"

💡 Hint

Remember t requires two 'A's, one 'B', and one 'C'.

The minimum window substring containing two 'A's, one 'B', and one 'C' is "AABBC".

🔧 Debug

advanced

Identify the Error in Minimum Window Substring Code

What error will this code raise when executed?

DSA Python

def min_window(s, t):
    from collections import Counter
    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
        if missing == 0:
            while left < right and need[s[left]] < 0:
                need[s[left]] += 1
                left += 1
            if end == 0 or right - left < end - start:
                start, end = left, right
            need[s[left]] += 1
            missing += 1
            left += 1
    return s[start:end]

print(min_window("ADOBECODEBANC", "ABC"))

A Returns empty string

B TypeError

C IndexError

D Returns incorrect substring "BANC"

💡 Hint

Check how the right pointer is used in slicing the string.

The code uses zero-based indexing for right but slices s[start:end] where end is zero-based, missing the last character. Also, right is zero-based but the loop does not add 1, so the substring is empty.

🧠 Conceptual

advanced

Why Use a Sliding Window in Minimum Window Substring?

Why is the sliding window technique suitable for solving the Minimum Window Substring problem?

A Because it sorts the string to find the substring faster

B Because it allows checking all substrings by expanding and contracting the window efficiently

C Because it uses recursion to explore all substring possibilities

D Because it converts the string into a graph for traversal

💡 Hint

Think about how to find the smallest substring containing all needed characters without checking every substring.

Sliding window expands to include needed characters and contracts to remove unnecessary ones, efficiently finding the minimum substring.

🚀 Application

expert

Minimum Window Substring with Unicode Characters

Given the string s = "𝔸𝔹𝔸𝔻𝔸𝔹𝔠" and t = "𝔸𝔠", what is the minimum window substring containing all characters of t?

A "𝔸𝔹𝔸𝔻𝔸𝔹𝔠"

B "𝔠"

C "𝔹𝔠"

D "𝔸𝔹𝔠"

💡 Hint

Unicode characters count as single characters; find the shortest substring containing both.

The substring "𝔸𝔹𝔠" is the shortest substring containing both "𝔸" and "𝔠" in the given string.

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