Minimum Window Substring in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the left pointer for the sliding window.

DSA C

int left = [1];

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Ds.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 instead of 0.

Using negative index which is invalid.

The left pointer starts at index 0 to begin scanning the string from the start.

2fill in blank

medium

Complete the code to check if the current character is in the target frequency map.

DSA C

if (target_freq[s[right]] [1] 0) {

Drag options to blanks, or click blank then click option'

A==

B<

C!=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which only checks equality, not presence.

Using '<' which is incorrect for frequency check.

We check if the frequency of the character is greater than zero to know if it is needed.

3fill in blank

hard

Fix the error in the condition to shrink the window when all characters are matched.

DSA C

while (formed [1] required) {

Drag options to blanks, or click blank then click option'

A==

B<

C>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' which causes incorrect shrinking.

Using '<' or '>' which do not represent the correct condition.

We shrink the window only when the number of formed characters equals the required count.

4fill in blank

hard

Fill both blanks to update the minimum window size and start index.

DSA C

if (right - left + 1 [1] min_len) {
    min_len = right - left + 1;
    min_start = [2];
}

Drag options to blanks, or click blank then click option'

A<

B>

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' causes wrong window size update.

Setting min_start to right instead of left.

We update min_len if the current window is smaller and record the start index as left.

5fill in blank

hard

Fill all three blanks to correctly update the frequency map and formed count when shrinking the window.

DSA C

char c = s[left];
window_freq[c]--;
if (window_freq[c] [1] target_freq[c]) {
    formed [2];
}
left [3] 1;

Drag options to blanks, or click blank then click option'

A<

B--

C+

D++

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' in frequency comparison.

Incrementing formed instead of decrementing.

Moving left pointer backward instead of forward.

When the window frequency falls below target, formed decreases; left pointer moves right.

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. Count the total number of substrings in a string

C. Find the longest substring without repeating characters

D. Sort the 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 includes all characters from another string.

2. Step 2: Compare options with the 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 = Smallest substring containing all chars [OK]

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

Common Mistakes:

• Confusing with longest substring problems
• Thinking it counts substrings instead of finding minimum
• Mixing sorting with substring search

2. Which of the following C code snippets correctly initializes a frequency array for characters in the Minimum Window Substring problem?

easy

A. int freq[256] = {0};

B. int freq[26]; for(int i=0; i<26; i++) freq[i] = -1;

C. int freq[128]; memset(freq, 1, sizeof(freq));

D. int freq[256]; freq = {0};

Solution

1. Step 1: Check correct array initialization syntax in C

   int freq[256] = {0}; uses correct syntax to initialize all 256 elements to zero using brace initialization.

2. Step 2: Analyze other options for errors

   int freq[26]; for(int i=0; i<26; i++) freq[i] = -1; does not initialize all elements and sets to -1 incorrectly. int freq[128]; memset(freq, 1, sizeof(freq)); uses memset with 1 which sets bytes to 1, not zero. int freq[256]; freq = {0}; tries to assign array which is invalid in C.

3. Final Answer:

   int freq[256] = {0}; -> Option A

4. Quick Check:

   Correct zero initialization = int freq[256] = {0}; [OK]

Hint: Use brace initialization to zero arrays in C [OK]

Common Mistakes:

• Using memset with wrong value
• Trying to assign arrays directly
• Not initializing all elements

3. What will be the output of the following C code snippet for Minimum Window Substring?

char s[] = "ADOBECODEBANC";
char t[] = "ABC";
// Assume function minWindow returns the minimum window substring
char* result = minWindow(s, t);
printf("%s", result);

Assuming minWindow is correctly implemented, what is printed?

medium

A. CODEBA

B. ADOBEC

C. BANC

D. DOBECODEBA

Solution

1. Step 1: Understand the input strings

   String s is "ADOBECODEBANC" and t is "ABC". We want the smallest substring of s containing A, B, and C.

2. Step 2: Identify the minimum window substring

   The substring "BANC" is the smallest substring containing all characters A, B, and C.

3. Final Answer:

   BANC -> Option C

4. Quick Check:

   Minimum window substring for "ABC" in s = "BANC" [OK]

Hint: Remember classic example: "BANC" is minimum window for "ABC" [OK]

Common Mistakes:

• Choosing first substring containing all chars instead of minimum
• Confusing order of characters
• Ignoring repeated characters

4. Consider this buggy C code snippet for Minimum Window Substring:

int left = 0, right = 0;
int count = 0;
while (right < strlen(s)) {
    if (freq[s[right]] > 0) count++;
    freq[s[right]]--;
    right++;
    while (count == t_len) {
        if (right - left < min_len) {
            min_len = right - left;
            start = left;
        }
        freq[s[left]]++;
        if (freq[s[left]] > 0) count--;
        left++;
    }
}
// Returns substring from start with length min_len

What is the main bug in this code?

medium

A. Using strlen(s) inside the loop condition causes repeated length calculation

B. Incorrect update of count when moving left pointer

C. The inner while condition should be count >= t_len

D. freq array is not initialized before use

Solution

1. Step 1: Analyze freq array usage

   The freq array is decremented and incremented but not initialized, so it contains garbage values causing incorrect logic.

2. Step 2: Check other options

   Using strlen(s) inside the loop condition causes repeated length calculation is inefficient but not a bug. Incorrect update of count when moving left pointer logic is correct for count update. The inner while condition should be count >= t_len is incorrect because count == t_len is the correct condition.

3. Final Answer:

   freq array is not initialized before use -> Option D

4. Quick Check:

   freq must be zero-initialized before use [OK]

Hint: Always initialize frequency arrays before use [OK]

Common Mistakes:

• Ignoring initialization of freq array
• Misunderstanding count update logic
• Confusing loop conditions

5. You want to find the minimum window substring of s = "aabbccdd" that contains all characters of t = "abc". Which approach correctly handles repeated characters in s and t to find the minimum window?

hard

A. Count unique characters in t and find first substring in s containing those unique chars

B. Use two frequency arrays: one for t counts and one for current window counts, then expand and contract window accordingly

C. Use a single frequency array for t and move right pointer until all chars found, then stop

D. Sort s and t and find common characters sequentially

Solution

1. Step 1: Understand the need for frequency tracking

   Since t can have repeated characters, we must track how many times each character is needed and how many times it appears in the current window.

2. Step 2: Use two frequency arrays and sliding window

   One array stores counts of characters in t, another tracks counts in the current window. Expand right pointer to include chars, contract left pointer to minimize window while keeping all required chars.

3. Final Answer:

   Use two frequency arrays: one for t counts and one for current window counts, then expand and contract window accordingly -> Option B

4. Quick Check:

   Two freq arrays + sliding window = correct approach [OK]

Hint: Track needed and current counts separately for repeated chars [OK]

Common Mistakes:

• Ignoring repeated characters in t
• Using only unique character counts
• Stopping window expansion too early