Minimum Window Substring in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Minimum Window Substring Function

What is the output of the following C function when called with s = "ADOBECODEBANC" and t = "ABC"?

DSA C

char* minWindow(char* s, char* t) {
    int need[128] = {0}, have[128] = {0};
    int required = 0, formed = 0;
    int left = 0, right = 0, start = 0, min_len = 100000;
    for (int i = 0; t[i]; i++) {
        if (need[t[i]] == 0) required++;
        need[t[i]]++;
    }
    while (s[right]) {
        char c = s[right];
        have[c]++;
        if (have[c] == need[c]) formed++;
        while (left <= right && formed == required) {
            if (right - left + 1 < min_len) {
                min_len = right - left + 1;
                start = left;
            }
            char d = s[left];
            have[d]--;
            if (have[d] < need[d]) formed--;
            left++;
        }
        right++;
    }
    if (min_len == 100000) return "";
    char* res = (char*)malloc(min_len + 1);
    strncpy(res, s + start, min_len);
    res[min_len] = '\0';
    return res;
}

int main() {
    char s[] = "ADOBECODEBANC";
    char t[] = "ABC";
    char* result = minWindow(s, t);
    printf("%s\n", result);
    free(result);
    return 0;
}

A "BANC"

B "ADOBEC"

C "CODEBA"

D "DOBECODEBA"

💡 Hint

Think about the smallest substring containing all characters of t.

The minimum window substring containing all characters 'A', 'B', and 'C' in s is "BANC".

🧠 Conceptual

intermediate

Understanding the Sliding Window Technique

In the Minimum Window Substring problem, why do we move the left pointer forward after finding a valid window?

A To reset the window and start searching from scratch.

B To increase the window size to include more characters.

C To try to find a smaller valid window by removing unnecessary characters from the left.

D To skip characters that are not in the target string t.

💡 Hint

Think about minimizing the window size while keeping it valid.

After finding a valid window, moving the left pointer forward helps to shrink the window and find the smallest possible valid substring.

🔧 Debug

advanced

Identify the Bug in Minimum Window Substring Code

What error will occur when running this code snippet for Minimum Window Substring?

DSA C

char* minWindow(char* s, char* t) {
    int need[128] = {0}, have[128] = {0};
    int required = 0, formed = 0;
    int left = 0, right = 0, start = 0, min_len = 100000;
    for (int i = 0; t[i]; i++) {
        need[t[i]]++;
        required++;
    }
    while (s[right]) {
        char c = s[right];
        have[c]++;
        if (have[c] == need[c]) formed++;
        while (left <= right && formed == required) {
            if (right - left + 1 < min_len) {
                min_len = right - left + 1;
                start = left;
            }
            char d = s[left];
            have[d]--;
            if (have[d] < need[d]) formed--;
            left++;
        }
        right++;
    }
    if (min_len == 100000) return "";
    char* res = (char*)malloc(min_len + 1);
    strncpy(res, s + start, min_len);
    res[min_len] = '\0';
    return res;
}

A Memory leak due to missing free() call for allocated memory.

B The variable 'required' counts total characters including duplicates, causing incorrect formed count.

C Segmentation fault due to accessing s[right] without checking null terminator.

D Syntax error because of missing semicolon after variable declarations.

💡 Hint

Check how 'required' is calculated and used.

The code increments 'required' for every character in t, including duplicates, but 'formed' counts unique characters matched. This mismatch causes the window to never be recognized as valid.

🚀 Application

advanced

Minimum Window Substring with Unicode Characters

How would you modify the Minimum Window Substring algorithm to handle Unicode characters beyond ASCII?

A Use a hash map or dictionary to count characters instead of fixed-size arrays.

B Increase the size of the fixed array to 256 to cover extended ASCII.

C Convert all characters to lowercase ASCII before processing.

D Ignore characters outside ASCII range to simplify the problem.

💡 Hint

Think about character sets larger than 128.

Unicode characters require dynamic data structures like hash maps to count frequencies, as fixed arrays for ASCII are insufficient.

🧠 Conceptual

expert

Time Complexity of Minimum Window Substring Algorithm

What is the time complexity of the optimal Minimum Window Substring algorithm using the sliding window technique?

A O(m^2), where m is length of t

B O(n * m), where n is length of s and m is length of t

C O(n^2), where n is length of s

D O(n + m), where n is length of s and m is length of t

💡 Hint

Consider how many times each pointer moves.

Each character in s is visited at most twice by the two pointers, and t is processed once, resulting in O(n + m) time.

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