Minimum Window Substring in DSA C - 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 of another string grows as the input size increases.

How does the algorithm's work change when the input strings get longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int minWindow(char* s, char* t) {
    int need[128] = {0}, window[128] = {0};
    int left = 0, right = 0, valid = 0;
    int start = 0, len = 0x3f3f3f3f;
    
    for (int i = 0; t[i]; i++) need[(unsigned char)t[i]]++;
    int total = 0;
    for (int i = 0; i < 128; i++) {
        if (need[i] > 0) total++;
    }
    
    while (s[right]) {
        char c = s[right];
        right++;
        if (need[(unsigned char)c] > 0) {
            window[(unsigned char)c]++;
            if (window[(unsigned char)c] == need[(unsigned char)c]) valid++;
        }
        while (valid == total) {
            if (right - left < len) {
                start = left;
                len = right - left;
            }
            char d = s[left];
            left++;
            if (need[(unsigned char)d] > 0) {
                if (window[(unsigned char)d] == need[(unsigned char)d]) valid--;
                window[(unsigned char)d]--;
            }
        }
    }
    return len == 0x3f3f3f3f ? 0 : len;
}

This code finds the smallest substring in s that contains all characters of t using a sliding window approach.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer while 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 window and once when shrinking it.

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 char 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 grows linearly with the size of both input strings, processing each character a limited number of times.

Common Mistake

[X] Wrong: "The nested while loops make this algorithm quadratic (O(n²)) because of the inner loop inside the outer loop."

[OK] Correct: Each character is visited at most twice, so the inner loop does not run fully for every outer loop iteration, keeping the total operations linear.

Interview Connect

Understanding this sliding window technique and its linear time complexity helps you solve many substring and array problems efficiently, a valuable skill in coding interviews.

Self-Check

"What if we changed the input to allow Unicode characters beyond ASCII? How would the time complexity change?"

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