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.
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beginner
Which data structure is commonly used to track character counts in the Minimum Window Substring problem?
A hash map or array is used to count characters needed and track characters found in the current window.
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intermediate
Explain the sliding window technique in the context of Minimum Window Substring.
Use two pointers to create a window that expands to include required characters and contracts to remove unnecessary ones, updating the minimum window found.
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intermediate
Why do we need to track the number of required characters matched in the current window?
To know when the current window contains all characters needed, so we can try to shrink it to find the smallest valid window.
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intermediate
What happens if the target string contains duplicate characters in Minimum Window Substring?
The window must include at least as many occurrences of each character as in the target string to be valid.
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What does the sliding window technique use to represent the current substring?
AA queue to hold characters
BA stack to store characters
CA single pointer moving through the string
DTwo pointers marking the start and end of the window
✗ Incorrect
The sliding window uses two pointers to mark the current substring's start and end positions.
In Minimum Window Substring, when do you try to shrink the window?
AWhen the window is empty
BWhen the window contains all required characters
CWhen the window has no required characters
DAt every step regardless of content
✗ Incorrect
Shrinking happens only when the window contains all required characters to find the smallest valid substring.
Which data structure helps track how many characters are still needed in the window?
AHash map or array counting required characters
BLinked list of characters
CStack of characters
DBinary tree of characters
✗ Incorrect
A hash map or array tracks counts of characters needed to know when the window is valid.
What is the time complexity of the optimal Minimum Window Substring solution using sliding window?
AO(n), where n is the length of the string
BO(n^2)
CO(log n)
DO(n!)
✗ Incorrect
The sliding window approach visits each character at most twice, resulting in O(n) time complexity.
If the target string is empty, what should the Minimum Window Substring function return?
ANull or None
BThe entire input string
CAn empty string
DThe first character of the input string
✗ Incorrect
If the target is empty, the smallest window containing all characters is an empty string.
Describe step-by-step how the sliding window technique finds the minimum window substring.
Think about expanding and contracting the window while checking character counts.
You got /6 concepts.
Explain why tracking character frequency is important in the Minimum Window Substring problem.
Focus on how character counts affect window validity.
You got /4 concepts.
Practice
(1/5)
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
Step 1: Understand the problem statement
The problem asks to find the smallest part of a string that includes all characters from another string.
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.
Final Answer:
Find the smallest substring in a string that contains all characters of another string -> Option A
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
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.
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.
Final Answer:
int freq[256] = {0}; -> Option A
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
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.
Step 2: Identify the minimum window substring
The substring "BANC" is the smallest substring containing all characters A, B, and C.
Final Answer:
BANC -> Option C
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
Step 1: Analyze freq array usage
The freq array is decremented and incremented but not initialized, so it contains garbage values causing incorrect logic.
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.
Final Answer:
freq array is not initialized before use -> Option D
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
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.
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.
Final Answer:
Use two frequency arrays: one for t counts and one for current window counts, then expand and contract window accordingly -> Option B
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
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