Minimum Window Substring in DSA C - Execution Trace

Concept Flow - Minimum Window Substring

Start

↓

Initialize frequency map of target chars

↓

Set two pointers: left=0, right=0

↓

Expand right pointer to include chars

↓

Update window counts and check if window valid

↓

While window valid: try to shrink from left

↓

Update minimum window if smaller

↓

Move left pointer to shrink window

↓

Repeat expand/shrink until right reaches end

↓

Return minimum window substring or empty if none

We move two pointers over the string to find the smallest substring containing all target chars, expanding right to include chars and shrinking left to minimize.

Execution Sample

DSA C

s = "ADOBECODEBANC";
t = "ABC";
// Find min window in s containing all chars of t

Find the smallest substring in s that contains all characters from t.

Execution Table

Step	Operation	Left Pointer	Right Pointer	Window Counts	Valid Count	Minimum Window	Visual State
1	Initialize frequency map for t	0	0	{"A":0,"B":0,"C":0}	0	null	""
2	Expand right to 'A'	0	0	{"A":1,"B":0,"C":0}	1	null	[A]
3	Expand right to 'D'	0	1	{"A":1,"B":0,"C":0}	1	null	[A D]
4	Expand right to 'O'	0	2	{"A":1,"B":0,"C":0}	1	null	[A D O]
5	Expand right to 'B'	0	3	{"A":1,"B":1,"C":0}	2	null	[A D O B]
6	Expand right to 'E'	0	4	{"A":1,"B":1,"C":0}	2	null	[A D O B E]
7	Expand right to 'C'	0	5	{"A":1,"B":1,"C":1}	3	Window 'ADOBEC'	[A D O B E C]
8	Shrink left from 'A'	1	5	{"A":0,"B":1,"C":1}	2	Window 'ADOBEC'	[D O B E C]
9	Expand right to 'O'	1	6	{"A":0,"B":1,"C":1}	2	Window 'ADOBEC'	[D O B E C O]
10	Expand right to 'D'	1	7	{"A":0,"B":1,"C":1}	2	Window 'ADOBEC'	[D O B E C O D]
11	Expand right to 'E'	1	8	{"A":0,"B":1,"C":1}	2	Window 'ADOBEC'	[D O B E C O D E]
12	Expand right to 'B'	1	9	{"A":0,"B":2,"C":1}	2	Window 'ADOBEC'	[D O B E C O D E B]
13	Expand right to 'A'	1	10	{"A":1,"B":2,"C":1}	3	Window 'ADOBEC'	[D O B E C O D E B A]
14	Shrink left from 'D'	2	10	{"A":1,"B":2,"C":1}	3	Window 'DOBECODEBA'	[O B E C O D E B A]
15	Shrink left from 'O'	3	10	{"A":1,"B":2,"C":1}	3	Window 'OBECODEBA'	[B E C O D E B A]
16	Shrink left from 'B'	4	10	{"A":1,"B":1,"C":1}	3	Window 'BECODEBA'	[E C O D E B A]
17	Shrink left from 'E'	5	10	{"A":1,"B":1,"C":1}	3	Window 'ECODEBA'	[C O D E B A]
18	Shrink left from 'C'	6	10	{"A":1,"B":1,"C":0}	2	Window 'ECODEBA'	[O D E B A]
19	Expand right to 'N'	6	11	{"A":1,"B":1,"C":0}	2	Window 'ECODEBA'	[O D E B A N]
20	Expand right to 'C'	6	12	{"A":1,"B":1,"C":1}	3	Window 'ECODEBANC'	[O D E B A N C]
21	Shrink left from 'O'	7	12	{"A":1,"B":1,"C":1}	3	Window 'CODEBANC'	[D E B A N C]
22	Shrink left from 'D'	8	12	{"A":1,"B":1,"C":1}	3	Window 'ODEBANC'	[E B A N C]
23	Shrink left from 'E'	9	12	{"A":1,"B":0,"C":1}	2	Window 'ODEBANC'	[B A N C]
24	End: right reached end, no smaller valid window	9	12	{"A":1,"B":0,"C":1}	2	Window 'BANC'	[B A N C]

💡 Right pointer reached end of string; minimum window found is 'BANC'

Variable Tracker

Variable	Start	After Step 7	After Step 13	After Step 20	Final
left	0	0	1	6	9
right	0	5	10	12	12
window_counts	{}	{"A":1,"B":1,"C":1}	{"A":1,"B":2,"C":1}	{"A":1,"B":1,"C":1}	{"A":1,"B":0,"C":1}
valid_count	0	3	3	3	2
min_window	null	"ADOBEC"	"ADOBEC"	"ECODEBANC"	"BANC"

Key Moments - 3 Insights

Why do we shrink the window from the left only when the window is valid?

Because shrinking the window when it is not valid would lose required characters. See steps 7 to 8 where shrinking starts only after valid_count reaches 3.

Why do we keep track of valid_count instead of just checking window_counts every time?

valid_count tracks how many required characters meet their needed frequency, making it efficient to know when the window is valid without checking all counts each time. See column 'Valid Count' in execution_table.

Why does the window_counts for 'B' increase beyond the needed count?

Because the window can have extra occurrences of characters. We only care if counts meet or exceed required counts. See step 13 where 'B' count is 2 but valid_count remains 3.

Visual Quiz - 3 Questions

Look at the execution table at step 7, what is the minimum window substring found?

A "ADOBEC"

B "DOBEC"

C "BANC"

D "CODEBA"

💡 Hint

Check the 'Minimum Window' column at step 7 in the execution_table.

At which step does the valid_count first reach 3?

A Step 5

B Step 7

C Step 13

D Step 20

💡 Hint

Look at the 'Valid Count' column in execution_table to find when it first becomes 3.

If we did not shrink the window from the left, what would happen to the minimum window substring?

A It would remain the same

B It would become larger or not minimal

C It would become empty

D It would contain fewer characters

💡 Hint

Refer to steps 7 to 17 where shrinking left reduces window size and updates minimum window.

Concept Snapshot

Minimum Window Substring:
- Use two pointers (left, right) to slide over string
- Expand right to include chars until window valid
- Shrink left to minimize window while valid
- Track counts of chars and valid matches
- Return smallest valid window or empty if none

Full Transcript

Minimum Window Substring finds the smallest substring in a string that contains all characters of another string. We use two pointers, left and right, to create a sliding window. We expand the right pointer to include characters and update counts. When the window contains all required characters, we try to shrink it from the left to find the smallest valid window. We track counts of characters and how many required characters are fully matched. The process continues until the right pointer reaches the end. The smallest valid window found is returned, or empty if none exists.

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