Minimum Window Substring in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

int left = [1];

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

A1

B0

C-1

Ds.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 skips the first character.

Using -1 causes index errors.

Setting left to s.length() is out of bounds.

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 window contains all required characters.

DSA Java

if (formed == required) [1]

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

A{

B[

C(

D;

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or brackets instead of braces causes syntax errors.

Missing braces leads to only one statement being conditionally executed.

The if statement requires curly braces to start the block of code to execute when condition is true.

3fill in blank

hard

Fix the error in updating the character count in the window.

DSA Java

windowCounts.put(c, windowCounts.getOrDefault(c, 0) [1] 1);

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

A-

B/

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using minus decreases count incorrectly.

Multiplying or dividing does not update count properly.

We add 1 to the current count of character c in the window.

4fill in blank

hard

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

DSA Java

if (right - left + 1 < [1]) {
    minLen = right - left + 1;
    [2] = left;
}

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

AminLen

BmaxLen

Cstart

Dend

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxLen instead of minLen reverses logic.

Updating end instead of start causes wrong window tracking.

We update minLen with the smaller window size and start with the current left index.

5fill in blank

hard

Fill all three blanks to correctly shrink the window and update counts.

DSA Java

char c = s.charAt([1]);
windowCounts.put(c, windowCounts.getOrDefault(c, 0) [2] 1);
if (windowCounts.get(c) < dictT.getOrDefault(c, 0)) {
    [3]--;
}

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

Aleft

B-

Cformed

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using right instead of left causes wrong character removal.

Using plus instead of minus causes incorrect count update.

Decreasing wrong variable instead of formed breaks logic.

We get the character at left, subtract 1 from its count, and decrease formed when the count is less than required.

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. Find the longest substring without repeating characters

C. Count the number of substrings that contain a specific character

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 about sorting instead of substring search
• Ignoring the requirement to include all characters

2. Which Java data structure is most suitable to keep track of character counts when solving the Minimum Window Substring problem?

easy

A. ArrayList<Character>

B. LinkedList<Character>

C. HashMap

D. Stack<Character>

Solution

1. Step 1: Identify the need for character frequency tracking

   We need to count how many times each character appears in the target string and current window.

2. Step 2: Choose a data structure that maps characters to counts

   HashMap allows storing character keys with their counts efficiently.

3. Final Answer:

   HashMap -> Option C

4. Quick Check:

   Use HashMap for char count tracking [OK]

Hint: Use HashMap to map chars to counts quickly [OK]

Common Mistakes:

• Using ArrayList which doesn't map keys to counts
• Using Stack or LinkedList which don't store counts
• Trying to use arrays without knowing character range

3. What will be the output of the following Java code snippet?

String s = "ADOBECODEBANC";
String t = "ABC";
// Assume minWindow(s, t) returns the minimum window substring
System.out.println(minWindow(s, t));

Given that minWindow is correctly implemented for the Minimum Window Substring problem.

medium

A. "BANC"

B. "ADOBEC"

C. "CODEBA"

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 shortest substring containing all characters A, B, and C.

3. Final Answer:

   "BANC" -> Option A

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 longer substrings that contain all chars
• Confusing order of characters in substring
• Assuming first occurrence is minimum window

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

public String minWindow(String s, String t) {
    Map map = new HashMap<>();
    for (char c : t.toCharArray()) {
        map.put(c, map.getOrDefault(c, 0) + 1);
    }
    int left = 0, right = 0, count = t.length();
    int minLen = Integer.MAX_VALUE, start = 0;
    while (right < s.length()) {
        char c = s.charAt(right);
        if (map.containsKey(c)) {
            map.put(c, map.get(c) - 1);
            if (map.get(c) >= 0) count--;
        }
        while (count == 0) {
            if (right - left + 1 < minLen) {
                minLen = right - left + 1;
                start = left;
            }
            char leftChar = s.charAt(left);
            if (map.containsKey(leftChar)) {
                map.put(leftChar, map.get(leftChar) + 1);
                if (map.get(leftChar) > 0) count++;
            }
            left++;
        }
        right++;
    }
    return minLen == Integer.MAX_VALUE ? "" : s.substring(start, start + minLen);
}

What is the main bug in this code?

medium

A. The map is not updated correctly when shrinking the window

B. The substring returned uses wrong indices

C. The while loop condition should be count > 0 instead of count == 0

D. The count variable is initialized incorrectly

Solution

1. Step 1: Analyze the substring return statement

   The substring is returned using s.substring(start, start + minLen). In Java, substring's end index is exclusive, so this is correct.

2. Step 2: Check map and count updates

   The map and count updates when expanding and shrinking the window are correct.

3. Step 3: Check loop conditions and initialization

   Count initialization and while loop conditions are correct.

4. Step 4: Identify the actual bug

   The main bug is that the code does not handle the case when the minimum window substring is not found, returning an empty string is correct, so no bug there. The code is logically correct.

5. Step 5: Re-examine options

   Options A, B, and C are incorrect. The substring returned uses wrong indices is also not a bug. Therefore, no bug exists in the code.

6. Step 6: Final Answer

   No bug found in the code snippet

Hint: Check carefully map and count updates and substring indices [OK]

Common Mistakes:

• Confusing loop conditions for count
• Incorrect substring indices
• Ignoring map updates when shrinking window

5. You want to find the minimum window substring of s = "aabbccdd" that contains all characters of t = "abc". Which of the following substrings is the correct minimum window substring?

hard

A. "abbc"

B. "bcca"

C. "abc"

D. "aabbcc"

Solution

1. Step 1: Identify all substrings containing 'a', 'b', and 'c'

   Check substrings of s that include at least one 'a', one 'b', and one 'c'.

2. Step 2: Find the smallest such substring

   "aabbcc" (positions 0 to 5) contains 'a', 'b', 'c' and is the smallest continuous substring containing all characters. "abbc" is not a continuous substring in s. "abc" does not appear as a continuous substring. "bcca" is not a substring in order.

3. Final Answer:

   "aabbcc" -> Option D

4. Quick Check:

   Minimum window substring = "aabbcc" [OK]

Hint: Check all substrings, pick shortest with all chars [OK]

Common Mistakes:

• Assuming "abc" must appear in order
• Choosing shorter substrings that are not continuous
• Ignoring repeated characters in s