Minimum Window Substring in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

left = [1]

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

A1

B0

C-1

Dlen(s)

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 skips the first character.

Using -1 causes index errors.

Setting left to length of string is out of bounds.

The left pointer starts at index 0 to begin the sliding window from the start of the string.

2fill in blank

medium

Complete the code to update the count of characters in the current window.

DSA Python

window_counts[character] = window_counts.get(character, 0) [1] 1

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

A+

B-

C*

D//

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction decreases the count incorrectly.

Multiplication or floor division are not suitable for counting.

We add 1 to the count of the current character to track its frequency in the window.

3fill in blank

hard

Fix the error in the condition to check if the current window contains all required characters.

DSA Python

if formed [1] required:

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

A>

B!=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' causes the condition to fail when window is valid.

Using '<' or '>' does not correctly check completeness.

The window is valid only when the number of formed characters equals the required number.

4fill in blank

hard

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

DSA Python

if right - left + 1 < [1]:
    min_len = right - left + 1
    min_substr = s[[2]:right + 1]

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

Amin_len

Bleft

Cright

Dmax_len

Attempts:

3 left

💡 Hint

Common Mistakes

Using max_len instead of min_len causes wrong comparisons.

Using right as start index slices wrong substring.

We compare the current window size with the minimum length found so far and update if smaller. The substring is sliced from left to right+1.

5fill in blank

hard

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

DSA Python

character = s[[1]]
window_counts[character] -= 1
if window_counts[character] < dict_t.get(character, 0):
    formed -= [2]
left += [3]

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

Aleft

B1

Cright

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using right instead of left causes wrong character removal.

Increasing left by 0 does not shrink the window.

We get the character at the left pointer, decrease its count, check if formed decreases, and then move left pointer by 1 to shrink the window.