Minimum Window Substring in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Rust

let mut left = [1];

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

A1

B0

C-1

Ds.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left pointer at 1 skips the first character.

Using negative index is invalid in Rust.

Setting left to string length causes out-of-bound errors.

The left pointer should start at 0 to begin scanning from the start of the string.

2fill in blank

medium

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

DSA Rust

window_counts.insert(c, window_counts.get(&c).unwrap_or(&0) + [1]);

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

A0

B2

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 0 does not update the count.

Adding -1 decreases the count incorrectly.

Adding 2 counts characters twice.

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

3fill in blank

hard

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

DSA Rust

if formed == required [1] {

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

A==

B<

C!=

D== false

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using < misses windows that exactly match requirements.

Using == false reverses the logic incorrectly.

We check if formed equals required to confirm the window has all needed characters.

4fill in blank

hard

Fill both blanks to correctly shrink the window from the left and update counts.

DSA Rust

let c = s.chars().nth(left).unwrap();
window_counts.insert(c, window_counts.get(&c).unwrap() [1] 1);
if window_counts.get(&c).unwrap() [2] target_counts.get(&c).unwrap_or(&0) [3] 0 {
    formed -= 1;
}

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

A-

B<

C>

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Adding instead of subtracting increases count incorrectly.

Using > instead of < reverses the condition logic.

Not updating formed causes incorrect window validation.

We subtract 1 from the count of c, then check if it is less than the required count to reduce formed.

5fill in blank

hard

Fill all three blanks to update the minimum window substring result correctly.

DSA Rust

if right - left + 1 < ans.[1] {
    ans.[2] = right - left + 1;
    ans.[3] = left;
}

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

A0

B1

C2

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing tuple indices causes wrong updates.

Using 'length' instead of numeric indices causes errors.

Not updating ans correctly leads to wrong final substring.

ans is a tuple where ans.0 stores the window length, ans.1 the start index, and ans.2 the end index.

Practice

1. What is the main goal of the Minimum Window Substring problem?

easy

A. Reverse the entire string

B. Sort a string in ascending order

C. Count the number of vowels in a string

D. Find the smallest substring in a string that contains all characters of another string

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: Identify the goal

   We want the minimum length substring that contains all required characters, not sorting or counting.

3. Final Answer:

   Find the smallest substring in a string that contains all characters of another string -> Option D

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 sorting or reversing strings
• Thinking it counts characters instead of finding substring
• Ignoring the requirement to include all characters

2. Which Rust data structure is most suitable to track character counts when solving Minimum Window Substring?

easy

A. Vec<char>

B. HashMap

C. LinkedList<char>

D. BinaryHeap<char>

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

   A HashMap allows storing counts for each character efficiently.

3. Final Answer:

   HashMap -> Option B

4. Quick Check:

   Use HashMap for char counts [OK]

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

Common Mistakes:

• Using Vec<char> which doesn't track counts
• Choosing LinkedList which is inefficient here
• Using BinaryHeap which is for priority, not counts

3. What is the output of the following Rust code snippet for Minimum Window Substring?

let s = "ADOBECODEBANC";
let t = "ABC";
let result = min_window(s, t);
println!("{}", result);

fn min_window(s: &str, t: &str) -> String {
    use std::collections::HashMap;
    let mut need = HashMap::new();
    for c in t.chars() {
        *need.entry(c).or_insert(0) += 1;
    }
    let (mut left, mut right, mut count) = (0, 0, 0);
    let mut min_len = usize::MAX;
    let mut start = 0;
    let mut window = HashMap::new();
    let s_chars: Vec<char> = s.chars().collect();
    while right < s_chars.len() {
        let c = s_chars[right];
        if need.contains_key(&c) {
            *window.entry(c).or_insert(0) += 1;
            if window[&c] == need[&c] {
                count += 1;
            }
        }
        while count == need.len() {
            if right - left + 1 < min_len {
                min_len = right - left + 1;
                start = left;
            }
            let left_char = s_chars[left];
            if need.contains_key(&left_char) {
                *window.entry(left_char).or_insert(0) -= 1;
                if window[&left_char] < need[&left_char] {
                    count -= 1;
                }
            }
            left += 1;
        }
        right += 1;
    }
    if min_len == usize::MAX {
        "".to_string()
    } else {
        s_chars[start..start + min_len].iter().collect()
    }
}

medium

A. "BANC"

B. "ADOBEC"

C. "CODEBA"

D. "ABC"

Solution

1. Step 1: Understand the input strings

   We want the smallest substring of "ADOBECODEBANC" containing all chars of "ABC".

2. Step 2: Trace the sliding window algorithm

   The code finds "BANC" as the minimum window containing 'A', 'B', and 'C'.

3. Final Answer:

   "BANC" -> Option A

4. Quick Check:

   Minimum window substring = "BANC" [OK]

Hint: Trace sliding window to find smallest valid substring [OK]

Common Mistakes:

• Choosing first found substring instead of minimum
• Ignoring character counts in window
• Confusing substring boundaries

4. Identify the bug in this Rust code snippet for Minimum Window Substring:

fn min_window(s: &str, t: &str) -> String {
    use std::collections::HashMap;
    let mut need = HashMap::new();
    for c in t.chars() {
        *need.entry(c).or_insert(0) += 1;
    }
    let (mut left, mut right, mut count) = (0, 0, 0);
    let mut min_len = usize::MAX;
    let mut start = 0;
    let mut window = HashMap::new();
    let s_chars: Vec<char> = s.chars().collect();
    while right <= s_chars.len() {
        let c = s_chars[right];
        if need.contains_key(&c) {
            *window.entry(c).or_insert(0) += 1;
            if window[&c] == need[&c] {
                count += 1;
            }
        }
        while count == need.len() {
            if right - left + 1 < min_len {
                min_len = right - left + 1;
                start = left;
            }
            let left_char = s_chars[left];
            if need.contains_key(&left_char) {
                *window.entry(left_char).or_insert(0) -= 1;
                if window[&left_char] < need[&left_char] {
                    count -= 1;
                }
            }
            left += 1;
        }
        right += 1;
    }
    if min_len == usize::MAX {
        "".to_string()
    } else {
        s_chars[start..start + min_len].iter().collect()
    }
}

medium

A. The function returns a &str instead of String

B. The count variable is never incremented

C. The loop condition should be right < s_chars.len() instead of <= to avoid out-of-bounds

D. The window HashMap is not initialized

Solution

1. Step 1: Check the while loop boundary

   The loop uses right <= s_chars.len(), which causes out-of-bounds access at s_chars[right].

2. Step 2: Correct the loop condition

   Changing it to right < s_chars.len() prevents accessing beyond the vector length.

3. Final Answer:

   The loop condition should be right < s_chars.len() instead of <= to avoid out-of-bounds -> Option C

4. Quick Check:

   Loop boundary must prevent out-of-bounds [OK]

Hint: Check loop limits to avoid indexing errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring vector length causing panic
• Not incrementing count correctly

5. Given the strings s = "aabbccdd" and t = "abc", what is the minimum window substring that contains all characters of t? Consider the Rust sliding window approach.

hard

A. "aabbc"

B. "abc"

C. "abbc"

D. "bcca"

Solution

1. Step 1: Identify required characters and counts

   String t requires 'a', 'b', and 'c' once each.

2. Step 2: Find smallest substring in s containing all

   "aabbc" (positions 0 to 4) contains 'a', 'a', 'b', 'b', 'c' and is the smallest window covering all required chars.

3. Final Answer:

   "aabbc" -> Option A

4. Quick Check:

   Minimum window substring = "aabbc" [OK]

Hint: Slide window to cover all chars, then shrink to minimum [OK]

Common Mistakes:

• Choosing substring missing one required character
• Picking longer substring than necessary
• Ignoring character counts in window