Minimum Window Substring in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main 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.

beginner

Which data structure is commonly used to track character frequencies in the Minimum Window Substring problem?

A hash map or array is used to count the frequency of characters needed and currently found in the sliding window.

intermediate

Explain the sliding window technique in the context of Minimum Window Substring.

It uses two pointers to create a window that expands to include required characters and contracts to find the smallest valid window containing all needed characters.

intermediate

Why do we need to check if the current window contains all required characters before trying to shrink it?

Because shrinking the window before it contains all required characters would lose necessary characters, making the substring invalid.

advanced

What is the time complexity of the optimal Minimum Window Substring solution using sliding window?

O(n + m), where n is the length of the main string and m is the length of the target string, since each character is visited at most twice.

In the Minimum Window Substring problem, what does the sliding window represent?

A A pointer to the start of the target string

B A fixed-size substring of the target string

C A substring of the main string that may contain all required characters

D The entire main string

The sliding window is a substring of the main string that is expanded and contracted to find the smallest substring containing all required characters.

Which of these is NOT a step in the sliding window approach for Minimum Window Substring?

A Sort the characters inside the window

B Shrink the window to remove unnecessary characters

C Expand the window to include required characters

D Check if the window contains all required characters

Sorting characters inside the window is not needed; the approach relies on frequency counts and pointers.

What data structure is best to keep track of how many required characters are still missing in the current window?

A Stack

B Hash map or array

C Queue

D Linked list

A hash map or array efficiently tracks character counts needed and found.

When shrinking the window, what condition must be true to continue shrinking?

A The window contains no required characters

B The window is empty

C The window size is zero

D The window still contains all required characters

Shrinking only continues while the window remains valid, meaning it still contains all required characters.

What is the main reason the Minimum Window Substring solution is efficient?

A Each character is visited at most twice using two pointers

B It sorts the string first

C It uses brute force to check all substrings

D It uses recursion

The two-pointer sliding window approach ensures each character is visited at most twice, leading to linear time complexity.

Describe the sliding window technique used to solve the Minimum Window Substring problem.

Think about how you move the start and end pointers to find the smallest valid substring.

Explain why frequency counting is important in the Minimum Window Substring problem.

Consider how you know if the current substring has all characters from the target string.

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