Minimum Window Substring in DSA Rust - Build from Scratch

Minimum Window Substring

📖 Scenario: Imagine you are working on a text search feature for a document editor. You want to find the smallest part of the document that contains all the letters of a given keyword.

🎯 Goal: Build a Rust program that finds the minimum window substring in a given string that contains all characters of another string.

📋 What You'll Learn

Create a string variable s with the value "ADOBECODEBANC"

Create a string variable t with the value "ABC"

Create a function min_window that takes s: &str and t: &str and returns a String

Use a sliding window approach with two pointers left and right to find the minimum window

Print the result of min_window(&s, &t)

💡 Why This Matters

🌍 Real World

Text search features in editors, search engines, and data filtering tools often need to find the smallest snippet containing all keywords.

💼 Career

Understanding sliding window algorithms and HashMaps is essential for software engineers working on search, text processing, and performance optimization.

1

Create the input strings

Create a string variable called s with the value "ADOBECODEBANC" and a string variable called t with the value "ABC".

DSA Rust

# Create the input strings
# Your code here

Hint

Use let s = String::from("ADOBECODEBANC") and let t = String::from("ABC").

2

Set up character frequency map for t

Create a mutable HashMap called need to count the frequency of each character in t. Use use std::collections::HashMap; at the top.

DSA Rust

fn main() {
    let s = String::from("ADOBECODEBANC");
    let t = String::from("ABC");

    use std::collections::HashMap;

    // Create a HashMap called need to count characters in t
    // Your code here
}

Hint

Use a for loop over t.chars() and update need with entry and or_insert(0).

3

Implement the minimum window substring logic

Create a function called min_window that takes s: &str and t: &str and returns a String. Use two pointers left and right, a HashMap called window to track characters in the current window, and variables valid, start, and length to track the minimum window. Implement the sliding window logic to find the smallest substring of s containing all characters of t. Return the substring or an empty string if none found.

DSA Rust

fn main() {
    let s = String::from("ADOBECODEBANC");
    let t = String::from("ABC");

    use std::collections::HashMap;

    let mut need = HashMap::new();
    for c in t.chars() {
        *need.entry(c).or_insert(0) += 1;
    }

    // Define min_window function below
    // Your code here
}

Hint

Use two pointers left and right to expand and shrink the window. Use window HashMap to count characters in the current window. Increase valid when a character count matches need. Update start and length when a smaller valid window is found.

4

Print the minimum window substring

Call the min_window function with &s and &t and print the result.

DSA Rust

fn main() {
    let s = String::from("ADOBECODEBANC");
    let t = String::from("ABC");

    use std::collections::HashMap;

    let mut need = HashMap::new();
    for c in t.chars() {
        *need.entry(c).or_insert(0) += 1;
    }

    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 window = HashMap::new();
        let (mut left, mut right) = (0, 0);
        let mut valid = 0;
        let mut start = 0;
        let mut length = usize::MAX;
        let s_chars: Vec = s.chars().collect();

        while right < s_chars.len() {
            let c = s_chars[right];
            right += 1;

            if need.contains_key(&c) {
                *window.entry(c).or_insert(0) += 1;
                if window[&c] == need[&c] {
                    valid += 1;
                }
            }

            while valid == need.len() {
                if right - left < length {
                    start = left;
                    length = right - left;
                }

                let d = s_chars[left];
                left += 1;

                if need.contains_key(&d) {
                    if window[&d] == need[&d] {
                        valid -= 1;
                    }
                    *window.entry(d).or_insert(0) -= 1;
                }
            }
        }

        if length == usize::MAX {
            String::new()
        } else {
            s_chars[start..start + length].iter().collect()
        }
    }

    // Print the minimum window substring
    // Your code here
}

Hint

Use println!("{}", min_window(&s, &t)); to print the result.

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