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

Recall & Review

beginner

What is the goal of the Minimum Window Substring problem?

To find the smallest substring in a given string that contains all characters of another string, including duplicates.

beginner

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

A hash map (or dictionary) is used to store the frequency of characters needed and to track characters in the current window.

intermediate

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

It uses two pointers to represent a window in the string. The window expands to include needed characters and contracts to remove unnecessary ones, aiming to find the smallest valid window.

intermediate

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

Because shrinking the window too early might remove required characters, making the window invalid. We only shrink when the window still contains all needed characters.

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 source string and m is the length of the target string, because each character is visited at most twice.

What does the sliding window technique use to represent the current substring?

A Two pointers marking the start and end of the window

B A stack to store characters

C A queue to process characters

D A single pointer moving through the string

The sliding window uses two pointers to mark the current substring's start and end.

In Minimum Window Substring, what do we track with a hash map?

A Number of unique characters in the string

B Positions of vowels only

C Frequency of characters needed and found

D Length of the longest substring

We track the frequency of characters needed and how many have been found in the current window.

When do we try to shrink the window in the Minimum Window Substring problem?

A When the window contains all required characters

B When the window is empty

C When the window has no required characters

D At every step regardless of content

We shrink the window only when it contains all required characters to try to find a smaller valid window.

What is the main challenge in the Minimum Window Substring problem?

A Sorting the string alphabetically

B Finding the smallest substring containing all characters of another string

C Counting vowels in the string

D Reversing the string

The challenge is to find the smallest substring that contains all characters of the target string.

What is the best time complexity achievable for Minimum Window Substring using sliding window?

A O(m^2)

B O(n^2)

C O(n * m)

D O(n + m)

The sliding window approach visits each character at most twice, resulting in O(n + m) time complexity.

Describe the sliding window approach 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 we need to track character frequencies when finding the Minimum Window Substring.

Consider what happens if the target string has repeated characters.

Practice

1. What does the "Minimum Window Substring" problem ask you to find given two strings s and t?

easy

A. The number of times t appears as a substring in s

B. The longest substring in s that does not contain any character from t

C. The smallest substring in s that contains all characters of t

D. The substring of s that is lexicographically smallest

Solution

1. Step 1: Understand the problem goal

   The problem asks to find the smallest part of string s that includes every character from string t at least once.

2. Step 2: Compare options with the problem goal

   The smallest substring in s that contains all characters of t matches the problem exactly. Other options describe different tasks unrelated to the minimum window substring.

3. Final Answer:

   The smallest substring in s that contains all characters of t -> Option C

4. Quick Check:

   Minimum window substring = smallest substring containing all chars [OK]

Hint: Look for smallest substring covering all chars of t [OK]

Common Mistakes:

• Confusing minimum window with longest substring
• Thinking about counting occurrences instead of substring
• Ignoring the need to include all characters of t

2. Which of the following Go code snippets correctly initializes a map to count characters for the Minimum Window Substring problem?

easy

A. counts := map[int]string{}

B. counts := make(map[rune]int)

C. counts := []int{}

D. counts := make([]rune, 0)

Solution

1. Step 1: Identify the correct data structure for counting characters

   We need a map with keys as characters (rune) and values as counts (int), so map[rune]int is correct.

2. Step 2: Check each option's type

   counts := make(map[rune]int) creates a map from rune to int, perfect for counting. Options B, C, and D use wrong types or slices, not suitable for counting characters.

3. Final Answer:

   counts := make(map[rune]int) -> Option B

4. Quick Check:

   Use map[rune]int for char counts [OK]

Hint: Use map[rune]int to count characters in Go [OK]

Common Mistakes:

• Using slice instead of map for counts
• Using wrong key or value types in map
• Not initializing the map before use

3. What will be the output of the following Go code snippet that tries to find the minimum window substring?

package main
import "fmt"
func main() {
    s := "ADOBECODEBANC"
    t := "ABC"
    result := minWindow(s, t)
    fmt.Println(result)
}

func minWindow(s, t string) string {
    need := make(map[rune]int)
    for _, c := range t {
        need[c]++
    }
    have := make(map[rune]int)
    left, right := 0, 0
    count := 0
    minLen := len(s) + 1
    start := 0
    for right < len(s) {
        c := rune(s[right])
        if need[c] > 0 {
            have[c]++
            if have[c] <= need[c] {
                count++
            }
        }
        for count == len(t) {
            if right-left+1 < minLen {
                minLen = right - left + 1
                start = left
            }
            leftChar := rune(s[left])
            if need[leftChar] > 0 {
                have[leftChar]--
                if have[leftChar] < need[leftChar] {
                    count--
                }
            }
            left++
        }
        right++
    }
    if minLen > len(s) {
        return ""
    }
    return s[start : start+minLen]
}

medium

A. "BANC"

B. "ADOBEC"

C. "CODEBA"

D. ""

Solution

1. Step 1: Understand the algorithm logic

   The code uses two pointers to find the smallest substring in s containing all chars of t. It updates the minimum length and start index when a valid window is found.

2. Step 2: Trace the example input

   For s = "ADOBECODEBANC" and t = "ABC", the minimum window substring containing 'A', 'B', and 'C' is "BANC".

3. Final Answer:

   "BANC" -> Option A

4. Quick Check:

   Minimum window substring for example = "BANC" [OK]

Hint: Trace pointers to find smallest valid substring [OK]

Common Mistakes:

• Not updating minimum length correctly
• Counting characters incorrectly
• Returning empty string when valid window exists

4. Identify the bug in this Go function for Minimum Window Substring and select the fix:

func minWindow(s, t string) string {
    need := make(map[rune]int)
    for _, c := range t {
        need[c]++
    }
    have := make(map[rune]int)
    left, right := 0, 0
    count := 0
    minLen := len(s) + 1
    start := 0
    for right < len(s) {
        c := rune(s[right])
        if need[c] > 0 {
            have[c]++
            if have[c] == need[c] {
                count++
            }
        }
        for count == len(t) {
            if right-left+1 < minLen {
                minLen = right - left + 1
                start = left
            }
            leftChar := rune(s[left])
            if need[leftChar] > 0 {
                have[leftChar]--
                if have[leftChar] < need[leftChar] {
                    count--
                }
            }
            left++
        }
        right++
    }
    if minLen > len(s) {
        return ""
    }
    return s[start : start+minLen]
}

medium

A. Initialize count to 1 instead of 0

B. Remove the if need[c] > 0 check inside the loop

C. Use for right <= len(s) instead of right < len(s)

D. Change count == len(t) to count == len(need) in the inner loop condition

Solution

1. Step 1: Understand the role of count and need map

   The variable count tracks how many required characters are matched. The map need holds unique characters from t. Using len(t) counts total characters including duplicates, which is incorrect.

2. Step 2: Identify the correct condition for a valid window

   The condition should check if count equals the number of unique characters needed, which is len(need). Changing the condition fixes the bug.

3. Final Answer:

   Change count == len(t) to count == len(need) in the inner loop condition -> Option D

4. Quick Check:

   Use unique char count for window check [OK]

Hint: Match count with unique chars, not total chars [OK]

Common Mistakes:

• Confusing total chars with unique chars count
• Incorrect loop boundary causing index errors
• Ignoring character presence check before counting

5. Given strings s = "aabbccdd" and t = "abc", what is the minimum window substring that contains all characters of t at least once? Consider the following approach: - Use two pointers to slide over s - Count characters needed from t - Shrink window when all chars matched What is the correct minimum window substring?

hard

A. "abbc"

B. "abc"

C. "aabbc"

D. "bcca"

Solution

1. Step 1: Identify characters needed and their counts

   Characters needed are 'a', 'b', and 'c'. We want the smallest substring containing at least one of each.

2. Step 2: Find the smallest substring in s containing 'a', 'b', and 'c'

   Checking substrings: "abbc" (positions 1-4) contains 'a', 'b', 'c' and is smaller than "aabbc" or others. "abc" does not appear as continuous substring. "bcca" is not in order and does not appear as substring.

3. Final Answer:

   "abbc" -> Option A

4. Quick Check:

   Smallest substring with all chars = "abbc" [OK]

Hint: Slide window and shrink when all chars matched [OK]

Common Mistakes:

• Assuming substring must be exact order of t
• Choosing longer substrings unnecessarily
• Ignoring repeated characters in s