Why Minimum Window Substring in DSA Go?

The Big Idea

What if you could instantly find the tiniest piece of text that holds all your important clues?

The Scenario

Imagine you have a long paragraph and you want to find the shortest part that contains all the important keywords you need. Doing this by reading every possible part one by one is like searching for a needle in a haystack.

The Problem

Manually checking every possible substring to see if it contains all keywords is very slow and tiring. It wastes time and can easily miss the shortest part because there are so many combinations to check.

The Solution

The Minimum Window Substring method smartly slides through the paragraph, keeping track of keywords found, and quickly narrows down the smallest part that has all the keywords. It avoids unnecessary checks and finds the answer efficiently.

Before vs After

✗ Before

for i := 0; i < len(s); i++ {
  for j := i; j < len(s); j++ {
    if containsAllKeywords(s[i:j+1]) {
      updateMinWindow(i, j)
    }
  }
}

✓ After

left, right := 0, 0
for right < len(s) {
  addChar(s[right])
  for windowHasAllKeywords() {
    updateMinWindow(left, right)
    removeChar(s[left])
    left++
  }
  right++
}

What It Enables

This technique lets you quickly find the smallest section containing all needed items, making searches in big data or texts fast and practical.

Real Life Example

Finding the shortest snippet in a news article that mentions all the key people and places you want to know about.

Key Takeaways

Manual search is slow and error-prone.

Sliding window technique efficiently finds the minimum substring.

Enables fast and practical searching in large texts or data.

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