Minimum Window Substring in DSA Go - Time & Space Complexity

Time Complexity: Minimum Window Substring

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to find the smallest substring containing all characters grows as the input string gets longer.

How does the algorithm's work increase when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func minWindow(s string, t string) string {
    need := map[byte]int{}
    for i := 0; i < len(t); i++ {
        need[t[i]]++
    }
    missing := len(t)
    left := 0
    start, end := 0, len(s)+1
    for right := 0; right < len(s); right++ {
        if need[s[right]] > 0 {
            missing--
        }
        need[s[right]]--
        for missing == 0 {
            if right-left+1 < end-start {
                start, end = left, right+1
            }
            need[s[left]]++
            if need[s[left]] > 0 {
                missing++
            }
            left++
        }
    }
    if end <= len(s) {
        return s[start:end]
    }
    return ""
}
    

This code finds the smallest substring in s that contains all characters of t.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The outer loop moves the right pointer over the string s.
• Secondary operation: The inner loop moves the left pointer to shrink the window when all needed characters are found.
• How many times: Each character in s is visited at most twice: once by right and once by left.

How Execution Grows With Input

As the input string s grows, the pointers move through it mostly once each.

Input Size (n)	Approx. Operations
10	About 20 moves (each pointer moves through string)
100	About 200 moves
1000	About 2000 moves

Pattern observation: The total work grows roughly in a straight line with input size, doubling because of two pointers moving.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the size of the input string s and the target string t.

Common Mistake

[X] Wrong: "The inner loop makes this algorithm quadratic because it is nested inside the outer loop."

[OK] Correct: Each character is visited at most twice, so the total steps add up linearly, not multiplying to a square.

Interview Connect

Understanding this sliding window approach and its time complexity shows you can handle efficient substring problems, a common skill interviewers look for.

Self-Check

"What if the target string t contains many repeated characters? How would that affect the time complexity?"

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