Minimum Window Substring in DSA Javascript - 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 the characters of another given string, including duplicates.

beginner

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

A hash map (or object in JavaScript) is used to store the frequency of characters needed and to track characters found 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 characters until all needed characters are found, then it contracts to find the smallest valid window.

beginner

Why do we need to check if the current window contains all characters of the target string?

Because only when the window contains all required characters can it be considered a valid candidate for the minimum window substring.

intermediate

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 hold 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 positions.

In Minimum Window Substring, what do we track in the hash map?

A Number of substrings

B Positions of vowels only

C Length of the string

D Frequency of characters needed and found

We track how many times each character is needed and how many times it appears 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 vowels

D When the window is larger than the string

We shrink the window to find the smallest valid substring once all required characters are included.

What is the main reason for using a sliding window instead of checking all substrings?

A To sort the string

B To use more memory

C To reduce time complexity from O(n^2) to O(n)

D To reverse the string

Sliding window helps avoid checking all substrings, reducing time complexity significantly.

What happens if the target string contains duplicate characters?

A Only one occurrence is needed

B The window must contain those duplicates as well

C Duplicates are ignored

D The problem cannot be solved

The window must include all duplicates to be valid.

Describe how the sliding window technique works to find the minimum window substring.

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

Explain why a hash map is essential in solving the Minimum Window Substring problem.

Consider how you know if the current substring contains all needed 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 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 exactly equal to t

Solution

1. Step 1: Understand the problem goal

   The problem asks to find the smallest part of string s that has all characters from string t, including duplicates.

2. Step 2: Compare options with the problem statement

   The smallest substring in s that contains all characters of t matches this goal exactly. Other options describe different problems.

3. Final Answer:

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

4. Quick Check:

   Minimum Window Substring = smallest substring with all chars [OK]

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

Common Mistakes:

• Confusing longest substring with smallest
• Ignoring duplicate characters in t
• Thinking it counts occurrences instead of substring

2. Which of the following JavaScript code snippets correctly initializes a frequency map for characters in string t?

easy

A. const freq = {}; for (const char of t) { freq[char] = (freq[char] || 0) + 1; }

B. const freq = new Map(); for (const char of t) { freq.set(char, freq.get(char) + 1); }

C. const freq = {}; for (let i = 0; i < t.length; i++) { freq[t[i]] = freq[t[i]] + 1; }

D. const freq = []; for (const char of t) { freq[char] += 1; }

Solution

1. Step 1: Check initialization and increment logic

   const freq = {}; for (const char of t) { freq[char] = (freq[char] || 0) + 1; } initializes an empty object and increments counts safely using freq[char] = (freq[char] || 0) + 1, which works even if freq[char] is undefined.

2. Step 2: Identify errors in other options

   const freq = new Map(); for (const char of t) { freq.set(char, freq.get(char) + 1); } tries to use Map but calls freq.get(char) + 1 without checking if freq.get(char) is undefined, causing NaN. const freq = {}; for (let i = 0; i < t.length; i++) { freq[t[i]] = freq[t[i]] + 1; } tries to add 1 to undefined without default. const freq = []; for (const char of t) { freq[char] += 1; } uses an array but accesses with string keys, which is invalid.

3. Final Answer:

   const freq = {}; for (const char of t) { freq[char] = (freq[char] || 0) + 1; } -> Option A

4. Quick Check:

   Safe frequency map init = const freq = {}; for (const char of t) { freq[char] = (freq[char] || 0) + 1; } [OK]

Hint: Use (freq[char] || 0) + 1 to safely count chars [OK]

Common Mistakes:

• Not initializing counts before adding
• Using Map without checking undefined values
• Using array with string keys

3. What will be the output of the following JavaScript code snippet?

const s = "ADOBECODEBANC";
const t = "ABC";

function minWindow(s, t) {
  const need = {};
  for (const c of t) need[c] = (need[c] || 0) + 1;
  let left = 0, right = 0, count = t.length;
  let minLen = Infinity, start = 0;

  while (right < s.length) {
    if (need[s[right]] > 0) count--;
    need[s[right]] = (need[s[right]] || 0) - 1;
    right++;

    while (count === 0) {
      if (right - left < minLen) {
        minLen = right - left;
        start = left;
      }
      need[s[left]]++;
      if (need[s[left]] > 0) count++;
      left++;
    }
  }
  return minLen === Infinity ? "" : s.substring(start, start + minLen);
}

console.log(minWindow(s, t));

medium

A. "DOBECODEBA"

B. "ADOBEC"

C. "CODEBA"

D. "BANC"

Solution

1. Step 1: Understand the sliding window logic

   The code uses two pointers expanding and shrinking a window over s. It tracks needed characters from t and shrinks window when all are found.

2. Step 2: Trace the window to find minimum substring

   For s = "ADOBECODEBANC" and t = "ABC", the smallest substring containing all chars is "BANC" starting at index 9 with length 4.

3. Final Answer:

   "BANC" -> Option D

4. Quick Check:

   Minimum window substring = "BANC" [OK]

Hint: Trace sliding window to find smallest valid substring [OK]

Common Mistakes:

• Confusing substring start and end indices
• Not updating minimum length correctly
• Ignoring character counts when shrinking window

4. Identify the bug in this JavaScript function for Minimum Window Substring:

function minWindow(s, t) {
  const need = {};
  for (const c of t) need[c] = (need[c] || 0) + 1;
  let left = 0, right = 0, count = t.length;
  let minLen = Infinity, start = 0;

  while (right < s.length) {
    if (need[s[right]] > 0) count--;
    need[s[right]]--;
    right++;

    while (count === 0) {
      if (right - left < minLen) {
        minLen = right - left;
        start = left;
      }
      need[s[left]]++;
      if (need[s[left]] > 0) count++;
      left++;
    }
  }
  return minLen === Infinity ? "" : s.substring(start, start + minLen);
}

medium

A. The count variable should be initialized to 0 instead of t.length

B. The decrement of need[s[right]] should use (need[s[right]] || 0) - 1 to avoid undefined

C. The inner while loop condition should be count > 0 instead of count === 0

D. The substring returned should use s.substring(start, right) instead of s.substring(start, start + minLen)

Solution

1. Step 1: Check how need map is updated

   The code does need[s[right]]-- without ensuring need[s[right]] is defined. If s[right] is not in t, need[s[right]] is undefined, causing NaN on decrement.

2. Step 2: Fix by defaulting undefined to 0 before decrement

   Use need[s[right]] = (need[s[right]] || 0) - 1 to avoid NaN and keep counts correct.

3. Final Answer:

   The decrement of need[s[right]] should use (need[s[right]] || 0) - 1 to avoid undefined -> Option B

4. Quick Check:

   Safe decrement with default 0 avoids NaN [OK]

Hint: Always default undefined to 0 before decrementing counts [OK]

Common Mistakes:

• Not handling characters not in t
• Wrong initialization of count variable
• Incorrect substring slicing

5. Given strings s = "aabbccdd" and t = "abc", what is the minimum window substring that contains all characters of t considering duplicates in t only once?

hard

A. "abbc"

B. "abc"

C. "aabbc"

D. "bcca"

Solution

1. Step 1: Understand character requirements from t

   String t = "abc" has characters 'a', 'b', 'c' each once. So the window must contain at least one 'a', one 'b', and one 'c'.

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

   In s = "aabbccdd", the substring "abc" starting at index 1 (characters at indices 1:'a', 2:'b', 3:'b', 4:'c') is not contiguous as "abc" is not directly present, but the smallest substring containing all characters once is actually "abc" starting at index 1 or 2. However, the substring "abc" is present starting at index 1 to 3 ("abb") is missing 'c', so the smallest substring containing all is "abc" at indices 1 to 3 is invalid. The substring "abbc" (indices 1 to 4) contains all characters and is minimal. But the substring "abc" (indices 1 to 3) is missing 'c'. The substring "abc" (indices 2 to 4) is "bbc" missing 'a'. The substring "abc" (indices 0 to 2) is "aab" missing 'c'. The substring "abc" (indices 3 to 5) is "bcc" missing 'a'. The substring "abc" (indices 1 to 4) is "abbc" which contains all. The substring "abc" (indices 1 to 3) is "abb" missing 'c'. The substring "abc" (indices 2 to 4) is "bbc" missing 'a'. The substring "abc" (indices 0 to 4) is "aabbc" which is longer. The substring "abc" (indices 1 to 4) "abbc" is the smallest substring containing all characters once. However, the option "abc" is present as "abc", but it is not a substring of s. So the correct minimal substring is "abbc".

3. Final Answer:

   "abbc" -> Option A

4. Quick Check:

   Minimum window with all chars once = "abbc" [OK]

Hint: Count chars in t once; find smallest substring covering them [OK]

Common Mistakes:

• Counting duplicates in t multiple times
• Choosing longer substrings unnecessarily
• Thinking it counts occurrences instead of substring