Substring

Problem Highlights

• 🔗 Leetcode Link: Substring
• 💡 Problem Difficulty: Easy
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Strings, Hash
• 🗒️ Similar Questions: Number of Matching Subsequences, Shortest Way to Form String

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Are both inputs only strings?
• Can elements of the string be both characters and numbers?
• Can the second string be larger than the first?
• What are the Time & Space considerations?
  • Best Case Time Complexity: O(n * m)
  • Best Case Space Complexity: O(1)

Run through a set of example cases:

HAPPY CASE
Input: laboratory, rat
Output: true

Input: cat, meow
Output: false

EDGE CASE
Input: "CATDOG", ""
Output: True

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Strings, common solution patterns include:

• Two pointer solutions (left and right pointer variables) or a pointer on each string
• Storing the characters of the string in a HashMap or a Set
• Traversing the string with a sliding window

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Identify each s character in string t in relative order

1. Identify each `s` character in string t in relative order

⚠️ Common Mistakes

• Some students may want to hash components of the strings, but this does not preserve order.
• We can solve this problem without any extra allocated space.

4: I-mplement

Implement the code to solve the algorithm.

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        # Identify each `s` character in string t in relative order
        for c in s:
            i = t.find(c)
            if i == -1:    
                return False
            else:   t = t[i+1:]
        return True

class Solution {
    public boolean isSubsequence(String s, String t) {
        // Identify each `s` character in string t in relative order
        int i = 0, j = 0;
        while(i < s.length() && j < t.length()){
            if(s.charAt(i) == t.charAt(j)) i++;
            j++;
        }
        return i == s.length();
    }
}

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(n*m), where n: length of first string, m: length of second string
• Space Complexity: O(1), where no data structures are needed in this problem