Is Subsequence

Here’s how to implement the isSubsequence function in Python using the two-pointer technique:

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class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        i, j = 0, 0  # Initialize two pointers at the beginning of s and t

        while i < len(t) and j < len(s):
            if t[i] == s[j]:  # If the characters match, move the pointer in s
                j += 1
            i += 1  # Always move the pointer in t

        return j == len(s)  # Check if we've found all characters of s in t

Key Concepts:

Two-pointer Technique: We use two pointers, i and j, to traverse the strings t and s, respectively.
Character Matching: We check if the characters at the pointers match. If they do, we move the pointer in s to the next character.
Subsequence Validation: We use the value of j after the loop to determine if s is a subsequence of t.

Problem Understanding

The problem asks you to check whether string s is a subsequence of string t. A subsequence is formed by deleting zero or more characters from t without changing the order of the remaining characters.

Initial Breakdown

You have two strings s and t.
You need to iterate over each string to compare their characters.
The objective is to check whether s can be obtained by deleting some characters from t.

Iterate over string t to find matching characters in s.
Track the index of the last found character in s.
If all characters in s are found in t in the same order, return true. Otherwise, return false.

Task Identification

String Traversal: Loop through each character in the string.
Character Matching: Compare characters in s and t.
Subsequence Check: Confirm whether s is a subsequence of t.

Task Abstraction

traverseString(string str): iter
matchChar(char a, char b): bool
isSubsequence(string s, string t): bool

Subproblem Interactions

You begin by traversing string t.
As you traverse t, you use matchChar to find matching characters in s.
isSubsequence takes the result of these operations to return the final answer.

Method Naming

traverseString: Iterates through each character of a string.
matchChar: Compares two characters for equality.
isSubsequence: Checks if s is a subsequence of t based on matchChar.

Algorithm

Initialize two pointers, i and j, to 0.
Loop over t using i.
- If s[j] matches t[i], increment j.
If j equals the length of s, return true. Otherwise, return false.

Time and Space Complexity

The time complexity of this algorithm is O(n), where n is the length of string t. This is because we only traverse t once. The space complexity is O(1) as we’re not using any additional data structures.

Key Takeaways

String traversal and character matching are the key operations.
Two-pointer technique efficiently checks for subsequences.
Always consider the simplest and most straightforward approach first. In this case, a single pass with two pointers suffices.

Define Problem

Problem: Determine if string s is a subsequence of string t.
Precondition: Two strings s and t are given, where s.length <= 100 and t.length <= 10^4. Both consist only of lowercase English letters.
Post-condition: Return true if s is a subsequence of t, otherwise false.

Invariant

The loop invariant in this problem is: At the start of each loop iteration, the first ( j ) characters of string ( s ) are a subsequence within the first ( i ) characters of string ( t ).

This invariant helps us keep track of how much of string ( s ) we have successfully found to be a subsequence within string ( t ) as we iterate through both strings. It provides a way to understand the progress and correctness of the algorithm.

Define Step

Compare the characters of s and t starting from the beginning, progressing through each string based on the match conditions.

Measure of Progress

The number of matched characters from s in t.

Define Loop Invariant

At the start of each loop iteration, j characters from s have been found in t.

Main Steps

Pseudocode for a single step within the loop:

IF t[i] == s[j]:
    j = j + 1
i = i + 1

Make Progress

In each iteration, we advance i by one, ensuring that we traverse t.
If we find a matching character, we also advance j, moving closer to confirming s as a subsequence of t.

Maintain Loop Invariant

The loop invariant is maintained by ensuring that j only increases when a match is found, ensuring correctness.

Establish the Loop Invariant

Pseudocode before entering the loop to establish the loop invariant:

i = 0
j = 0

Exit Condition

Exit the loop when i reaches the end of t or when j reaches the end of s.

Ending

The exit condition ensures that if j equals the length of s, all characters in s are in t, making it a subsequence.
Pseudocode after the loop ends and to return the required output:

IF j == LENGTH(s):
    RETURN true
ELSE:
    RETURN false

For the problem of checking if string s is a subsequence of string t, let’s establish a loop invariant and show its validity through Initialization, Maintenance, and Termination.

Consider a loop in which we traverse both s and t strings character by character. We’ll use indices i and j for t and s respectively.

Loop Invariant:

At the start of each loop iteration, the first j characters of s are a subsequence in the first i characters of t.

Initialization:

Prior to the first iteration of the loop, both i and j are initialized to 0. At this point, the first j=0 characters from s trivially form a subsequence in the first i=0 characters of t. This satisfies our loop invariant.

Maintenance:

To see that each iteration maintains the loop invariant, let’s consider what happens within an iteration. The loop compares t[i] with s[j]. If they are equal, j is incremented by 1. Regardless of the match, i is incremented by 1.

Here, we perform two actions that maintain the loop invariant:

If a match is found, j is incremented, extending the subsequence match from s in t.
i is incremented, extending the part of t in which we look for the subsequence.

The loop invariant is preserved.

Termination:

The loop can terminate for one of two reasons:

We reach the end of string t (i == len(t)), or
We find all characters of string s in t (j == len(s)).

At termination, the loop invariant tells us that the first j characters of s are a subsequence in the first i characters of t. If j == len(s), this implies that all of s is a subsequence of t, and we return true. Otherwise, we return false.

Thus, the loop invariant provides a useful property for showing the correctness of the algorithm when the loop terminates.

Identifying Problem Isomorphism

“Longest Common Subsequence” is a more complex variant of the problem “Is Subsequence”.

In “Is Subsequence”, you are given two strings, s and t, and the task is to check if s is a subsequence of t. A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

In “Longest Common Subsequence”, you are given two strings, text1 and text2, and asked to find the length of their longest common subsequence. A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

The concept in both problems is identifying a subsequence, but “Longest Common Subsequence” is more complex because it requires not only identifying a common subsequence but also optimizing it to be the longest possible. “Is Subsequence” is simpler as it only requires determining the existence of one string as a subsequence of the other.

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class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        for c in s:
            i = t.find(c)
            if i == -1:    
              return False
            else:   
              t = t[i+1:]

        return True

Brute-Force Solution

In a brute-force approach, you would start by iterating through each character in string t. For each character in t, you could generate all possible subsequences and check if s is among them.

Inefficiencies:

Time Complexity: Generating all subsequences has exponential time complexity, ( O(2^N) ), where ( N ) is the length of string t.
Space Complexity: You’d also need to store all these subsequences, which also takes exponential space.

Optimized Solution

Instead of generating all subsequences, you can iterate through both strings s and t at the same time and check whether each character in s appears in sequence in t.

Step 1: Initialize Pointers

Initialize two pointers, ( i ) and ( j ), to 0. Pointer ( i ) will iterate through string ( s ), and pointer ( j ) will iterate through string ( t ).

Step 2: Loop Through Strings

Iterate through both strings. For each character ( t[j] ):

If ( s[i] = t[j] ), then increment ( i ).
Increment ( j ).

Step 3: Check Pointer Value

If ( i ) becomes equal to the length of ( s ), then ( s ) is a subsequence of ( t ).

Time and Space Complexity:

Time Complexity: ( O(N) ), where ( N ) is the length of string ( t ). In the worst case, you’d iterate through all characters of ( t ).
Space Complexity: ( O(1) ), as you’re only using two pointers.

Comparison with Brute-Force

Time Complexity: Reduced from exponential to linear.
Space Complexity: Reduced from exponential to constant.

Loop Invariant in Optimized Solution

At the start of each loop iteration, the first ( i ) characters of ( s ) are a subsequence within the first ( j ) characters of ( t ).

Initialization: Initially, both ( i ) and ( j ) are 0, which means an empty string is a subsequence of another empty string, so the invariant holds.
Maintenance: During each loop, if ( s[i] = t[j] ), ( i ) is incremented. ( j ) is always incremented. This maintains the invariant because we’re only incrementing ( i ) when we find a matching character in ( t ).
Termination: When the loop terminates, if ( i ) has reached the length of ( s ), then ( s ) is indeed a subsequence of ( t ), proving the algorithm’s correctness.

By focusing on what’s essential (matching characters in order), we’ve significantly optimized the initial brute-force solution.

Problem Classification

String Manipulation: The problem is based on manipulating and comparing strings, which falls under the category of string manipulation.
Subsequence: The task involves determining whether one string is a subsequence of another. This places it in the realm of problems involving subsequences and sequence alignment.
Two-Pointers: Though it’s not explicitly stated, a common approach to solve such problems is using a two-pointer technique to traverse through both strings. This makes it a two-pointer problem as well.
Comparative Analysis: The problem requires comparing elements of two strings, so it can be classified under comparative analysis problems.

These categorizations frame how to think about the problem and what techniques might be relevant for finding a solution.

Language Agnostic Coding Drills

String Manipulation: Understand how to manipulate and parse strings. Learn the basic string operations like slicing, finding characters, and string traversal. In this problem, string traversal and locating a character in a string are key elements.
Subsequences: Understand what a subsequence is. It’s a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. Learn how to generate and check for subsequences.
Two-Pointers Technique: Get familiar with the two-pointer technique. In this problem, the two-pointer technique can be applied by keeping one pointer on the ’s’ string and another on the ’t’ string.
Conditional Statements: Understand how to use conditional statements to guide the program’s logic. In this problem, they are used to check if a character is found in the ’t’ string and decide what to do next.

Step by step problem-solving approach:

a. Start by looping over each character of the string ’s’. b. For each character, try to find its position in the string ’t’. c. If the character is found (i.e., the find function doesn’t return -1), move on to the next character in ’s’ and start the search from the position after the found character in ’t’. d. If the character is not found, it means that ’s’ is not a subsequence of ’t’, so return False. e. If all characters in ’s’ are found in order in ’t’, return True, meaning ’s’ is a subsequence of ’t'.

Targeted Drills in Python

String Manipulation:
Drill: Write a Python function that accepts a string and a character as parameters. The function should find the position of the character in the string and return it. If the character is not found, return -1.
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def find_char_position(string, char): return string.find(char) print(find_char_position("hello", "l")) # Output: 2 print(find_char_position("hello", "z")) # Output: -1

Subsequences:

Drill: For understanding purposes, write a Python function that accepts a string as a parameter and prints all subsequences of the string.

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def print_subsequences(string):
    from itertools import combinations
    for i in range(len(string)+1):
        for subset in combinations(string, i):
            print(''.join(subset))

print_subsequences("abc")  # Output: a, b, c, ab, ac, bc, abc

Two-Pointers Technique:

Drill: Write a Python function that accepts a string as a parameter. The function should print each character of the string using two-pointer technique.

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def two_pointers(string):
    left, right = 0, len(string) - 1
    while left <= right:
        print(string[left], string[right])
        left += 1
        right -= 1

two_pointers("hello")  # Output: h o, e l, l l

Conditional Statements:

Drill: Write a Python function that accepts two integers as parameters. The function should print “equal” if they are equal, “less” if the first number is less than the second one, and “greater” otherwise.

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def compare_numbers(a, b):
    if a == b:
        print("equal")
    elif a < b:
        print("less")
    else:
        print("greater")

compare_numbers(5, 10)  # Output: less
compare_numbers(10, 5)  # Output: greater
compare_numbers(5, 5)  # Output: equal

By understanding and implementing these drills, one can build up to the solution of the problem by integrating these concepts.

Here are 10 problems that use similar underlying concepts:

Two Sum: This problem also deals with iterating through an array to find specific elements. The essence is similar: traversing a data structure to find a condition that satisfies the problem requirements.
Longest Common Subsequence: Similar to finding if one string is a subsequence of another, this problem extends to finding the length of the longest common subsequence.
Implement strStr(): This problem requires you to find where a smaller string occurs in a larger string, which also involves iterating and checking for matching sequences.
Reverse String: Although simpler, this problem also requires iterating through a string or an array to perform specific checks or operations.
Find All Anagrams in a String: Here, you have to find all the start indices of a string’s anagrams in another string. You iterate through the longer string while maintaining a count of characters, similar to the subsequence problem.
Valid Anagram: This problem asks to determine if two strings are anagrams. It also involves string manipulation and condition checking, albeit in a more straightforward manner.
Minimum Window Substring: This is a more complex problem that also involves scanning through two strings to find a condition that satisfies the problem requirements.
Longest Substring Without Repeating Characters: This involves traversing a string to find a substring satisfying certain conditions. It’s about understanding the structure of the string and using pointers or indices smartly.
Valid Parentheses: While not about strings specifically, this problem involves iterating through a data structure (string or array) and maintaining a stack to check for conditions, similar to checking for subsequences.
Palindrome Linked List: Although this problem is about linked lists, the underlying concept of traversing a data structure to check for specific conditions (in this case, palindromicity) is similar.

Each of these problems involves iterating through data structures like strings or arrays and evaluating conditions at each step, similar to what we did for checking if a string is a subsequence of another string.