LeetCode 392: Is Subsequence — Python Solution

Solve LeetCode 392: Is Subsequence in Python with a subsequence pointer approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

What the problem is testing

Advance the target pointer only when the current source character matches it.

Algorithm

1. Advance the target pointer only when the current source character matches it.
2. Maintain this invariant: The matched prefix is the longest prefix of s that can be formed from the processed part of t.
3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def isSubsequence(self, s, t):
        i = 0
        for char in t:
            if i < len(s) and s[i] == char:
                i += 1
        return i == len(s)

Why this is correct

The proof follows the maintained state: The matched prefix is the longest prefix of s that can be formed from the processed part of t. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

O(n) time and O(1) extra space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.

Edge cases

An empty s always succeeds; repeated characters must preserve order.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.

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