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Leetcode #392: Is Subsequence

In this guide, we solve Leetcode #392 Is Subsequence in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given two strings s and t, return true if s is a subsequence of t, or false otherwise. A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters.

Quick Facts

  • Difficulty: Easy
  • Premium: No
  • Tags: Two Pointers, String, Dynamic Programming

Intuition

The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.

Moving one pointer at a time keeps the invariant intact and avoids nested loops.

Approach

Place pointers at the left and right ends and move them based on the comparison or target condition.

This yields a clean linear pass after any required sorting.

Steps:

  • Set left and right pointers.
  • Move a pointer based on the condition.
  • Update the best answer while scanning.

Example

Input: s = "abc", t = "ahbgdc" Output: true

Python Solution

class Solution: def isSubsequence(self, s: str, t: str) -> bool: i = j = 0 while i < len(s) and j < len(t): if s[i] == t[j]: i += 1 j += 1 return i == len(s)

Complexity

The time complexity is O(m+n)O(m + n)O(m+n), where mmm and nnn are the lengths of the strings sss and ttt respectively. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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