392. Is Subsequence

At a Glance

Topic: two-pointers
Pattern: Two Pointers
Difficulty: Easy
Companies: Amazon, Google, Meta, Bloomberg, Microsoft
Frequency: High
LeetCode: 392

Problem (one-liner)

Given strings source and stream, return whether source can be obtained by deleting some (possibly zero) characters from stream without reordering the rest.

Recognition Cues

"Subsequence" vs substring
Scan longer string once while matching shorter
Greedy: match source left-to-right in stream

Diagram

At-a-glance flow (replace with problem-specific Mermaid as you refine this note). camelCase node IDs; no spaces in IDs.

Loading diagram…

Approaches

Brute force — DP LCS-style — O(m*n) time — overkill.
Better — two pointers — O(m+n) time / O(1) space — one pass.
Optimal — same two-pointer greedy — proven for subsequence check.

Optimal Solution

Go

func isSubsequence(source string, stream string) bool {
	if len(source) == 0 {
		return true
	}
	match := 0
	for index := 0; index < len(stream) && match < len(source); index++ {
		if stream[index] == source[match] {
			match++
		}
	}
	return match == len(source)
}

JavaScript

function isSubsequence(source, stream) {
	if (source.length === 0) {
		return true;
	}
	let matched = 0;
	for (let index = 0; index < stream.length && matched < source.length; index++) {
		if (stream[index] === source[matched]) {
			matched++;
		}
	}
	return matched === source.length;
}