LeetCode 392: Determine if a String is a Subsequence Using the Two‑Pointer Technique

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Determine Subsequence

Problem (LeetCode 392) : Given strings s and t, determine whether s is a subsequence of t. A subsequence is formed by deleting zero or more characters without changing the order of the remaining characters.

Approach

Initialize two pointers i and j at the start of s and t respectively.

Traverse t with j. If s[i] equals t[j], increment i. In any case, increment j.

After the loop, if i has reached the end of s, s is a subsequence of t; otherwise it is not.

The time complexity of this method is O(n+m), where n and m are the lengths of s and t respectively.

Demonstration Process

The following images illustrate each step of the two‑pointer traversal for the first example (s = "abc", t = "ahbgdc").

Initial state: i points to the first character of s, j points to the first character of t.

Both characters match, move i and j.

Characters do not match, move only j.

Match again, move both pointers.

Continue until i reaches the end of s, confirming it is a subsequence.

Reference Code

C++

#include <iostream>
using namespace std;

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int i = 0; // s index
        int j = 0; // t index
        while (i < s.size() && j < t.size()) {
            if (s[i] == t[j]) {
                i++; // move s pointer on match
            }
            j++; // always move t pointer
        }
        return i == s.size(); // true if s fully matched
    }
};

int main() {
    Solution solution;
    string s = "abc";
    string t = "ahbgdc";
    cout << "Input: s = \"" << s << "\", t = \"" << t << "\"" << endl;
    cout << "Output: " << (solution.isSubsequence(s, t) ? "true" : "false") << endl;
    return 0;
}

Java

public class Solution {
    public boolean isSubsequence(String s, String t) {
        int i = 0; // s index
        int j = 0; // t index
        while (i < s.length() && j < t.length()) {
            if (s.charAt(i) == t.charAt(j)) {
                i++; // move s pointer on match
            }
            j++; // always move t pointer
        }
        return i == s.length(); // true if s fully matched
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        String s = "abc";
        String t = "ahbgdc";
        System.out.println("Input: s = \"" + s + "\", t = \"" + t + "\"");
        System.out.println("Output: " + (solution.isSubsequence(s, t) ? "true" : "false"));
    }
}

Python

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        i = 0  # s index
        j = 0  # t index
        # traverse t with two pointers
        while i < len(s) and j < len(t):
            if s[i] == t[j]:
                i += 1  # move s pointer on match
            j += 1  # always move t pointer
        return i == len(s)  # true if s fully matched

if __name__ == "__main__":
    solution = Solution()
    s = "abc"
    t = "ahbgdc"
    print(f"Input: s = \"{s}\", t = \"{t}\"")
    print("Output:", solution.isSubsequence(s, t))