Interview Classic 150 Questions P392 Is Subsequence

---

Timeline

timeline

2025-10-15

init

two pointers, dynamic programming

---

Problem:

P392 Is Subsequence

https://leetcode.cn/problems/is-subsequence/description/?envType=study-plan-v2&envId=top-interview-150

Two Pointers

Two pointers point to s and t respectively. If they match, both pointers move forward; if not, the pointer pointing to t moves forward. Note that empty strings need to be handled separately.

#include <string>
using std::string;

class Solution {
    public:
	bool isSubsequence(string s, string t)
	{
		// Determine if s is a subsequence of t
		int i = 0, j = 0;
		int s_len = s.size();
		int t_len = t.size();

		if (s_len == 0) {
			return true;
		}
		while (i < s_len && j < t_len) {
			if (s[i] == t[j]) {
				i++;
			}
			j++;
		}
		if (i == s_len && s[i - 1] == t[j - 1]) {
			return true;
		}
		return false;
	}
};

#include <stdio.h>
int main()
{
	Solution solution;
	string s = "abc";
	string t = "ahbgdc";
	printf("%d\n", solution.isSubsequence(s, t));
}

Dynamic Programming

Considering the previous two-pointer approach, we notice that we spend a lot of time finding the next matching character in t. Thus, we can preprocess for each position in t the first occurrence of each character from that position onward. We can use dynamic programming to implement the preprocessing:

Let $f[i][j]$ denote the first occurrence of character j starting from position i in string t.

During state transition: if the character at position i in t is j, then$f[i][j]=i$, otherwise j appears starting from position i+1 onwards, i.e.,$f[i][j]=f[i+1][j]$, therefore we need to construct dp from back to front

The state transition equation is:

$$f[i][j] = \begin{cases} i & t[i] = j \\ f[i+1][j] & t[i] \neq j \end{cases}$$

The boundary conditions are:

$$f[m][j] = m \quad \forall j \in \{0, 1, \dots, 25\}$$

where:

• $i$denotes the string$t$position in (starting from 0, 0 ≤ i ≤ m)
• $j$denotes the character index (usually ‘a’ corresponds to 0, ‘b’ to 1, …, ‘z’ to 25)
• $f[i][j] = m$denotes from position$i$onwards there is no character$j$

#include <vector>
#include <string>

using std::vector;
using std::string;

class Solution {
    public:
        bool isSubsequence(string s, string t)
        {
                int n = s.size(), m = t.size();
                // m+1 is to handle boundary cases
                vector<vector<int> > f(m + 1, vector<int>(26, 0));
                for (int i = 0; i < 26; i++) {
                        f[m][i] = m;
                }

                for (int i = m - 1; i >= 0; i--) {
                        for (int j = 0; j < 26; j++) {
                                if (t[i] == j + 'a')
                                        f[i][j] = i;
                                else
                                        f[i][j] = f[i + 1][j];
                        }
                }
                int add = 0;
                for (int i = 0; i < n; i++) {
                        if (f[add][s[i] - 'a'] == m) {
                                return false;
                        }
                        add = f[add][s[i] - 'a'] + 1;
                }
                return true;
        }
};