Is Subsequence LeetCode Solution

Problem – Is Subsequence

Given two strings s and t, return true if s is a subsequence of t, or false otherwise.

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. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:

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

Example 2:

Input: s = "axc", t = "ahbgdc"
Output: false

Constraints:

  • 0 <= s.length <= 100
  • 0 <= t.length <= 104
  • s and t consist only of lowercase English letters.

Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 109, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?

Is Subsequence LeetCode Solution in Python

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        if len(s) > len(t):return False
        if len(s) == 0:return True
        subsequence=0
        for i in range(0,len(t)):
            if subsequence <= len(s) -1:
                print(s[subsequence])
                if s[subsequence]==t[i]:

                    subsequence+=1
        return  subsequence == len(s)

Is Subsequence LeetCode Solution in C++

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int n = s.length(),m=t.length();
        int j = 0; 
    // For index of s (or subsequence
 
    // Traverse s and t, and
    // compare current character
    // of s with first unmatched char
    // of t, if matched
    // then move ahead in s
    for (int i = 0; i < m and j < n; i++)
        if (s[j] == t[i])
            j++;
 
    // If all characters of s were found in t
    return (j == n);
    }
};

Is Subsequence LeetCode Solution in Java

public class Solution {
    public boolean isSubsequence(String s, String t) {
        if (s.length() == 0) return true;
        int indexS = 0, indexT = 0;
        while (indexT < t.length()) {
            if (t.charAt(indexT) == s.charAt(indexS)) {
                indexS++;
                if (indexS == s.length()) return true;
            }
            indexT++;
        }
        return false;
    }
}
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