Leetcode 392. Is Subsequence

Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and sis a short string (<=100).

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:
s = "abc", t = "ahbgdc"

Return true.

Example 2:
s = "axc", t = "ahbgdc"

Return false.

Follow up:
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

Solution:
The simplest solution is to use two pointers. If s.charAt(i) == t.charAt(j), move both of the pointers and check if i reach to the endo f s. Otherwise, just move the j pointer.


Code (Java):

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class Solution {     public boolean isSubsequence(String s, String t) {         if (s == null || s.length() == 0) {             return true;         }                   int i = 0;         for (int j = 0; j < t.length(); j++) {             if (s.charAt(i) == t.charAt(j)) {                 i++;                                   if (i == s.length()) {                     return true;                 }             }         }                   return false;     } }

Followup solution:

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class Solution {     public boolean isSubsequence(String s, String t) {         // step 1: save all the index for the t         //         Map<Character, List<Integer>> map = new HashMap<>();         for (int i = 0; i < t.length(); i++) {             char c = t.charAt(i);             if (!map.containsKey(c)) {                 List<Integer> pos = new ArrayList<>();                 pos.add(i);                 map.put(c, pos);             } else {                 List<Integer> pos = map.get(c);                 pos.add(i);                 map.put(c, pos);             }         }                   // step 2: for each char in s, find the first index         //         int prev = -1;         for (int i = 0; i < s.length(); i++) {             char c = s.charAt(i);             List<Integer> pos = map.get(c);             if (pos == null || pos.size() == 0) {                 return false;             }                           int curr = getNextSmall(pos, prev);                           if (curr == -1) {                 return false;             }                           prev = curr;         }                   return true;     }           // find next number greater than target     // if not found, return -1     //     private int getNextSmall(List<Integer> pos, int target) {         int lo = 0;         int hi = pos.size() - 1;                   while (lo + 1 <= hi) {             int mid = lo + (hi - lo) / 2;             if (pos.get(mid) == target) {                 lo = mid + 1;             } else if (pos.get(mid) > target) {                 hi = mid;             } else if (pos.get(mid) < target) {                 lo = mid + 1;             }         }                   if (pos.get(lo) > target) {             return pos.get(lo);         }                   if (pos.get(hi) > target) {             return pos.get(hi);         }                   return -1;     } }