package com.fishercoder.solutions;

/**
 * 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 s is 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?

 */
public class _392 {

    public static class Solution1 {
        public boolean isSubsequence(String s, String t) {
            int left = 0;
            for (int i = 0; i < s.length(); i++) {
                boolean foundI = false;
                int j = left;
                for (; j < t.length(); j++) {
                    if (s.charAt(i) == t.charAt(j)) {
                        left = j + 1;
                        foundI = true;
                        break;
                    }
                }
                if (j == t.length() && !foundI) {
                    return false;
                }
            }
            return true;
        }
    }
}

package com.fishercoder.solutions;

public class _392 {

    public static class Solution1 {
        public boolean isSubsequence(String s, String t) {
            int left = 0;
            for (int i = 0; i < s.length(); i++) {
                boolean foundI = false;
                int j = left;
                for (; j < t.length(); j++) {
                    if (s.charAt(i) == t.charAt(j)) {
                        left = j + 1;
                        foundI = true;
                        break;
                    }
                }
                if (j == t.length() && !foundI) {
                    return false;
                }
            }
            return true;
        }
    }
}

refactor 392