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// @saorav21994

// TC : O(length of t)

// SC : O(1)

class Solution {

public boolean isSubsequence(String s, String t) {

int ls = s.length();

int lt = t.length();

int i = 0, j = 0;

while (i < ls) {

char chs = s.charAt(i);

boolean match = false;

while (j < lt && chs != t.charAt(j)) {

j += 1;

}

if (j >= lt && i < ls)

return false;

j += 1;

i += 1;

}

return true;

}

class Solution {

// TC : O(len(s + T))

// SC : O(1)

public boolean isSubsequence(String s, String t) {

if(s.length() == 0){

return true;

}

int i = 0;

int j = 0;

while(j<t.length() && i<s.length()){

if(s.charAt(i) == t.charAt(j)){

i++;

}

j++;

}

return i == s.length();

}

// Author : @romitdutta10

// TC : O(len(s + t))

// SC : O(1)

// Problem : https://leetcode.com/problems/is-subsequence/

class Solution {

public boolean isSubsequence(String s, String t) {

int n = s.length();

if(n > t.length()) {

return false;

}

if(s == null || s.length() == 0) {

return true;

}

int index = 0;

for(char c : t.toCharArray()) {

if(c == s.charAt(index)) {

index++;

if(index == n) {

return true;

}

return false;

}

// Author : @romitdutta10

// TC : O(len(s + t))

// SC : O(1)

// Problem : https://leetcode.com/problems/is-subsequence/

// Another interesting approach but space and time complexity remains same

class Solution {

public boolean isSubsequence(String s, String t) {

for (char ch : s.toCharArray()) {

int find = t.indexOf(ch);

if (find == -1) {

return false;

} else {

t = t.substring(find + 1);

}

return true;

}

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