Is Subsequence | LeetCode 392 | Google Coding Interview Tutorial

[@TerribleWhiteboard]

Updated code that only uses one while loop:
let isSubsequence = function(s, t) {
if (s.length === 0) {
return true;
}
let pointer1 = 0;
let pointer2 = 0;
while (pointer1 < s.length && pointer2 < t.length) {
if (t.charAt(pointer2) === s.charAt(pointer1)) {
pointer1++;
}
pointer2++;
}
return pointer1 === s.length;
};
Also, if there are any videos you'd like me to make or if you have any ideas on how to optimize this solution, let me know!

[@sathishraj1]

If S is acb and T is abcd , S is not a subsequence of T right? But it will fail here right,?

[@TerribleWhiteboard]

Hm. Why would it fail?

[@sathishraj1]

@@TerribleWhiteboard My bad.. It will not fail..By the way , u are an awesome problem solver and tutor.. Thank you !

[@TerribleWhiteboard]

Thanks!

[@mileicampeonmundial]

@@TerribleWhiteboard but what if abc and abdxi you return true cuz pointer1 === s.length but it is not

[@followerOfChrist212-x5n]

Great Explanation...thanks buddy

[@TerribleWhiteboard]

Thanks!

[@jillfleming9742]

Thanks for the detail.

[@TerribleWhiteboard]

No problem!

[@lilycooper7672]

Great solution!

[@TerribleWhiteboard]

Thanks!

[@ajaywadhwa3398]

Thanks for your explanations .Really helped alot.

[@TerribleWhiteboard]

You're welcome!

[@BryantCabrera]

I find your videos so helpful, thank you!

[@dharmiknaik1772]

Thanks for the explanation.

[@TerribleWhiteboard]

You're welcome!

[@NareshKumar-dw9xp]

Thanks for all this . Easy and straight forward :)

[@TerribleWhiteboard]

You're welcome!

[@sumitnegi2708]

that nested while loop was unnecessary.

[@TerribleWhiteboard]

True. I updated the code in the description. Thanks.

[@valikonen]

Your example it is to complicated...
function isSub(t, str) {
if(t === '') {
return true;
}
if(t.length > str.length) {
return false;
}
for(let i = 0; i < t.length; i++) {
if(str.indexOf(t[i]) === -1) {
return false;
}
}
return true;
}

[@shadmanmartinpiyal4057]

var isSubsequence = function(s, t) {
let pS = 0, pT = 0;
while(!!s[pS] && !!t[pT]){
if(s[pS] === t[pT]){
pS++;
}
pT++;
}
return (pS - 1) === (s.length - 1)
};