Somtimes you gats cram answer. This answer dey very crammable.

#leetcode

I posted the wrong screenshot for the last message.

So i am posting the correct screenshot for correction.

I actually did yesterday's question yesterday but I was too tired to post here.

First, you create an array that represent each letter in english alphabet as digits. That is, an array of length 26 and give all a value of false.

Then for every char in allowed, mark their corresponding values as true based on their index. I mean, a will be index 0, b index 1, c index 2 etc.

So if allowed is 'ac', mark index 0 and c as true.

Then make a 2 level loop. The first loops through the array of words and the second loops through each char in each words.

Now check if the word is mark as false, if it is, increment a counter. The initial value of the counter is 0.

Lastly, return the difference between the length of words array and the counter.

That magically gives you the answer. ;D

#leetcode

Today's leetcode is an easy one and quite simple to be frank.

so you have 2 binary numbers, say 11011 and 10110. Loop through the two binaries and check if the last bits in the binaries are different, if they are, increment count. That's all.

continuation