Which identifier is an example of camel...

Transcript

00:01 Okay, so for this exercise we need to consider well the variables that are involved in the java language so the point is that the java language are some variables that have length between one and 65 ,535 okay, so really big strings of characters and these variables can have upper case letters lower case letters the dollar symbol the underscore symbol and the digits so we need to count how many possibilities for not for for for variables we have in java so the point here is that first we have all this possible symbols for the tractors of the of these strings so need to count all the possible outcomes right so all the possible characters that we can put on the string for the name of the variables.

01:04 So uppercase letters we have 26 possible letters, lowercase, again 26.

01:12 We have one symbol of dollar, one underscore symbol and we have 10 digits.

01:20 So in total if we sum all of them, all these possible, all these possibilities, we have 64 characters for a java string okay so let's focus on this part so we have 64 characters for our string and the the name of the variables have lent k and k is restricted from 1 to 65 ,500 if we want to count all the possibilities we need to count all the possibilities of different lengths of strings for the name of the variables that means that we need to involve a sum from k equals to 1 to 65 ,535 okay and this is only because this sum is inclusive in case that it is not inclusive we need to start from 2 and from up to 34 but in this case the the restriction the interval is inclusive so we need to consider the borders of the interval but what we need to sum well let's suppose that k is equal to two that means an string of two a two length string in each position of this string we have 64 possible characters that we can put there between the upper k letters lower case letters the dollar symbol the underscore and the digits okay but we need to consider one more thing and is that at the beginning okay and that's important the the first position cannot have any digits so in the first position no digit is allowed no digits so the first position is going to be we're going to have 54 possibilities okay between upper case letters lower case letters and the two symbols that we have that the dollar and the underscore and for the rest well we have the 64 possible characters to put on the string okay so this is more or less the structure of the strings let's see what happened when k is equal to three the length of the string is three so in that case we have here three positions and the first position we have 54 possible characters and for the rest we have 64 and 64 so you can see that we have a pattern that depends of course on the value of the length of the string first we have here the 54 for all these strings because all of them need to start with a character different from a digit but for the rest we have 64 possible characters and those are rise to the power of k minus 1 okay to the power of k minus 1 and you can verify here 3 minus 1 is 64 to the to square that count the two remaining places of the string and we can put any character of the one that that i listed before.

05:05 So to count all the possible strings we need to solve this sum here from k equals to 1 up to 65 ,535...

Question

Which identifier is an example of camel case? oneNumber _one_number _one_number one_number

Question

Which identifier is an example of camel case? oneNumber _one_number _one_number one_number

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