Developer Interview Questions

sorry u need to give input ;) ------ len(re.findall('[a-zA-Z]', "The quick brown fox jumps over the lazy dog"))

It depends on what has to be considered a word. For example, if we consider a word any string between spaces, we can write it in a more compact way: public static int countWords(String str){ if(str == null || str.isEmpty()) return 0; return str.split(" ").length; }

The question is how to count the characters in the string. If we can assume that the input is ASCII - always clarify first - then we know that the character range is 0-255. void count(char* str, int counts[255]) { if (str == 0) return; if (counts == 0) return; for (char *c = str; c != 0; c++) { counts[*c]++; } } First we assert the input is valid. Note that we take an additional parameter - an array of the count of each ASCII character. We loop through the string until we reach the null terminator, and we iterate a char pointer through each character in the string. For each iteration, we increment the counter in the array. Memory complexity is O(1), runtime complexity is O(n). If the input must be unicode, then we may consider alternatively using a hash table. void count(wchar_t* str, std::unordered_map& counts) { if (str == 0) return; for (char *c = str; c != 0; c++) { counts[*c]++; } } We still arrive at an O(1) memory complexity and O(n) runtime complexity (although it is worth noting that despite a hash table lookup takes O(1) like an array, the fixed cost of each lookup is higher for the hash table due to the hashing function, and in the worst case a lookup can be O(n)).

The question asks to count the characters, without using String.length() you can do this: public static Pair countLetters(String s) { //If the string is null or it is empty then it will have no character if (s == null || s.isEmpty()) { //So return the pair with 0 and 0 return new Pair(0, 0); } //If we should still run the loop. boolean run = true; //The count of characters int count = 0; //The count of characters without spaces int countWithoutSpace = 0; //While we are still running (we are at a valid index) while (run) { //Then try to try { //Get the character at the current count char c = s.charAt(count); //Add one to the count as we have a valid character count++; //If the character is not a space if (c != ' ') { //Then add one to the count without spaces. countWithoutSpace++; } //If we get a StringIndexOutOfBooundsException it means that the current count is outside the length of the string. } catch (StringIndexOutOfBoundsException e) { //So stop the loop. run = false; } } //And return the pair with the count and the count without spaces. return new Pair(count, countWithoutSpace); } This returns both the count with and without spaces and does not use a built in length function.

Wait..Am I missing something? string.Length will do the job?

Yeah i think string.length() will do the job.

failed! String.length returns the length of the String. Google asked you how many letters - in other words you cannot count the spaces. String. length returns number of unicode characters -and so includes spaces

I would do this: public static int countWords(String sentence){ String noSpace; //REMOVE SPACE noSpace = sentence.replaceAll(" ", ""); return noSpace.length }

anonymous is right, also wants to mention i=15 returns 15 and is true

bit lenghty but 1)both 3 & 5l are empty. 2) fill 3l, pour in 5l, fill 3l again and pour in 5l, you'll have 1 l in 3 l. now pour this 1l somewhere then fill 3l one more time.

Add 2 full measures using the 5l, remove 2 full measures with the 3l, you are left with 4l.

Very inappropriate indeed. If you are a female candidate, it is doubly inappropriate as it's rude to ask a lady her age.

Summation(i=0,floor(n/2))[(n-i)C(i)]. I'm sure this can be further simplified though.

Correction on the above: Summation(i=0,floor(n/2))[(n-i)P(i)].

Summation(i=0,floor(n/2))[(n-i)C(i)] is correct. I need to sleep. Sorry.

You should spin. Spinning resets probablity of live chamber to 1/3. Without spinning you are in a 1/2 chance of getting a bullet: 1 2 3 0 x x Can simplify it to a three chamber problem. Interviewer must have been on chamber 2 or 3 - so when it advances it will be on 3 or 4 - giving a 50:50 chance of having a bullet.

I'm not in the mood of doing maths-work, but it's obviusly an harmonic series (actually a p-series) and probability problem. As you don't know what's the chamber you start from, but you know that every time you pull the trigger your probability to live decreases. You have to choose wisely as you write your algorithm, then you can compare your convergence/divergence test with randomness and maybe even point out a game strategy (I may not roll the cylinder to force opponent to do so, or shooting and get killed). It must have been a very interesting way to see your reasoning process and coding skills.