↳

using golang ---- package main import ( "fmt" ) func main() { input := [][]string{{"hotel_1234", "Sheraton", "Amsterdam"}, {"hotel_1000", "Sheraton", "Buenos Aires"}, {"hotel_1001", "Hilton", "Amsterdam"}, {"hotel_1002", "Royal Palace", "Bogota"}, {"hotel_1003", "Hilton", "Amsterdam"}, {"hotel_1004", "Sheraton", "Buenos Aires"}, {"hotel_1005", "Sheraton", "Buenos Aires"}, } confuse_cities := []string{} Cities := make(map[string]map[string]int) for _, line := range input { if _, ok := Cities[line[2]]; ok { if val, ok2 := Cities[line[2]][line[1]]; ok2 { Cities[line[2]][line[1]] = val + 1 if Cities[line[2]][line[1]] == 3 { confuse_cities = append(confuse_cities, line[2]) } } } else { Cities[line[2]] = map[string]int{} Cities[line[2]][line[1]] = 1 } } fmt.Println(confuse_cities) } Less

↳

from collections import Counter Input = [ ("hotel_1234", "Sheraton", "Amsterdam") , ("hotel_1000", "Sheraton", "Buenos Aires") , ("hotel_1001", "Hilton", "Amsterdam") , ("hotel_1002", "Royal Palace", "Bogota") , ("hotel_1003", "Hilton", "Amsterdam") , ("hotel_1004", "Sheraton", "Buenos Aires") , ("hotel_1005", "Sheraton", "Buenos Aires"), ("hotel_1232", "Sheraton", "Amsterdam"), ("hotel_123222", "Sheraton", "Amsterdam") ] confusing_cities = [] for i,v in Counter((elem[1],elem[2]) for elem in Input).items(): if v >= 3: confusing_cities.append(i[1]) print(confusing_cities) Less

↳

4 giga

↳

The number that will overflow a variable of type unsigned 32-bit int. const MAX_U32INT = 2^32 -1 Less

↳

P2P is the best solution. If P2P is not allowed. We can use broadcast, for example: server1-> server2, then we have 2 sources, server1->server3, server2-server4; then it will take about (time to transfer one copy)*log(10000)= 3*3.32192809489*(time to transfer one copy). For fault tolerance, redundancy or retry. Or deploy a distributed file system so that file can be accessed. Central server solution is like NFS, however, NFS's server could be bottleneck. Less

↳

We can take checksum of the file when it is correct and compare it with the checksum of the same file in future. If both checksum are found to be same then the file isn't corrupted otherwise it is corrupted. Less

↳

It's a question to check if you understand behind the scenes of a terminal when a command is issued. I would answer it in that way: when I type a command and press enter, the Shell has a command line parser that checks if the text entered is a available command in the PATH or not, if not found it checks the system libs to see if it can find the definition of it, if nothing is found, it throws error ": command not found". In the successful case, where the command is found it loads the binary from the defined path and a system call is executed accordingly to fetch the process data from the OS. Less

↳

see list of all processes

↳

context switch - 3

↳

context switching is costlier than RAM access. I think we all have read this in those operating system lectures. Less

↳

This is what I came up with, probably not the smartest way : #!/usr/bin/env python def sum_to_n(numbers, n): """ Pick the first two numbers in the list first, then add the third one. Return if the sum is equal to n. Then remove the first two, since you are done with them. Do the same thing as long as you have at least 3 elements in the list """ numbers = numbers.split(",") output = [] first = int(numbers[0]) second = int(numbers[1]) while len(numbers) > 2: for item in numbers[2:]: if first + second + int(item) == n: output.append((first, second, item)) numbers.pop(0) numbers.pop(1) first = int(numbers[0]) second = int(numbers[1]) return output numbers = raw_input("Enter a comma separated list of integers:") n = raw_input("Enter an integer:") print sum_to_n(numbers, int(n)) Less

↳

public static void printPairSums(int[] array, int sum) { Arrays.sort(array); // 2 pointeurs int first = 0; int last = array.length - 1; while (first < last) { int s = array[first] + array[last]; if (s == sum) { System.out.println("( "+array[first] + " ," + array[last]+ " )"); ++first; --last; } else { if (s < sum) ++first; else --last; } } } //Exemple public static void main(String[] args) { // TODO Auto-generated method stub int[] myIntArray = new int[]{1,2,3,-1,4,5,-2,7,1}; printPairSums (myIntArray, 3); } Less

↳

“acrecar” is a palindrome by that definition of the word, “racecar”, “dad”, “pop”, “poop”. These I feel like may be better examples. “454” would be an example of a palindrome that is a number. Less

↳

Count the occurrence of each character; if more than one char has odd number of occurrences, it’s false Less

↳

or the number 15.

↳

sigkill

↳

standard fizzbuzz algorithm. I answered in python.

↳

for i in range(1,101): a=i if(i%4==0): a='linked' if(i%6==0): a='in' if((i%4==0) and (i%6==0)): a="linkedin" print a Less