The clock shows h hours (0 <= h <= 23), m minutes (0 <= m <= 59) and s seconds (0 <= s <= 59) after midnight. Your task is to write a function which returns the time since midnight in milliseconds.

Examples:

h = 0, m = 0, s = 0 -> res = 0
h = 0, m = 1, s = 1 -> res = 61000
h = 1, m = 0, s = 1 -> res = 3601000

public class Clock {
  public static int Past(int h, int m, int s) {
    // Your solution
  }
}

public class Clock {
  public static int Past(int h, int m, int s) {
    return ((h * 60 * 60) + (m * 60) + s) * 1000;
  }
}

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The first century spans from the year 1 up to and including the year 100, The second - from the year 101 up to and including the year 200, etc. Given a year, return the century it is in.

public class Solution {
  public static int century(int number) {
    // Your solution
  }
}

public class Solution {
  public static int century(int number) {
    return (int)(Math.ceil(number / 100.0)); 
  }
}

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Create a function that takes an integer as an argument and returns "Even" for even numbers or "Odd" for odd numbers.

public class EvenOrOdd {
  public static String even_or_odd(int number) {
    // Your solution
  }
}

public class EvenOrOdd {
  public static String even_or_odd(int number) {
    return number % 2 == 0 ? "Even" : "Odd";
  }
}

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Given an array of integers, find the one that appears an odd number of times. There will always be only one integer that appears an odd number of times.

public class FindOdd {
  public static int findIt(int[] a) {
    // Your solution
  }
}

import static java.util.Arrays.stream;

public class FindOdd {
  public static int findIt(int[] arr) {
    return stream(arr).reduce(0, (a, b) -> a ^ b);
  }
}

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Create a function that checks if a number n is divisible by two numbers x AND y. All inputs are positive, non-zero digits.

public class DivisibleNb {
  public static boolean isDivisible(long n, long x, long y) {
    // Your solution
  }
}

public class DivisibleNb {
  public static boolean isDivisible(long n, long x, long y) {
    return (n % x == 0) && (n % y == 0);
  }
}

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Nathan loves cycling. Because Nathan knows it is important to stay hydrated, he drinks 0.5 litres of water per hour of cycling. You get given the time in hours and you need to return the number of litres Nathan will drink, rounded to the smallest value.

public class KeepHydrated {
  public int Liters(double time) {
    // Your solution
  }
}

public class KeepHydrated {
  public int Liters(double time) {
    return (int)(time / 2);
  }
}

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If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Finish the solution so that it returns the sum of all the multiples of 3 or 5 below the number passed in.

Note: If the number is a multiple of both 3 and 5, only count it once. Also, if a number is negative, return 0.

public class Solution {
  public int solution(int number) {
    // Your solution
  }
}

public class Solution {
  public int solution(int number) {
    int sum = 0;
    for (int i = 3; i < number; i++) {
      if (i % 3 == 0 || i % 5 == 0) {
        sum += i;
      }
    }
    return sum;
  }
}

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This code does not execute properly. Try to figure out why.

public class Multiply {
  public static Double multiply(Double a, Double b) {
    return a * b
  }
}

public class Multiply {
  public static Double multiply(Double a, Double b) {
    return a * b;
  }
}

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Return the number (count) of vowels (a, e, i, o, u) in the given string. The input string will only consist of lower case letters and/or spaces.

public class Vowels {
  public static int getCount(String str) {
    // Your solution
  }
}

public class Vowels {
  public static int getCount(String str) {
    int vowelsCount = 0;
    String vowels = "aeiou";
    for (int i = 0; i < str.length(); i++) {
      if (vowels.contains(String.valueOf(str.charAt(i)))) {
        vowelsCount++;
      }
    }
    return vowelsCount;
  }
}

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