There is no other way to pass the Java coding tests other than practicing. I have failed Java coding tests due to lack of practice. How often do you get work with tree and graph data structures? Ho often do you work with algorithms?

#### 50+ Java coding practice problems Links:

__Can you write code in Java?__ | __Designing your classes & interfaces in Java__ | __Java Data Structures & Algorithms__ | __ Passing the unit tests__ | __What is wrong with this code?__

**Q1**. Can you write code to check if a given number is prime?

**A1**. A prime number is a number that is divisible only by **itself** and of course 1.

#### More facts about prime numbers

- 1 is not a prime number. Prime numbers start from 2
- 2 is the first and only
**even**prime number - all the other prime numbers apart from 2 are odd numbers starting from 3.
**E.g.**3, 5, 7

#### A naive solution

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public static boolean isPrimeNaive(int number) { if (number % 2 == 0 || number == 1) { return (number == 2); } //if the number is divisible by other than itself (i.e number) //it is not a prime for (int i = 3; i < number; i++) { if(number % i == 0){ return false; } } return true; } |

**Q**. What is wrong with the above solution?

**A**. It is not optimal. if the number is 97, you will end up dividing 97 by from 3 to 96. Goes through the loop 93 times.

#### A better solution

Two improvements can be made

**1)** Instead of i++, do **i+2**, to check for only odd numbers as 2 is the only even prime number. All the others are odd

**2)** Instead of i < number, do **i*i < number** because if you look at factors of 99, the factors are repeated **half way mark**

1 |
99 = 1 * 99 == 3 * 33 == 9 * 11 == 11 * 9 == 33 * 3 == 99 * 1 |

You can see repeated factors –> 9 * 11 and 11 * 9, 3 * 33 and 33 * 3, etc. So, the revised solution will take advantage of this finding. So, for 97, it will only loop through 4 times for numbers 3, 5, 7, 9. When it gets to 11, it exits the loop as 11 * 11 = 121 which is > 97.

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public static boolean isPrime(int number) { // even numbers stop here. // Only 2 is an even prime number & the first prime number // 1 is not a prime number if (number % 2 == 0 || number == 1) { return (number == 2); } // odd numbers from 3 to n get here // goes inside a loop only if i*i <= number // so, numbers 3, 5, 7 skip this for loop // 9, 11, 13, 15, etc get in as 3*3 = 9, 3*3 < 11, and so on for (int i = 3; i * i <= number; i += 2) { // divisible by other than itself if (number % i == 0){ return false; } } // if gets here, it is a prime return true; } |

Another more effective approach for isPrime(int number) is:

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private static boolean isPrime(int number) { // even numbers stop here. // Only 2 is an even prime number & the first prime number // 1 is not a prime number if (number % 2 == 0 || number == 1) { return (number == 2); } int i = 2; while (i < Math.sqrt(number)) { if (number % i == 0) { return false; } i++; } // if gets here, it is a prime return true; } |

**Q2**. Can you write a method that gives a list of prime numbers within a given range?

**A2**. The range will be supplied via “from” and “to”.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 |
import java.util.List; public class PrimeNumber { public static void main(String[] args) { System.out.println("Prime numbers=" + getPrimeNumbers(1, 200)); } public static List<Integer> getPrimeNumbers(int from, int to) { List<Integer> primeNumbers = new ArrayList<Integer>(); for (int number = from; number <= to; number++) { if(isPrime(number)){ primeNumbers.add(number); } } return primeNumbers; } private static boolean isPrime(int number) { // even numbers stop here. // Only 2 is an even prime number & the first prime number // 1 is not a prime number if (number % 2 == 0 || number == 1) { return (number == 2); } int count = 0; // odd numbers from 3 to n get here // goes inside a loop only if i*i <= number // so, numbers 3, 5, 7 skip this for loop // 9, 11, 13, 15, etc get in as 3*3 = 9, 3*3 < 11, and so on for (int i = 3; i * i <= number; i += 2) { System.out.println("i=" + i); count++; // divisible by other than itself if (number % i == 0){ return false; } } System.out.println("count=" + count); // if gets here, it is a prime return true; } } |

**Output:**

1 2 |
Prime numbers=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199] |

**Q3**. Can you list “powerful numbers” between a given range, where the definition of a powerful number is — A positive integer **m** which is for __every__” **p**” that divides “m”, “p*p” must also divide m, where “p” is a prime number

**The powerful numbers from 1 to 40 are:** 1, 4, 8, 9, 16, 25, 27, 32, 36, …

**12** is not a powerful number because: 3 divides 12 but 3*3=9 does not.

**18** is not a powerful number because: 2 divides 18, but 2*2=4 does not.

**1** is a powerful number because: neither p nor p*p divides it.

**A3**.

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package algorithms; import java.util.ArrayList; import java.util.List; public class PowerfulNumber { public static void main(String[] args) { System.out.println("powerfulNums= " + getPowerfulNumbers(1, 40)); } public static List<Integer> getPowerfulNumbers(int from, int to) { List<Integer> powerfulNums = new ArrayList<Integer>(); List<Integer> primeNumbers = getPrimeNumbers(from, to); for (int m = from; m <= to; m++) { boolean isPowerfulNumber = true; for (Integer p : primeNumbers) { // every p that divides m, p*p must also divide m. if(m % p == 0 && m % (p*p) != 0){ isPowerfulNumber = false; break; } } if(isPowerfulNumber){ powerfulNums.add(m); } } return powerfulNums; } private static List<Integer> getPrimeNumbers(int from, int to) { List<Integer> primeNumbers = new ArrayList<Integer>(); for (int number = from; number <= to; number++) { if(isPrime(number)){ primeNumbers.add(number); } } return primeNumbers; } private static boolean isPrime(int number) { if (number % 2 == 0 || number == 1) { return (number == 2); } for (int i = 3; i * i <= number; i += 2) { // divisible by other than itself if (number % i == 0){ return false; } } return true; } } |

**Output:**

1 |
powerfulNums= [1, 4, 8, 9, 16, 25, 27, 32, 36] |

#### 50+ Java coding practice problems Links:

__Can you write code in Java?__ | __Designing your classes & interfaces in Java__ | __Java Data Structures & Algorithms__ | __ Passing the unit tests__ | __What is wrong with this code?__

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