remove ProjectEuler

DIRECTORY.md

@@ -220,17 +220,6 @@
   * [TwoPointers](https://github.com/TheAlgorithms/Java/blob/master/Others/TwoPointers.java)
   * [WorstFit](https://github.com/TheAlgorithms/Java/blob/master/Others/WorstFit.java)
 
-## ProjectEuler
-  * [Problem01](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem01.java)
-  * [Problem02](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem02.java)
-  * [Problem03](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem03.java)
-  * [Problem04](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem04.java)
-  * [Problem06](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem06.java)
-  * [Problem07](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem07.java)
-  * [Problem09](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem09.java)
-  * [Problem10](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem10.java)
-  * [Problem12](https://github.com/TheAlgorithms/Java/blob/master/ProjectEuler/Problem12.java)
-
 ## Searches
   * [BinarySearch](https://github.com/TheAlgorithms/Java/blob/master/Searches/BinarySearch.java)
   * [InterpolationSearch](https://github.com/TheAlgorithms/Java/blob/master/Searches/InterpolationSearch.java)

ProjectEuler/Problem01.java

@@ -1,51 +0,0 @@
-package ProjectEuler;
-
-/**
- * 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.
- *
- * <p>Find the sum of all the multiples of 3 or 5 below 1000.
- *
- * <p>Link: https://projecteuler.net/problem=1
- */
-public class Problem01 {
-  public static void main(String[] args) {
-    int[][] testNumber = {
-      {3, 0},
-      {4, 3},
-      {10, 23},
-      {1000, 233168},
-      {-1, 0}
-    };
-
-    for (int[] ints : testNumber) {
-      assert solution1(ints[0]) == ints[1];
-      assert solution2(ints[0]) == ints[1];
-    }
-  }
-
-  private static int solution1(int n) {
-    int sum = 0;
-    for (int i = 3; i < n; ++i) {
-      if (i % 3 == 0 || i % 5 == 0) {
-        sum += i;
-      }
-    }
-    return sum;
-  }
-
-  private static int solution2(int n) {
-    int sum = 0;
-
-    int terms = (n - 1) / 3;
-    sum += terms * (6 + (terms - 1) * 3) / 2;
-
-    terms = (n - 1) / 5;
-    sum += terms * (10 + (terms - 1) * 5) / 2;
-
-    terms = (n - 1) / 15;
-    sum -= terms * (30 + (terms - 1) * 15) / 2;
-
-    return sum;
-  }
-}

ProjectEuler/Problem02.java

@@ -1,43 +0,0 @@
-package ProjectEuler;
-
-/**
- * Each new term in the Fibonacci sequence is generated by adding the previous two terms. By
- * starting with 1 and 2, the first 10 terms will be:
- *
- * <p>1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
- *
- * <p>By considering the terms in the Fibonacci sequence whose values do not exceed four million,
- * find the sum of the even-valued terms.
- *
- * <p>Link: https://projecteuler.net/problem=2
- */
-public class Problem02 {
-  public static void main(String[] args) {
-    int[][] testNumbers = {
-      {10, 10}, /* 2 + 8 == 10 */
-      {15, 10}, /* 2 + 8 == 10 */
-      {2, 2},
-      {1, 0},
-      {89, 44} /* 2 + 8 + 34 == 44 */
-    };
-
-    for (int[] ints : testNumbers) {
-      assert solution1(ints[0]) == ints[1];
-    }
-  }
-
-  private static int solution1(int n) {
-    int sum = 0;
-    int first = 1;
-    int second = 2;
-    while (second <= n) {
-      if (second % 2 == 0) {
-        sum += second;
-      }
-      int temp = first + second;
-      first = second;
-      second = temp;
-    }
-    return sum;
-  }
-}

ProjectEuler/Problem03.java

@@ -1,62 +0,0 @@
-package ProjectEuler;
-
-/**
- * The prime factors of 13195 are 5, 7, 13 and 29.
- *
- * <p>What is the largest prime factor of the number 600851475143 ?
- *
- * <p>Link: https://projecteuler.net/problem=3
- */
-public class Problem03 {
-
-  /**
-   * Checks if a number is prime or not
-   *
-   * @param n the number
-   * @return {@code true} if {@code n} is prime
-   */
-  public static boolean isPrime(int n) {
-    if (n == 2) {
-      return true;
-    }
-    if (n < 2 || n % 2 == 0) {
-      return false;
-    }
-    for (int i = 3, limit = (int) Math.sqrt(n); i <= limit; i += 2) {
-      if (n % i == 0) {
-        return false;
-      }
-    }
-    return true;
-  }
-
-  /**
-   * Calculate all the prime factors of a number and return the largest
-   *
-   * @param n integer number
-   * @return the maximum prime factor of the given number
-   */
-  static long maxPrimeFactor(long n) {
-    for (int i = 2; i < n / 2; i++) {
-      if (isPrime(i))
-        while (n % i == 0) {
-          n /= i;
-        }
-    }
-    return n;
-  }
-
-  public static void main(String[] args) {
-    int[][] testNumbers = {
-      {87, 29},
-      {10, 5},
-      {21, 7},
-      {657, 73},
-      {777, 37}
-    };
-    for (int[] num : testNumbers) {
-      assert Problem03.maxPrimeFactor(num[0]) == num[1];
-    }
-    assert Problem03.maxPrimeFactor(600851475143L) == 6857;
-  }
-}

ProjectEuler/Problem04.java

@@ -1,41 +0,0 @@
-package ProjectEuler;
-
-/**
- * A palindromic number reads the same both ways. The largest palindrome made from the product of
- * two 2-digit numbers is 9009 = 91 × 99.
- *
- * <p>Find the largest palindrome made from the product of two 3-digit numbers.
- *
- * <p>link: https://projecteuler.net/problem=4
- */
-public class Problem04 {
-  public static void main(String[] args) {
-
-    assert solution1(10000) == -1;
-    assert solution1(20000) == 19591; /* 19591 == 143*137 */
-    assert solution1(30000) == 29992; /* 29992 == 184*163 */
-    assert solution1(40000) == 39893; /* 39893 == 287*139 */
-    assert solution1(50000) == 49894; /* 49894 == 494*101 */
-    assert solution1(60000) == 59995; /* 59995 == 355*169 */
-    assert solution1(70000) == 69996; /* 69996 == 614*114 */
-    assert solution1(80000) == 79897; /* 79897 == 733*109 */
-    assert solution1(90000) == 89798; /* 89798 == 761*118 */
-    assert solution1(100000) == 99999; /* 100000 == 813*123 */
-  }
-
-  private static int solution1(int n) {
-    for (int i = n - 1; i >= 10000; --i) {
-      String strNumber = String.valueOf(i);
-
-      /* Test if strNumber is palindrome */
-      if (new StringBuilder(strNumber).reverse().toString().equals(strNumber)) {
-        for (int divisor = 999; divisor >= 100; --divisor) {
-          if (i % divisor == 0 && String.valueOf(i / divisor).length() == 3) {
-            return i;
-          }
-        }
-      }
-    }
-    return -1; /* not found */
-  }
-}

ProjectEuler/Problem06.java

@@ -1,42 +0,0 @@
-package ProjectEuler;
-
-/**
- * The sum of the squares of the first ten natural numbers is, 1^2 + 2^2 + ... + 10^2 = 385 The
- * square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)^2 = 552 = 3025 Hence
- * the difference between the sum of the squares of the first ten natural numbers and the square of
- * the sum is 3025 − 385 = 2640. Find the difference between the sum of the squares of the first N
- * natural numbers and the square of the sum.
- *
- * <p>link: https://projecteuler.net/problem=6
- */
-public class Problem06 {
-  public static void main(String[] args) {
-    int[][] testNumbers = {
-      {10, 2640},
-      {15, 13160},
-      {20, 41230},
-      {50, 1582700}
-    };
-
-    for (int[] testNumber : testNumbers) {
-      assert solution1(testNumber[0]) == testNumber[1]
-          && solutions2(testNumber[0]) == testNumber[1];
-    }
-  }
-
-  private static int solution1(int n) {
-    int sum1 = 0;
-    int sum2 = 0;
-    for (int i = 1; i <= n; ++i) {
-      sum1 += i * i;
-      sum2 += i;
-    }
-    return sum2 * sum2 - sum1;
-  }
-
-  private static int solutions2(int n) {
-    int sumOfSquares = n * (n + 1) * (2 * n + 1) / 6;
-    int squareOfSum = (int) Math.pow((n * (n + 1) / 2.0), 2);
-    return squareOfSum - sumOfSquares;
-  }
-}

ProjectEuler/Problem07.java

@@ -1,60 +0,0 @@
-package ProjectEuler;
-
-/**
- * By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is
- * 13.
- *
- * <p>What is the 10 001st prime number?
- *
- * <p>link: https://projecteuler.net/problem=7
- */
-public class Problem07 {
-  public static void main(String[] args) {
-    int[][] testNumbers = {
-      {1, 2},
-      {2, 3},
-      {3, 5},
-      {4, 7},
-      {5, 11},
-      {6, 13},
-      {20, 71},
-      {50, 229},
-      {100, 541}
-    };
-    for (int[] number : testNumbers) {
-      assert solution1(number[0]) == number[1];
-    }
-  }
-
-  /***
-   * Checks if a number is prime or not
-   * @param number the number
-   * @return {@code true} if {@code number} is prime
-   */
-  private static boolean isPrime(int number) {
-    if (number == 2) {
-      return true;
-    }
-    if (number < 2 || number % 2 == 0) {
-      return false;
-    }
-    for (int i = 3, limit = (int) Math.sqrt(number); i <= limit; i += 2) {
-      if (number % i == 0) {
-        return false;
-      }
-    }
-    return true;
-  }
-
-  private static int solution1(int n) {
-    int count = 0;
-    int number = 1;
-
-    while (count != n) {
-      if (isPrime(++number)) {
-        count++;
-      }
-    }
-    return number;
-  }
-}

ProjectEuler/Problem09.java

@@ -1,28 +0,0 @@
-package ProjectEuler;
-
-/**
- * A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
- *
- * <p>a^2 + b^2 = c^2 For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
- *
- * <p>There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
- *
- * <p>link: https://projecteuler.net/problem=9
- */
-public class Problem09 {
-  public static void main(String[] args) {
-    assert solution1() == 31875000;
-  }
-
-  private static int solution1() {
-    for (int i = 0; i <= 300; ++i) {
-      for (int j = 0; j <= 400; ++j) {
-        int k = 1000 - i - j;
-        if (i * i + j * j == k * k) {
-          return i * j * k;
-        }
-      }
-    }
-    return -1; /* should not happen */
-  }
-}

ProjectEuler/Problem10.java

@@ -1,55 +0,0 @@
-package ProjectEuler;
-
-/**
- * The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
- *
- * <p>Find the sum of all the primes below two million.
- *
- * <p>link: https://projecteuler.net/problem=10
- */
-public class Problem10 {
-  public static void main(String[] args) {
-    long[][] testNumbers = {
-      {2000000, 142913828922L},
-      {10000, 5736396},
-      {5000, 1548136},
-      {1000, 76127},
-      {10, 17},
-      {7, 10}
-    };
-
-    for (long[] testNumber : testNumbers) {
-      assert solution1(testNumber[0]) == testNumber[1];
-    }
-  }
-
-  /***
-   * Checks if a number is prime or not
-   * @param n the number
-   * @return {@code true} if {@code n} is prime
-   */
-  private static boolean isPrime(int n) {
-    if (n == 2) {
-      return true;
-    }
-    if (n < 2 || n % 2 == 0) {
-      return false;
-    }
-    for (int i = 3, limit = (int) Math.sqrt(n); i <= limit; i += 2) {
-      if (n % i == 0) {
-        return false;
-      }
-    }
-    return true;
-  }
-
-  private static long solution1(long n) {
-    long sum = 0;
-    for (int i = 2; i < n; ++i) {
-      if (isPrime(i)) {
-        sum += i;
-      }
-    }
-    return sum;
-  }
-}

ProjectEuler/Problem12.java

@@ -1,54 +0,0 @@
-package ProjectEuler;
-/**
- * The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle
- * number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
- *
- * <p>1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
- *
- * <p>Let us list the factors of the first seven triangle numbers:
- *
- * <p>1: 1 3: 1,3 6: 1,2,3,6 10: 1,2,5,10 15: 1,3,5,15 21: 1,3,7,21 28: 1,2,4,7,14,28 We can see
- * that 28 is the first triangle number to have over five divisors.
- *
- * <p>What is the value of the first triangle number to have over five hundred divisors?
- *
- * <p>link: https://projecteuler.net/problem=12
- */
-public class Problem12 {
-
-  /** Driver Code */
-  public static void main(String[] args) {
-    assert solution1(500) == 76576500;
-  }
-
-  /* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
-  public static int triangleNumber(int n) {
-    int sum = 0;
-    for (int i = 0; i <= n; i++) sum += i;
-    return sum;
-  }
-
-  public static int solution1(int number) {
-    int j = 0; // j represents the jth triangle number
-    int n = 0; // n represents the triangle number corresponding to j
-    int numberOfDivisors = 0; // number of divisors for triangle number n
-
-    while (numberOfDivisors <= number) {
-
-      // resets numberOfDivisors because it's now checking a new triangle number
-      // and also sets n to be the next triangle number
-      numberOfDivisors = 0;
-      j++;
-      n = triangleNumber(j);
-
-      // for every number from 1 to the square root of this triangle number,
-      // count the number of divisors
-      for (int i = 1; i <= Math.sqrt(n); i++) if (n % i == 0) numberOfDivisors++;
-
-      // 1 to the square root of the number holds exactly half of the divisors
-      // so multiply it by 2 to include the other corresponding half
-      numberOfDivisors *= 2;
-    }
-    return n;
-  }
-}