Hackerrank – Problem description The problem description – Hackerrank. Solution We need to sum all given numbers. The catch is to use a feature of chosen programming language, which allows us operation with big numbers (e. g. BigInteger in Java) I created solution in: Java All solutions are also available on my GitHub profile. Java […]
Hackerrank – Problem description The problem description – Hackerrank. Solution I check all numbers from 1 to N. I need to do a prime factorization of each number. In the end there is a following state: . To calculate the prime numbers I use for example Sieve of Eratosthenes. I calculate the exponents […]
Hackerrank – Problem description The problem description – Hackerrank. Solution We have to calculate products of number foursomes for all directions in a grid and take the greatest product value. I created solution in: Java All solutions are also available on my GitHub profile. Java
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import java.util.Scanner; public class LargestProductGrid { private static final int N = 20; private static final int ADJACENT = 4; public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int[][] visited = new int[N][N]; int[][] matrix = new int[N][N]; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { matrix[i][j] = scanner.nextInt(); } } System.out.println(max(matrix, visited)); scanner.close(); } private static long max(int[][] matrix, int[][] visited) { long max = 0; for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { long product = productToRight(matrix, i, j); if (product > max) { max = product; } product = productToDown(matrix, i, j); if (product > max) { max = product; } product = productToDiagRight(matrix, i, j); if (product > max) { max = product; } product = productToDiagLeft(matrix, i, j); if (product > max) { max = product; } } } return max; } private static long productToRight(int[][] matrix, int i, int j) { long product = 1; if (j <= N - ADJACENT) { for (int r = j; r < j + ADJACENT; r++) { product *= matrix[i][r]; } } return product; } private static long productToDown(int[][] matrix, int i, int j) { long product = 1; if (i <= N - ADJACENT) { for (int r = i; r < i + ADJACENT; r++) { product *= matrix[r][j]; } } return product; } private static long productToDiagRight(int[][] matrix, int i, int j) { long product = 1; int s = j; if (i <= N - ADJACENT && j <= N - ADJACENT) { for (int r = i; r < i + ADJACENT; r++) { product *= matrix[r][s++]; } } return product; } private static long productToDiagLeft(int[][] matrix, int i, int j) { long product = 1; if (i <= N - ADJACENT && j >= ADJACENT - 1) { for (int r = i; r < i + ADJACENT; r++) { product *= matrix[r][j--]; } } return product; } } |
Hackerrank – Problem description The problem description – Hackerrank. Solution First, I created a list of all prime numbers from 2 to 106 (upper constraint). I used Sieve of Eratosthenes algorithm. Then, I sum each element from a subset of prime numbers gained from test case. I created solution in: Java All solutions are also […]
Hackerrank – Problem description The problem description – Hackerrank. Solution I found the solution on StackOverflow. One could use the Wikipedia as a source. I created solution in: Java All solutions are also available on my GitHub profile. Java
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import java.util.Scanner; public class PythagoreanTriplet { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int tests = scanner.nextInt(); for (int i = 0; i < tests; i++) { int n = scanner.nextInt(); System.out.println(findTripletDirty(n)); } scanner.close(); } static long findTriplet(int number) { long maxTriplet = -1; for (int a = 1; a <= number / 3; a++) { int aSquare = a * a; for (int b = a + 1; b <= number / 2; b++) { int c = number - a - b; if(c > 0 && c * c == aSquare + b * b) { long product = (long) a * b * c; if(product > maxTriplet) { maxTriplet = product; } } } } return maxTriplet; } // http://stackoverflow.com/questions/2817848/find-pythagorean-triplet-for-which-a-b-c-1000 static long findTripletDirty(int number) { long max = -1; // sum of integer Pythagorean triplet is always even if(number % 2 == 0) { for (int b = 1; b < number / 2; b++) { if(number * (number / 2 - b) % (number - b) == 0) { int a = number * (number / 2 - b) / (number - b); int c = number - a - b; long product = (long) c * a * b; if(product > max) { max = product; } } } } return max; } } |
Hackerrank – Problem description The problem description – Hackerrank. Solution I took k following numbers and I calculate a product of these k numbers. Compare the product with actual maximum. Print the maximum if there are no more k following numbers. I created solution in: Java All solutions are also available on my GitHub profile. […]
Hackerrank – Problem description The problem description – Hackerrank. Solution The challenge has upper constraint that we could have at most 10 001 prime numbers from the smallest (2). I found out that the upper bound is the number 104 743. There exist first 10 001 prime numbers. We calculate list all of them and […]
Hackerrank – Problem description The problem description – Hackerrank. Solution We calculate 3 equations for every number from 1 to N. Sum of squares: Square of sum: Result – Difference between sums: I created solution in: Java All solutions are also available on my GitHub profile. Java
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import java.util.Scanner; public class SumSquareDifference { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int tests = Integer.parseInt(scanner.nextLine()); for (int i = 0; i < tests; i++) { int number = Integer.parseInt(scanner.nextLine()); System.out.println(sumDifference(number)); } scanner.close(); } private static long sumDifference(int number) { long squareSum = 0; long sumSqares = 0; for (int i = 1; i <= number; i++) { squareSum += i; sumSqares += (i * i); } squareSum *= squareSum; return squareSum - sumSqares; } } |
Hackerrank – Problem description The problem description – Hackerrank. Solution We need to make a prime factorization of number (problem constraints), because some of the numbers are divisors of the other numbers. We reduce unnecessary multiplications and divisions. We could use Sieve of Eratosthenes to calculate all prime numbers from 2 to N. Let’s calculate […]
Hackerrank – Problem description The problem description – Hackerrank. Solution We need to find out if the product of two 3-figured number is a palindrome and it is smaller than exercise upper constraint. My solution starts with initializing two numbers as 999. I continually decremented second number by 1, until it is equal to 100. […]