Hackerrank – Problem description The problem description – Hackerrank. Solution For detailed informations look at the implementation. I created basic number to word translations. The I find out how many times is a given number present in basic translations. E.g. 12 324 is 12 x 1000, than “twelve thousand”. The rest is without 12 000, […]
Hackerrank – Problem description The problem description – Hackerrank. Solution 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
|
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 |
import java.math.BigInteger; import java.util.Scanner; public class PowerDigitSum { 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++) { BigInteger base = new BigInteger("2"); int exp = Integer.parseInt(scanner.nextLine().trim()); base = base.pow(exp); long result = sumDigits(base); System.out.println(result); } scanner.close(); } private static long sumDigits(BigInteger num) { long result = 0; char[] digits = num.toString().toCharArray(); for (char digit : digits) { result += Character.digit(digit, 10); } return result; } } |
Hackerrank – Problem description The problem description – Hackerrank. Solution The solution is based on this article. I created solution in: Java All solutions are also available on my GitHub profile. Java
|
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 48 49 50 |
import java.math.BigInteger; import java.util.HashSet; import java.util.Scanner; import java.util.Set; public class LatticePaths { // http://www.robertdickau.com/lattices.html 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++) { String[] numbers = scanner.nextLine().split("\\s+"); int n = Integer.parseInt(numbers[0]); int m = Integer.parseInt(numbers[1]); Set<Integer> nFactorialMembers = new HashSet<>(); for (int j = 1; j <= n; j++) { nFactorialMembers.add(j); } Set<Integer> mFactorialMembers = new HashSet<>(); for (int j = 1; j <= m; j++) { mFactorialMembers.add(j); } Set<Integer> mnFactorialMembers = new HashSet<>(); for (int j = 1; j <= m + n; j++) { mnFactorialMembers.add(j); } // (m+n) chose m === (m+n)!/m!n! // divided n! mnFactorialMembers.removeAll(nFactorialMembers); BigInteger result = BigInteger.ONE; for (Integer k : mnFactorialMembers) { result = result.multiply(new BigInteger(k.toString())); } for (Integer k : mFactorialMembers) { result = result.divide(new BigInteger(k.toString())); } System.out.println(result.mod(new BigInteger("1000000007"))); } scanner.close(); } } |
Hackerrank – Problem description The problem description – Hackerrank. Solution There are given 2 rules for Collatz sequence: if n is even, then n/2 if n is odd, then 3n + 1 The given problem set of possibilities is to big and it would take too much time. My solution is to precalculate sequence values […]
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
|
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 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 |
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
|
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 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 |
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. […]