Algorithms - Discrete Mathematics - K Candy Store

K Candy Store

The following is the solution to Hacker Rank problem K Candy Store using Java.  For solutions to other Hacker Rank Problem visit my page HackerRank, alternatively try searching for the problem in my blog.

Score: 5/5

/**  *  */ import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.math.BigInteger; import java.util.StringTokenizer; /**  * @author Arun.G  *  */ public class Solution {        BufferedReader reader;        StringTokenizer tokenizer;        PrintWriter out;        public BigInteger factorial(BigInteger  n)        {               if(n.compareTo(BigInteger.ONE)==0                           || n.compareTo(BigInteger.ZERO)==0)                      return BigInteger.ONE;               else                      return (n.multiply(                         factorial(n.subtract(BigInteger.ONE))                         ));        }        public void solve() throws IOException {               int T = nextInt();               for (int count = 0; count < T; count++) {                      int N = nextInt();                      int K = nextInt();                      BigInteger nCk = BigInteger.ONE;                      nCk = factorial(new BigInteger((N+K-1)+"")).divide(factorial(new BigInteger(K+"")).multiply(factorial( BigInteger.valueOf(N-1))));                                          if(nCk.toString().length()>9)                      {                       System.out.println(new BigInteger(nCk.toString().substring(nCk.toString().length()-9,nCk.toString().length())));                      }                      else                      {                      System.out.println(nCk);                      }                                    }        }        /**         * @param args         */        public static void main(String[] args) {               new KCandyStore().run();        }        public void run() {               try {                      reader = new BufferedReader(                         new InputStreamReader(System.in));                      tokenizer = null;                      out = new PrintWriter(System.out);                      solve();                      reader.close();                      out.close();               } catch (Exception e) {                      e.printStackTrace();                      System.exit(1);               }        }        int nextInt() throws IOException {               return Integer.parseInt(nextToken());        }        long nextLong() throws IOException {               return Long.parseLong(nextToken());        }        double nextDouble() throws IOException {               return Double.parseDouble(nextToken());        }        String nextToken() throws IOException {               while (tokenizer == null || !tokenizer.hasMoreTokens()) {                      tokenizer = new StringTokenizer(reader.readLine());               }               return tokenizer.nextToken();        } }

Algorithms - Discrete Mathematics - nCr table

nCr table

The following is the solution to Hacker Rank problem nCr Table using Java.  For solutions to other Hacker Rank Problem visit my page HackerRank, alternatively try searching for the problem in my blog.

Score: 5/5

/**  *  */ import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.math.BigInteger; import java.util.StringTokenizer; /**  * @author Arun.G  *  */ public class Solution {        BufferedReader reader;        StringTokenizer tokenizer;        PrintWriter out;        public void solve() throws IOException {               int T = nextInt();               for (int count = 0; count < T; count++) {                      int N = nextInt();                      BigInteger nCk = BigInteger.ONE;                      int i=N;                            for (int j = 0; j <= N; j++) {                                   System.out.print(nCk.mod(new BigInteger("1000000000")) + " ");                                   nCk = nCk.multiply(new BigInteger((i - j)+"")).divide(new BigInteger(((j + 1))+""));                            }                            System.out.println();                                    }        }        /**         * @param args         */        public static void main(String[] args) {               new nCrtable().run();        }        public void run() {               try {                      reader = new BufferedReader(new InputStreamReader(System.in));                      tokenizer = null;                      out = new PrintWriter(System.out);                      solve();                      reader.close();                      out.close();               } catch (Exception e) {                      e.printStackTrace();                      System.exit(1);               }        }        int nextInt() throws IOException {               return Integer.parseInt(nextToken());        }        long nextLong() throws IOException {               return Long.parseLong(nextToken());        }        double nextDouble() throws IOException {               return Double.parseDouble(nextToken());        }        String nextToken() throws IOException {               while (tokenizer == null || !tokenizer.hasMoreTokens()) {                      tokenizer = new StringTokenizer(reader.readLine());               }               return tokenizer.nextToken();        } }

Algorithms - Discrete Mathematics - Diwali Lights

Diwali Lights

The following is the solution to Hacker Rank problem Diwali Lights using Java.  For solutions to other Hacker Rank Problem visit my page HackerRank, alternatively try searching for the problem in my blog.

The Approach I'm following is given below:

1. The problem is easy, if we have 2 bulbs atleast 1 bulb should be set at any time so there are 2²-1 possibilities eliminating the condition where all the bulbs are turned off.
2. So if we have N bulbs then we have 2^N-1 possibilities which is large  number to compute.
3. We can do this easily with Java's BigInteger , we can calculate the power using pow function.
4. Also we must note since this number is large we are asked to print the result mod 10⁵ we can also do this easily with Java's BigInteger mod function.

Score: 5/5

/**  *  */ import java.io.BufferedReader; import java.io.InputStreamReader; import java.math.BigInteger; /**  * @author Arun.G  *  */ public class Solution {        /**         * @param args         */        static BufferedReader in = new BufferedReader(new InputStreamReader(                      System.in));        public static BigInteger ExponentBySquare(long x, long n)     {               BigInteger big = BigInteger.valueOf(x)                    .pow((int)n)                    .subtract(BigInteger.ONE).mod(new BigInteger("100000"));                             return big;     }        public static void main(String[] args) throws NumberFormatException,                      Exception {                             int T = Integer.parseInt(in.readLine());                 for(int i=0;i<T;i++)           {           long noOfBulbs = Long.parseLong(in.readLine());           System.out.println(ExponentBySquare(2,noOfBulbs).toString());         }                 in.close();                                    } }

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Algorithms - Math - Ice Cream Parlor

Ice Cream Parlor

The following is the solution to Hacker Rank problem Ice Cream Parlor using Java.  For solutions to other Hacker Rank Problem visit my page HackerRank, alternatively try searching for the problem in my blog.

The Approach I'm following is given below:

1. I have used easy approach, i.e a Complete Search to find if any two items can be bought for given cost and print the indexes.
2. The indexes in the answer starts from 1 to N instead of 0 to N-1;

Score: 30/30

/**  *  */ import java.util.Scanner; /**  * @author Arun.G  *  */ public class Solution {        /**         * @param args         */        public static void main(String[] args) {               // TODO Auto-generated method stub               Scanner sc = new Scanner(System.in);               int T = sc.nextInt();               for (int k = 0; k < T; k++) {                      int M = sc.nextInt();                      int N = sc.nextInt();                      int[] Cost = new int[N];                      // get the cost                      for (int index = 0; index < N; index++)                            Cost[index] = sc.nextInt();                      for (int i = 0; i < N; i++) {                            for (int j = i; j < N; j++) {                                   if (i == j)                                          continue;                                   else {                                          if (Cost[i] + Cost[j] > M)                                                 continue;                                          else if (Cost[i] + Cost[j] == M) {                                          System.out.println((i+1) + " " + (j+1));                                          }                                   }                            }                      }               }               sc.close();        } }

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