Monday, 4 June 2018

[Algorithm] Codility Lesson 4-1 Counting Elements - FrogRiverOne

Problem


A small frog wants to get to the other side of a river. The frog is initially located on one bank of the river (position 0) and wants to get to the opposite bank (position X+1). Leaves fall from a tree onto the surface of the river.
You are given an array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in seconds.
The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X (that is, we want to find the earliest moment when all the positions from 1 to X are covered by leaves). You may assume that the speed of the current in the river is negligibly small, i.e. the leaves do not change their positions once they fall in the river.
For example, you are given integer X = 5 and array A such that:
A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3 A[6] = 5 A[7] = 4
In second 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.
Write a function:
int solution(int X, int A[], int N);
that, given a non-empty array A consisting of N integers and integer X, returns the earliest time when the frog can jump to the other side of the river.
If the frog is never able to jump to the other side of the river, the function should return −1.
For example, given X = 5 and array A such that:
A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3 A[6] = 5 A[7] = 4
the function should return 6, as explained above.
Assume that:
  • N and X are integers within the range [1..100,000];
  • each element of array A is an integer within the range [1..X].
Complexity:
  • expected worst-case time complexity is O(N);
  • expected worst-case space complexity is O(X) (not counting the storage required for input arguments).
Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.


Solution Code



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package io.dorbae.study.algorithm.codility.countingelements;

import java.util.HashSet;
import java.util.Set;

/*****************************************************************
 * 
 * FrogRiverOne.java
 * 
 
 A small frog wants to get to the other side of a river. The frog is initially located on one bank of the river (position 0) and wants to get to the opposite bank (position X+1). Leaves fall from a tree onto the surface of the river.

You are given an array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in seconds.

The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X (that is, we want to find the earliest moment when all the positions from 1 to X are covered by leaves). You may assume that the speed of the current in the river is negligibly small, i.e. the leaves do not change their positions once they fall in the river.

For example, you are given integer X = 5 and array A such that:

  A[0] = 1
  A[1] = 3
  A[2] = 1
  A[3] = 4
  A[4] = 2
  A[5] = 3
  A[6] = 5
  A[7] = 4
In second 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.

Write a function:

class Solution { public int solution(int X, int[] A); }

that, given a non-empty array A consisting of N integers and integer X, returns the earliest time when the frog can jump to the other side of the river.

If the frog is never able to jump to the other side of the river, the function should return −1.

For example, given X = 5 and array A such that:

  A[0] = 1
  A[1] = 3
  A[2] = 1
  A[3] = 4
  A[4] = 2
  A[5] = 3
  A[6] = 5
  A[7] = 4
the function should return 6, as explained above.

Assume that:

N and X are integers within the range [1..100,000];
each element of array A is an integer within the range [1..X].
Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(X) (not counting the storage required for input arguments).
Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.


 *
 *****************************************************************
 *
 * @version 0.0.0 2018-06-04 23:12:05 dorbae 최초생성
 * @since 1.0
 * @author dorbae(dorbae.io@gmail.com)
 *
 */
public class FrogRiverOne {
 
 private static final int RESCODE_INVALID   = -1;
 private static final int RESCODE_CANT_CROSS   = -1;
 
 /**
  * 
  *
  * @version 1.1.0 2018-06-04 23:52:01 dorbae Array -> Set. and modify logic error
  * @version 1.0.0 2018-06-04 23:21:41 dorbae 최초생성
  * @since 1.0.0
  * @author dorbae(dorbae.io@gmail.com)
  *
  * @param X : the position 
  * @param A
  * @return
  */
 public int solution( int X, int[] A) {
  // Illegal Argument
  if ( A == null) {
   System.err.println( "Invalid array A.");
   return RESCODE_INVALID;
  }
  
  int N = A.length;
  if ( N < 1 || N > 100000) {
   System.err.println( "Invalid length of array");
   return RESCODE_INVALID;
  }
  
  if ( X < 1 || X > 100000) {
   System.err.println( "Invalid position X");
   return RESCODE_INVALID;
  }
  
  // Initial Check Set
  Set< Integer> leafExistsChecker = new HashSet< Integer>();
  for ( int ll = 1; ll <= X; ll++) {
   leafExistsChecker.add( ll);
  }
  
  for ( int ll = 0; ll < N; ll++) {
   if ( A[ ll] < 1 || A[ ll] > X) {
    System.err.println( "Invalid element in array");
    return RESCODE_INVALID;
   }
   
   if ( leafExistsChecker.remove( A[ ll])) { // New one
    if ( leafExistsChecker.isEmpty())
     return ll;
   }
  }
  
  return RESCODE_CANT_CROSS;
 }
 
 /**
  * 
  *
  * @version 1.0.0 2018-06-04 23:35:11 dorbae 최초생성
  * @since 1.0.0
  * @author dorbae(dorbae.io@gmail.com)
  *
  * @param leafExistChecker
  * @return
  */
 @Deprecated
 private static boolean isRelay( boolean[] leafExistChecker) {
  for ( int ll = 1; ll < leafExistChecker.length; ll++) {
   if ( !leafExistChecker[ ll]) {
    return false;
   }
  }
  
  return true;
 }
}



Result

ver.1.0.0






ver.1.1.0






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Friday, 1 June 2018

[Algorithm] Codility Lesson 3-3 Time Complexity - TapeEquilirium

Problem


A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
For example, consider array A such that:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3
We can split this tape in four places:
  • P = 1, difference = |3 − 10| = 7 
  • P = 2, difference = |4 − 9| = 5 
  • P = 3, difference = |6 − 7| = 1 
  • P = 4, difference = |10 − 3| = 7 
Write a function:
int solution(int A[], int N);
that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.
For example, given:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3
the function should return 1, as explained above.
Assume that:
  • N is an integer within the range [2..100,000];
  • each element of array A is an integer within the range [−1,000..1,000].
Complexity:
  • expected worst-case time complexity is O(N);
  • expected worst-case space complexity is O(N) (not counting the storage required for input arguments).
Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

Solution Code


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package io.dorbae.study.algorithm.codility.timecomplexity;

/*****************************************************************
 * 
 * TapeEquilibrium.java
 * 
 * 
 * 
A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

  A[0] = 3
  A[1] = 1
  A[2] = 2
  A[3] = 4
  A[4] = 3
We can split this tape in four places:

P = 1, difference = |3 − 10| = 7 
P = 2, difference = |4 − 9| = 5 
P = 3, difference = |6 − 7| = 1 
P = 4, difference = |10 − 3| = 7 
Write a function:

class Solution { public int solution(int[] A); }

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:

  A[0] = 3
  A[1] = 1
  A[2] = 2
  A[3] = 4
  A[4] = 3
the function should return 1, as explained above.

Assume that:

N is an integer within the range [2..100,000];
each element of array A is an integer within the range [−1,000..1,000].
Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N) (not counting the storage required for input arguments).
Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
 *
 *****************************************************************
 *
 * @version 0.0.0 2018-06-01 15:04:21 dorbae 최초생성
 * @since 1.0
 * @author dorbae(dorbae.io@gmail.com)
 *
 */
public class TapeEquilibrium {
 
 private static final int RESCODE_INVALID = 0;
 /**
  * 
  * @version 1.2.0 2018-06-02 15:05:10 dorbae Remove unnecessary logic and modify incorrect logic
  * @version 1.1.0 2018-06-02 14:45:06 dorbae Add exception about range of elements
  * @version 1.0.0 2018-06-01 15:10:34 dorbae 최초생성
  * @since 1.0.0
  * @author dorbae(dorbae.io@gmail.com)
  *
  * @param A : a non-empty array (the range of elements is -1000 ~ 1000)
  * 
  * @return : a minimal different between two arrays that be seperated by Pb
  */
 public int solution( int[] A) {
  if ( A == null) {
   System.err.println( "Invalid array A.");
   return RESCODE_INVALID;
  }
  
  /** N : the length of array A. 2 <= N < 100,000 */
  int N = A.length;
  if (  N < 2 || N > 100000) {
   System.err.println( "Invalid array A.");
   return RESCODE_INVALID;
  }
  
  int minimumValue = Integer.MAX_VALUE;
  int preValue = 0;
  int postValue = 0;
  int differntValue = 0;
  int ll = 0;
  preValue = 0;
  postValue = 0;
  
  /* Removed in ver 1.2.0 
  preValue = A[ 0];
  
  // Exception
  if ( A[ 0] < -1000 || A[ 0] > 1000) {
   System.err.println( "Invalid elment value.");
   return RESCODE_INVALID;
  }
    */
  
  for ( ll = 0; ll < N; ll++) {
   /* Added in ver 1.1.0 */
   // Exception
   if ( A[ ll] < -1000 || A[ ll] > 1000) {
    System.err.println( "Invalid elment value.");
    return RESCODE_INVALID;
   }
   /* End */
   
   postValue += A[ ll];
   
  }
  
  /* Removed in ver 1.2.0
  differntValue = Math.abs( preValue - postValue);
  if ( differntValue < minimumValue) {
   minimumValue = differntValue;
  }
  */
  
  /* Modified in ver 1.2.0 */
  /* P : a index to split array A. (seperated arrays must be not empty.). 0 < P < N */
  --N;
  for ( int P = 0; P < N; P++) {
   preValue += A[ P];
   postValue -= A[ P];
   differntValue = Math.abs( preValue - postValue);
   if ( differntValue < minimumValue) {
    minimumValue = differntValue;
   }
  }
  
  return minimumValue;
  
    }
}


Result

ver.1.0.0







ver.1.2.0


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Thursday, 24 May 2018

[Algorithm] Codility Lessson 3-2. Time Complexity - PermMissingElem

Problem


An array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.
Your goal is to find that missing element.
Write a function:
int solution(int A[], int N);
that, given an array A, returns the value of the missing element.
For example, given array A such that:
A[0] = 2 A[1] = 3 A[2] = 1 A[3] = 5
the function should return 4, as it is the missing element.
Assume that:
  • N is an integer within the range [0..100,000];
  • the elements of A are all distinct;
  • each element of array A is an integer within the range [1..(N + 1)].
Complexity:
  • expected worst-case time complexity is O(N);
  • expected worst-case space complexity is O(1) (not counting the storage required for input arguments).
Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.





Solution Code


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package io.dorbae.study.algorithm.codility.timecomplexity;

/*****************************************************************
 * 
 * PermMissingElem.java
 * 
 * ***************************************************************
 * 
 An array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.

Your goal is to find that missing element.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A, returns the value of the missing element.

For example, given array A such that:

  A[0] = 2
  A[1] = 3
  A[2] = 1
  A[3] = 5
the function should return 4, as it is the missing element.

Assume that:

N is an integer within the range [0..100,000];
the elements of A are all distinct;
each element of array A is an integer within the range [1..(N + 1)].
Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1) (not counting the storage required for input arguments).
Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
 
 
 *
 *****************************************************************
 *
 * @version 0.0.0 2018-05-24 20:30:39 dorbae 최초생성
 * @since 1.0
 * @author dorbae(dorbae.io@gmail.com)
 *
 */
public class PermMissingElem {
 private static final int RESCODE_INVALID = 0;
 private static final int RESCODE_NOTFOUND = 0;
 
 /**
  * 
  * @version 1.1.0 2018-05-24 21:04:11 dorbae empty_and_single 오류 수
  * @version 1.0.0 2018-05-24 20:32:54 dorbae 최초생성
  * @since 1.0.0
  * @author dorbae(dorbae.io@gmail.com)
  *
  * @param A : An array A consisting of N different integers is given
  * @return : Missing element
  */
 public int solution( int[] A) {
  if ( A == null) {
   System.err.println( "Illegal Argument.");
   return RESCODE_INVALID;
  }
  
  int N = A.length;
  
  if ( N < 0 || N > 100000 ) {
   System.err.println( "Illegal Argument.");
   return RESCODE_INVALID;
   
  /** 
   * version 1.1.0
   * Solve empty and single error
  */
  } else if ( N == 0) {
//   return 0; // Old (Error)
   return 1;
   
  }
  /**
   * End of version 1.1.0
   */
  
  boolean[] checkBitArray = new boolean[ N +2]; // Default primitive boolean value is false.
  
  // Checking
  for ( int ll = 0; ll < N; ll++) {
   try {
    if ( checkBitArray[ A[ ll]]) { // Already Checked
     System.err.println( "Already existing value.");
     return RESCODE_INVALID;
    
    } else {
     checkBitArray[ A[ ll]] = !checkBitArray[ A[ ll]];
    
    }
    
   } catch ( IndexOutOfBoundsException e) {
    System.err.println( "each element of array A is an integer within the range [1..(N + 1)]");
    return RESCODE_INVALID;
    
   }
  }
  
  if ( checkBitArray[ 0]) { // one of elements of array A is 0. Each element of array A is an integer within the range [1..(N + 1)]
   return RESCODE_INVALID;
  }
  
  // Find missing element
  for ( int ll = 1; ll < checkBitArray.length; ll++) {
   if ( !checkBitArray[ ll]) {
    return ll;
   }
  }
  
  return RESCODE_NOTFOUND; 
  
    }

}




Result

ver.1.0.0










ver.1.1.0




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Wednesday, 23 May 2018

[Algorithm] Codility Lessson 1-1. Iterations - BinaryGap

Problem


A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.
For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps. The number 32 has binary representation 100000 and has no binary gaps.
Write a function:
int solution(int N);
that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn't contain a binary gap.
For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5. Given N = 32 the function should return 0, because N has binary representation '100000' and thus no binary gaps.
Assume that:
  • N is an integer within the range [1..2,147,483,647].
Complexity:
  • expected worst-case time complexity is O(log(N));
  • expected worst-case space complexity is O(1).
Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.







Solution Code


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package io.dorbae.study.algorithm.codility.iterations;

/*****************************************************************
 * 
 * BinaryGap.java
 * 
 * Task description
 * 
 * A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.
 * For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps.
 * 
 * Write a function:
 * class Solution { public int solution(int N); }
 * that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn't contain a binary gap.
 * For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5.
 * Assume that:
 * N is an integer within the range [1..2,147,483,647].
 * 
 * Complexity:
 * expected worst-case time complexity is O(log(N));
 * expected worst-case space complexity is O(1).
 * Copyright 2009–2018 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
 *
 *****************************************************************
 *
 * @version 0.0.0 2018-05-14 23:12:59 dorbae 최초생성
 * @since 1.0
 * @author dorbae(dorbae.io@gmail.com)
 *
 */
public class BinaryGap {
 /**
  * 
  *
  * @version 1.1.0 2018-05-23 23:13:14 dorbae 예외처리. 오류 수
  * @version 1.0.0 2018-05-14 23:12:59 dorbae 최초생성
  * @since 1.0.0
  * @author dorbae(dorbae.io@gmail.com)
  *
  * @param N : positive integer
  * @return : the length of its longest binary gap
  */
 public int solution( int N) {
  if ( N < 1 || N > 2147483647) {
   System.out.println( "N is out of range.");
   return 0;
  }
  
  int maximumInterval = 0;
  int count = 0;
  String binary = Integer.toBinaryString( N);
//  System.out.println( "binary=" + binary);
  char[] checkData = binary.toCharArray();
  for ( int ll = 1; ll < checkData.length; ll++) {
   if ( checkData[ ll] == '1') {
    if ( count > maximumInterval)
     maximumInterval = count;
    count = 0;
    
   } else {
    ++count;
   }
  }
  
  return maximumInterval;
 }
 
}



Result






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[Apache Hive] Install Hive2

Requirements

Java Version

- Hive version 1.2부터 Java 1.7 이상 버전이 필요
- Hive version 0.14~1.1까지는 Java 1.6에서도 구동 가능
- Java 1.8을 사용할 것을 권장

Hadoop Version

- Hadoop 2.x 를 권장
- Hadoop 1.x 도 가능하나 Hive 2.0.0 부터 지원 X



1. Download Hive

#> cd /usr/local
#> wget http://mirror.apache-kr.org/hive/hive-2.3.0/apache-hive-2.3.0-bin.tar.gz




2. Unzip Hive Install File

#> tar zxf apache-hive-2.3.0-bin.tar.gz





3. Setup Environment Variables







4. Create Hive Meta Store Directory and Change Mode in HDFS

#> hadoop fs -mkdir /tmp
#> hadoop fs -mkdir /user
#> hadoop fs -mkdir /user/hive
#> hadoop fs -mkdir /user/hive/warehouse
#> hadoop fs -chmod g+w /tmp
#> hadoop fs -chmod g+w /user/hive/warehouse





5. Initialize Schema as DB Type 'derby'

#> schematool -dbType derby -initSchema





6. Start Hive Server

#> hiverserver2 &




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