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



  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
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
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






Posted by on No comments:
Labels: , ,

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


  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
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
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


Posted by on No comments:
Labels: ,
Subscribe to: Posts (Atom)