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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