TapeEquilibrium

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Task Score     :100
Correctness   :100
Performance : 100

Task description

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.

Write an efficient algorithm for the following assumptions:

• 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].

Algorithm

public class TapeEquilibrium {

 public static void main(String[] args) {
  int[] A ={3,1,2,4,3};
  System.out.println(new TapeEquilibrium().solution(A));
 }
 public int solution(int[] A) {

  if(A.length==0) {
   return 0;
  }
  if(A.length == 1) {
   return Math.abs(A[0]);
  }
  if(A.length == 2) {
   return Math.abs(A[0] - A[1]);
  }
  
  int currentMin =Integer.MAX_VALUE;
  
  int total = 0;
  
  for (int i = 0; i < A.length; i++) {
   total+=A[i];
  }
  
  int temp = 0;
  for (int i = 0; i < A.length-1 ; i++) {
   temp +=A[i];
   
   int second = total - temp;
   
   currentMin =Math.min(currentMin, Math.abs(temp-second));
   
  }
  
  return currentMin;
    }
}

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