Codility: Tape Equilibrium

Minimize the value |(A[0] + … + A[P-1]) – (A[P] + … + A[N-1])|.

Question

A non-empty zero-indexed 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 zero-indexed 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), beyond input storage (not counting the storage required for input arguments).

Solution

//A: ("-1000", 1000) expected 2000
//A: (5, 6, 2, 4, 1) expected 4
//A: ("-10", "-5", "-3", "-4", "-5") expected 3
- (int)tapeEquilibrium:(NSMutableArray*) A {
    int min = 1000000000; //outside bounds
    int difference;
    int addup = [A[0]intValue];
    int sum = 0;
    
    for(int i= 1; i < A.count; i++){
        sum += [A[i] integerValue];
    }
    
    for(int i= 0; i < A.count-1; i++){
        difference = abs(addup - sum);
        if(difference < min || !min){
            min = difference;
        }
        sum = sum -[A[i+1] intValue];
        addup = addup + [A[i+1] intValue];
    }
    
    return min;
}

Detected time complexity:
O(N)

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