TapeEquilibrium.go

package complexity

import "math"

/*
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:

func Solution(A []int) int
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).
 */

func TapeEquilibrium(A []int) int {
	arraySum := 0
	currentMin := 1 << 32 - 1

	for _, value := range A {
		arraySum += value
	}
	lhs := A[0]
	rhs := arraySum - lhs

	for i := 1; i < len(A); i++ {
		diff := int(math.Abs(float64(lhs) - float64(rhs)))

		if (diff < currentMin) {
			currentMin = diff
		}
		lhs += A[i]
		rhs -= A[i]
	}

	return currentMin
}