TapeEquilibrium

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

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:

func Solution(A []int) int

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].

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題目大意

解題思路

來源

• https://app.codility.com/programmers/lessons/3-time_complexity/tape_equilibrium/

解答

https://github.com/kimi0230/LeetcodeGolang/blob/master/Codility/Lesson/0003.Time-Complexity/TapeEquilibrium/TapeEquilibrium.go

package TapeEquilibrium

import "math"

func Solution(A []int) int {

    totalSum := 0
    for _, v := range A {
        totalSum += v
    }

    leftSum := A[0]
    rightSum := totalSum - leftSum
    result := math.MaxInt32
    for i := 1; i < len(A); i++ {
        tmpDiff := int(math.Abs(float64(rightSum) - float64(leftSum)))
        if tmpDiff < result {
            result = tmpDiff
        }
        rightSum -= A[i]
        leftSum += A[i]
    }

    if result == math.MaxInt32 {
        result = 0
    }

    return result
}