[codility] Time Complexity - TapeEquilibrium

문제

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:

public func solution(_ A : inout [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].Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

 

 

문제 해설

 

 

구현 해설

 

import Foundation
import Glibc

// you can write to stdout for debugging purposes, e.g.
// print("this is a debug message")

public func solution(_ A : inout [Int]) -> Int {
    // write your code in Swift 4.2.1 (Linux)
    //[3, 1, 2, 4, 3]
    
    let total = A.reduce(0, {$0+$1})
    var P = 0
    var arr: [Int] = []
    for (index, _) in A.enumerated() {
        
        if index == A.count-1 { break }
        
        P += A[index]
        
        let diff = abs(P - (total - P))
        arr.append(diff)
    }
    arr.sort()
    return arr[0]
}