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[Codility][Kotlin] TapeEquilibrium

jordancancode 2024. 10. 3. 23:43

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:

fun solution(A: IntArray): 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].

 

풀이1

import kotlin.math.*


fun solution(A: IntArray): Int {
    var result = Integer.MAX_VALUE
    var idx = 1
    while(idx < A.size) {
        var left = A.sliceArray(0 until idx).sum()
        var right = A.sliceArray(idx until A.size).sum()
        if(result < abs(left-right)) {
            return result 
        }
        result = abs(left-right)
        idx+=1
    }   
    return result
}

처참한 풀이.... wrong answer과 timeout error의 콜라보...

sum을 반복적으로 수행했기 때문에, 시간복잡도가 커질 수 밖에 없었다. 그리고, 최솟값 업데이트 로직이 잘못되었다...ㅎ...

import kotlin.math.*


fun solution(A: IntArray): Int {
    val sum = A.sum()
    val results = mutableListOf<Int>()
    var tmp = 0
    repeat(A.size) {
        tmp += A[it]
        results.add(abs(tmp - (sum-tmp)))
    }
    results.removeLast()
    return results.minOrNull() ?:0
}

이게 정답이다. 해당 코드를 참조하며 풀었다.

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