[codility] Counting Elements - FrogRiverOne

문제

A small frog wants to get to the other side of a river. The frog is initially located on one bank of the river (position 0) and wants to get to the opposite bank (position X+1). Leaves fall from a tree onto the surface of the river.

You are given an array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in seconds.

The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X (that is, we want to find the earliest moment when all the positions from 1 to X are covered by leaves). You may assume that the speed of the current in the river is negligibly small, i.e. the leaves do not change their positions once they fall in the river.

For example, you are given integer X = 5 and array A such that:

A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3 A[6] = 5 A[7] = 4

In second 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.

Write a function:

public func solution(_ X : Int, _ A : inout [Int]) -> Int

that, given a non-empty array A consisting of N integers and integer X, returns the earliest time when the frog can jump to the other side of the river.

If the frog is never able to jump to the other side of the river, the function should return −1.

For example, given X = 5 and array A such that:

A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3 A[6] = 5 A[7] = 4

the function should return 6, as explained above.

Write an efficient algorithm for the following assumptions:

• N and X are integers within the range [1..100,000];
• each element of array A is an integer within the range [1..X].Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

 

문제 해설

나뭇잎의 배열인 A [1,4,3,2,3,1,5,4], X = 5 가 주어지면 1부터 X까지 연속된 숫자가 도달되는 인덱스를 구하여 리턴, 없으면 -1 리턴

 

 

구현 해설

set을 이용하여 중복제거

set에 A 배열의 값을 인서트

set 사이즈와 X값과 같으면 리턴

 

import Foundation
import Glibc

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

public func solution(_ X : Int, _ A : inout [Int]) -> Int {
    // write your code in Swift 4.2.1 (Linux)

    var sets = Set<Int>()
    for (index, _) in A.enumerated() {
        sets.insert(A[index])
        if sets.count == X {
            return index
        }
    }
    return -1
}