[Codility] Lesson4. FrogRiverone (C++)

🔮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:

int solution(int X, vector<int> &A);

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].

🔮풀이

이해하는데 시간이 좀 걸렸던 문제.

요약하자면 개구리가 X로 점프할 수 있는 가장 빠른 시간을 찾아야 한다.

목적지를 나타내는 X와 각 요소의 범위가 1 ~ X로 이루어진 배열 A가 주어지는데, 배열 A를 탐색하며 1부터 X 까지의 수가 모두 나타났을 때의 index를 반환하면 된다.

따라서,

1) set 집합 배열에 A[idx]까지 나타난 값을 insert한다.

2) set 집합의 크기가 X와 같아진다면 목적지 X에 도달했다는 의미이므로

3) 그때의 index를 반환한다.

4) A배열 마지막까지 반복문을 돌았는데도 도달하지 못 한다면 -1 반환.

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🔮코드

#include <algorithm>
#include <set>
#include <vector>

int solution(int X, vector<int> &A) {
    int size = A.size();
    set<int> s;

    for(int i=0; i<size; i++){
        s.insert(A[i]);
        if(s.size() == X){
            return i;
        }
    }

    return -1;
}