[Lesson 4] 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:
class Solution { public int solution(int X, 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

 

첫번째 시도 score : 72

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(int X, vector<int> &A) {
    // Implement your solution here
    //ON/OFF

    int visited[100003];
    int check = (X+1)*X/2;

    int i = 0;
    while(check != 0 && i < A.size())
    {
        
        if(visited[A[i]] == 0)
        {
            visited[A[i]] = 1;
            check -= A[i];
        } 

        i++;      
    }

    return i-1;
}

 

문제를 제대로 안읽어서, 예외 상황에는 -1이 들어간다는 점을 간과했다.

 

두번째 시도 score : 90

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(int X, vector<int> &A) {
    // Implement your solution here
    //ON/OFF

    int visited[100003];
    int check = (X+1)*X/2;

    int i = 0;

    for(int i = 0; i < A.size(); i++)
    {
        if(visited[A[i]] == 0)
        {
            visited[A[i]] = 1;
            check -= A[i];
        }

        if(check == 0) return i;
    }
    //if count still not 0

    return -1;
}

아까는 큰범위에서 오류가 안났는데 이번에는 큰범위에서 오류가난다ㅠㅠ

 

세번째 시도 score : 100

 

잘 모르겠어서, 다른 사람들의 풀이를 봤는데, set을 사용했다.

set은 중복을 없애고, 정렬을 유지하는 stl로 레드블랙트리로 구현되어있다.

 

A의 사이즈 만큼 for문을 돌려서, set에 넣은 뒤, set의 크기가 X가 되면 for문의 i를 리턴해주고, 아니라면 -1를 기본값으로 리턴해준다.

// you can use includes, for example:
// #include <algorithm>
#include <set>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(int X, vector<int> &A) {
    // Implement your solution here
    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;
}