강 저편으로 가고 싶은 개구리가 있다.
이 개구리는 X 지점까지 가고 싶은데, X 지점까지 가기 위해서는 1 부터 X 까지의 지점을 모두 거쳐야한다.
1 부터 X 까지의 지점을 모두 거쳐서 X에 도달했다면 해당 인덱스를 리턴하면 된다.
예를들어, X 가 5 라고 하고, 배열이 아래와 같이 주어졌을 때
A[0] = 1 A[1] = 3 A[2] = 1 A[3] = 4 A[4] = 2 A[5] = 3 A[6] = 5 A[7] = 4
A[6]=5 이다. 그리고 A[0] 에서 A[6] 까지는 1 부터 X 까지 즉 1부터 5까지의 모든 지점이 존재하기 때문에
개구리가 1 부터 5 까지의 지점(1,2,3,4,5)을 모두 거쳤다고 볼 수 있으므로 해당 인덱스인 6을 리턴하면 된다.
문제 원문:
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.
Assume that:
- N and X are integers within the range [1..100,000];
- each element of array A is an integer within the range [1..X].
Complexity:
- expected worst-case time complexity is O(N);
- expected worst-case space complexity is O(X) (not counting the storage required for input arguments).
문제 이해하는데 20분 푸는데 20분 정도 걸려서 100% 점수로 문제를 풀었다.
정답:
// you can also use imports, for example: // import java.util.*; // you can write to stdout for debugging purposes, e.g. // System.out.println("this is a debug message"); class Solution { public int solution(int X, int[] A) { int N = A.length; boolean[] checkArr = new boolean[X]; int cnt = 0; for(int i = 0; i < N; i++) { if(A[i] <= X) { if(!checkArr[A[i] - 1]) cnt++; checkArr[A[i] - 1] = true; } if(cnt == X) return i; } return -1; } }