# Codility 1. FrogRiverOne

annmj·2021년 11월 15일
1

## Codility

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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 = 1
A = 3
A = 1
A = 4
A = 2
A = 3
A = 5
A = 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 = 1
A = 3
A = 1
A = 4
A = 2
A = 3
A = 5
A = 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

1. 1 에서 X 까지 이동하는데, 그 중에서 소요시간이 가장 짧은 경우를 찾는 문제이다.
2. i 까지 이동하는데 걸린 시간이 최소가 되는 배열을 minLeaf 라고 했고, 구하고자 하는 정답은 그 중에서 최대값이어야만 한다.
3. 문제의 조건을 생각해보면 1~K 까지 모든 낙엽이 존재해야만 건널 수 있다고 했으므로, 문제의 조건도 꼼꼼히 체크해야되겠다는 것을 알게 되었다.

### 📝 소스코드

// 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[] minLeaf = new int[X + 1];
int max = 0;
for (int j = 0; j < minLeaf.length; j++) {
minLeaf[j] = Integer.MAX_VALUE;
}

for (int i = 0; i < A.length; i++) {
minLeaf[A[i]] = Math.min(minLeaf[A[i]], i);
}

for(int k = 1; k < X + 1; k++) {
max = Math.max(max , minLeaf[k]); // X 번째가 1초, X - 1번째가 5초 이면 전체 5초걸리니까 max 로 생각
}
if(max == Integer.MAX_VALUE) return -1;
else return max;
}
}