[Codility Lessons] FrogRiverOne(8/n)

[Lesson4] Counting Elements

FrogRiverOne : Find the earliest time when a frog can jump to the other side of a river.

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Java Solution

import java.util.*;

class Solution {
    public int solution(int X, int[] A) {
        // A 배열을 arr HashSet에 넣고 HashSet 크기와 X 비교
        HashSet<Integer> arr = new HashSet<Integer>();
        for(int i = 0; i < A.length; i++){
            arr.add(A[i]);
            if(arr.size() == X){
                return i;    
            }
        }
        return -1;
    }
}

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PHP Solution

function solution($X, $A) {
    $arr = array();
    foreach($A as $index=>$value){
        if(!isset($arr[$value])){
            $arr[$value] = $value;    
        }
        if(count($arr) == $X){
            return $index;
        }
    }
    return -1;
}