java codility Frog-River-One

Question

I have been trying to solve a Java exercise on a Codility web page.

Below is the link to the mentioned exercise and my solution.

https://codility.com/demo/results/demoH5GMV3-PV8

Can anyone tell what can I correct in my code in order to improve the score?

Just in case here is the task description:

A small frog wants to get to the other side of a river. The frog is currently located at position 0, and wants to get to position X. Leaves fall from a tree onto the surface of the river.

You are given a non-empty zero-indexed array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in minutes.

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.

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 minute 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 zero-indexed 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), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

And here is my solution:

import java.util.ArrayList;
import java.util.List;

class Solution {

    public int solution(int X, int[] A) {
        int list[] = A;
        int sum = 0;
        int searchedValue = X;

        List<Integer> arrayList = new ArrayList<Integer>();

        for (int iii = 0; iii < list.length; iii++) {

            if (list[iii] <= searchedValue && !arrayList.contains(list[iii])) {
                sum += list[iii];
                arrayList.add(list[iii]);
            }
            if (list[iii] == searchedValue) {
                if (sum == searchedValue * (searchedValue + 1) / 2) {
                    return iii;
                }
            }
        }
        return -1;
    }
}

Answer 1

You are using arrayList.contains inside a loop, which will traverse the whole list unnecessarily.

Here is my solution (I wrote it some time ago, but I believe it scores 100/100):

    public int frog(int X, int[] A) {
        int steps = X;
        boolean[] bitmap = new boolean[steps+1];
        for(int i = 0; i < A.length; i++){
            if(!bitmap[A[i]]){
                bitmap[A[i]] = true;
                steps--;
                if(steps == 0) return i;
            }

        }
        return -1;
    }

Answer 2

Here is my solution. It got me 100/100:

public int solution(int X, int[] A)
{
     int[] B = A.Distinct().ToArray();
     return (B.Length != X) ? -1 : Array.IndexOf<int>(A, B[B.Length - 1]);
}

Answer 3

100/100

public static int solution (int X, int[] A){

    int[]counter = new int[X+1];
    int ans = -1;
    int x = 0;

    for (int i=0; i<A.length; i++){
        if (counter[A[i]] == 0){
            counter[A[i]] = A[i];
            x += 1;
            if (x == X){
                return i;
            }
        } 
    }

    return ans;
}

Answer 4

A Java solution using Sets (Collections Framework) Got a 100%

import java.util.Set;
import java.util.TreeSet;
public class Froggy {
    public static int solution(int X, int[] A){
    int steps=-1;
    Set<Integer> values = new TreeSet<Integer>();
    for(int i=0; i<A.length;i++){
        if(A[i]<=X){
            values.add(A[i]);
        }
        if(values.size()==X){
            steps=i;
            break;
        }
    }
        return steps;
    }