I've googled and asked people about this question, but can't get the meaning. I know my solution is incorrect, but I hope it can help (it's on JS):

let C = [1, 3, 1, 4, 2, 3, 5, 4],
  Y = 4;

function solution(X, A) {

  var leaves = [];
  var i = 0;
  var result = -1;

  for (i = 0; i < A.length; i++) {
    if (typeof leaves[A[i]] == 'undefined') {
      leaves[A[i]] = i;
    }
  }

  if (leaves.length <= X) {
    return -1;
  }

  for (i = 1; i <= X; i++) {
    if (typeof leaves[i] == 'undefined') {
      return -1;
    } else {
      result = Math.max(result, leaves[i]);
    }
  }

  return result;
}

console.log(solution(Y, C));

I've googled and asked people about this question, but can't get the meaning. I know my solution is incorrect, but I hope it can help (it's on JS):

let C = [1, 3, 1, 4, 2, 3, 5, 4],
  Y = 4;

function solution(X, A) {

  var leaves = [];
  var i = 0;
  var result = -1;

  for (i = 0; i < A.length; i++) {
    if (typeof leaves[A[i]] == 'undefined') {
      leaves[A[i]] = i;
    }
  }

  if (leaves.length <= X) {
    return -1;
  }

  for (i = 1; i <= X; i++) {
    if (typeof leaves[i] == 'undefined') {
      return -1;
    } else {
      result = Math.max(result, leaves[i]);
    }
  }

  return result;
}

console.log(solution(Y, C));

I don't need a solution to the task, I just need its clearer explanation, i.e what I have to do to solve it.

• javascript

• What is this code supposed to acplish? What would be the expected result for this input? What happens instead? Your variable names are pletely unhelpful. – Máté Safranka Commented Apr 23, 2018 at 9:42
• what is FrogRiverOne, and what should your code do? – Nina Scholz Commented Apr 23, 2018 at 9:43
• "I've googled and asked people about this question" what question? – Yury Tarabanko Commented Apr 23, 2018 at 9:43
• i dont understand the problem, but here is the description: app.codility./programmers/lessons/4-counting_elements/… – Nina Scholz Commented Apr 23, 2018 at 9:48
• answer1 and answer2, you can get answer. – Durga Commented Apr 23, 2018 at 9:50

12 Answers 12

Simple 100% solution using Set(). Being a Set can only contain unique values, you can simply iterate through array A, adding the values into a Set until the size of Set equals the integer X, at which point that array key is the solution.

function solution(X, A) {

    let holdValues = new Set();
    for(i=0;i<A.length;i++) {
        holdValues.add(A[i]);
        if(holdValues.size == X)return i;
    }

    return -1;

} 

Here is my solution using javascript:

function solution(X, A) {
    
    let set = new Set();
    
    for(let i = 0; i < A.length; i++){
        if(A[i] <= X){
            set.add(A[i]);
        }
        
        if(set.size === X){
            return i;
        }
    }
    
    return -1;
}

Okay, I figured out your problem for you. Your solution is actually almost correct, but you overplicated the evaluation. All you have to do is initialize a counter variable to 0, and as you iterate over A in the first loop, whenever leaves[A[i]] is undefined, increment this counter. This indicates that a leaf has fallen into a position where there is no leaf yet. If after incrementing the counter is equal to X, it means all positions now have leaves on them, so just return i. If your loop makes it all the way to the end of A, it means there are uncovered positions, so return -1.

My java Solution (100% Correctness, 100% Performance) it has:

Time plexity => O(N)
Space plexity => O(X)

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

    int[] flags = new int[X + 1];
    int sum = 0;

    //Equivalent to all the leafs have fallen. 
    int desiredSum = (X * (X + 1)) / 2; // 1 + 2 + ....n = (n * (n + 1)) / 2

    for (int i = 0; i < A.length; i++) {

        //Verify if the leaf [i] has fallen. 
        if (A[i] <= X && flags[A[i]] == 0) {
            flags[A[i]] = 1;
            sum += A[i];
        }

        //Verify if all the leafs have fallen.
        if (sum == desiredSum) {
            return i;
        }
    }
    return -1;
}

Dictionary is better than Set, the plexity of updating or getting count of dic is O(1).

public func solution(_ x : Int, _ array: [Int]) -> Int {
    let count = array.count
    if x > count { return -1 }
    
    var map = [Int: Bool]()
    for i in 0..<count {
        map[array[i]] = true

        if map.count == x {
            return i
        }
    }
    
    return -1
}

public int solution(int X, int[] A) {
        int sec = -1;
        Set<Integer> checkList = new HashSet<>();
        for(int i = 0; i < A.length; i++) {
            if(A[i] >= 1 && A[i] <= X && checkList.size() < X) {
                checkList.add(A[i]);
                if (sec < i) {
                    sec = i;
                }
            }
            if(checkList.size() == X) {
                break;
            }
        }
        System.out.println(checkList.toString());
        if(checkList.size() < X) {
            sec = -1;
        }
        return sec;
    }

C#:

    using System.Collections.Generic;
...
    public static int solution(int X, int[] A)
            {
                HashSet<int> numbers = new HashSet<int>();
                for (int i = 0; i < A.Length; i++)
                {
                    numbers.Add(A[i]);
    
                    if (numbers.Count==X)
                    {
                        return i;
                    }
                  }
                  return -1;
            }

Solution in Javascript, 100%, O(N), its a bit quicker than using sets on the performance tests. Longest part of this exercise was figuring out the question! Anyway, here's my solution:

function solution(X, A) {
  let newArray = new Array(X + 1);

  for (let i = 0; i < A.length; i++) {
    if (!newArray[A[i]]) {
      newArray[A[i]] = true;
      X--;

      if (!X) {
        return i;
      }
    }
  }
  return -1;
}

this is my solution, java score 100:

    HashSet<Integer> set = new HashSet<Integer>();
    int response = -1;
    for (int i = 0; i < A.length; i++){
        set.add(A[i]);
        if (set.size() == X){
            response = i;
            break;
        }

    }
    return response;

function solution(A) {
    let count = Math.floor(A.length / 2);

    let map = new Map();

    for (let i = 0; i < A.length; i++)
        map.set(A[i], map.has(A[i]) ? map.get(A[i]) + 1 : 1);

    for (const k of map.keys())
        if (map.get(k) > count)
            return A.indexOf(k);

    return -1;
}

Plain java solution

 class Solution {
        public int solution(int X, int[] A) {
            // write your code in Java SE 8
    
            int leaves = A.length;
            int counter=0;
            int [] path= new int [X+1];
    
            for (int t=0; t<leaves; t++){
                if(path[A[t]]!=A[t]){
                path[A[t]]=A[t];
                counter++;
                }
                if (counter==X)
                return t;
    
    
            }  
            return -1;
        }
    }

Swift 100% solution:

public func solution(_ X : Int, _ A : inout [Int]) -> Int {
    var occurs: [Int] = Array(repeating: -1, count: X + 1)

    for i in 0..<A.count {
        if occurs[A[i]] == -1 {
            occurs[A[i]] = i
        }
    }

    occurs.removeFirst()

    if (occurs.filter { $0 == -1 }).count == 0 {
        return occurs.max(by: <) ?? 0
    }

    return -1
}