Test results - Codility

Tasks Details

easy

1. Distinct

Compute number of distinct values in an array.

100%

Write a function

class Solution { public int solution(int[] A); }

that, given an array A consisting of N integers, returns the number of distinct values in array A.

For example, given array A consisting of six elements such that:

A[0] = 2 A[1] = 1 A[2] = 1 A[3] = 2 A[4] = 3 A[5] = 1

the function should return 3, because there are 3 distinct values appearing in array A, namely 1, 2 and 3.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [0..100,000];

each element of array A is an integer within the range [−1,000,000..1,000,000].

Solution

Programming language used Java 8

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 06:48:25 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

// 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[] A) {
		int N = A.length;
		if(0 == N) return 0;

		quickSort(A, 0, N-1);
		int distinctNum = 1;
		for (int i=1; i<N; i++){
			if(0 != (A[i -1] ^ A[i])) distinctNum++;
		}
		return distinctNum;
	}

	public void quickSort(int[] arr, int begin, int end) {
		int middle = (begin + end) / 2;
		int pivot = arr[middle];
		int left = begin;
		int right = end;
		int temp;

		while (left < right) { //left와 right가 만나면 루프 종료
			while (arr[left] < pivot) left++;       //left : pivot 값 보다 큰 값을 찾기 위해 이동
			while (arr[right] > pivot) right--;     //right : pivot 값 보다 작은 값을 찾기 위해 이동

			if (left <= right) { //left가 right보다 같거나 작으면 서로 값 교환해줌
				temp = arr[left];
				arr[left] = arr[right];
				arr[right] = temp;
				left++;
				right--;
			}
		}
		//부분 분할 실행
		if (begin < right) quickSort(arr, begin, right);
		if (end > left) quickSort(arr, left, end);
	}
}

Analysis

▶

example1
example test, positive answer

✔

Code: 06:48:35 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

// 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[] A) {
		int N = A.length;
		if(0 == N) return 0;

		quickSort(A, 0, N-1);
		int distinctNum = 1;
		for (int i=1; i<N; i++){
			if(0 != (A[i -1] ^ A[i])) distinctNum++;
		}
		return distinctNum;
	}

	public void quickSort(int[] arr, int begin, int end) {
		int middle = (begin + end) / 2;
		int pivot = arr[middle];
		int left = begin;
		int right = end;
		int temp;

		while (left < right) { //left와 right가 만나면 루프 종료
			while (arr[left] < pivot) left++;       //left : pivot 값 보다 큰 값을 찾기 위해 이동
			while (arr[right] > pivot) right--;     //right : pivot 값 보다 작은 값을 찾기 위해 이동

			if (left <= right) { //left가 right보다 같거나 작으면 서로 값 교환해줌
				temp = arr[left];
				arr[left] = arr[right];
				arr[right] = temp;
				left++;
				right--;
			}
		}
		//부분 분할 실행
		if (begin < right) quickSort(arr, begin, right);
		if (end > left) quickSort(arr, left, end);
	}
}

Analysis

▶

example1
example test, positive answer

✔

Code: 06:48:41 UTC, java, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

// 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[] A) {
		int N = A.length;
		if(0 == N) return 0;

		quickSort(A, 0, N-1);
		int distinctNum = 1;
		for (int i=1; i<N; i++){
			if(0 != (A[i -1] ^ A[i])) distinctNum++;
		}
		return distinctNum;
	}

	public void quickSort(int[] arr, int begin, int end) {
		int middle = (begin + end) / 2;
		int pivot = arr[middle];
		int left = begin;
		int right = end;
		int temp;

		while (left < right) { //left와 right가 만나면 루프 종료
			while (arr[left] < pivot) left++;       //left : pivot 값 보다 큰 값을 찾기 위해 이동
			while (arr[right] > pivot) right--;     //right : pivot 값 보다 작은 값을 찾기 위해 이동

			if (left <= right) { //left가 right보다 같거나 작으면 서로 값 교환해줌
				temp = arr[left];
				arr[left] = arr[right];
				arr[right] = temp;
				left++;
				right--;
			}
		}
		//부분 분할 실행
		if (begin < right) quickSort(arr, begin, right);
		if (end > left) quickSort(arr, left, end);
	}
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N*log(N)) or O(N)

▶

example1
example test, positive answer

✔

▶

extreme_empty
empty sequence

✔

▶

extreme_single
sequence of one element

✔

▶

extreme_two_elems
sequence of three distinct elements

✔

▶

extreme_one_value
sequence of 10 equal elements

✔

▶

extreme_negative
sequence of negative elements, length=5

✔

▶

extreme_big_values
sequence with big values, length=5

✔

▶

medium1
chaotic sequence of value sfrom [0..1K], length=100

✔

▶

medium2
chaotic sequence of value sfrom [0..1K], length=200

✔

▶

medium3
chaotic sequence of values from [0..10], length=200

✔

▶

large1
chaotic sequence of values from [0..100K], length=10K

✔

▶

large_random1
chaotic sequence of values from [-1M..1M], length=100K

✔

▶

large_random2
another chaotic sequence of values from [-1M..1M], length=100K

✔