Test results - Codility

Tasks Details

easy

1. Dominator

Find an index of an array such that its value occurs at more than half of indices in the array.

100%

An array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.

For example, consider array A such that

A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3

The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.

Write a function

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

that, given an array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.

For example, given array A such that

A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3

the function may return 0, 2, 4, 6 or 7, as explained above.

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 [−2,147,483,648..2,147,483,647].

Solution

Programming language used Java 8

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 09:58:50 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

// 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) {
        // write your code in Java SE 8
        int[] B = A.clone();
        Arrays.sort(A);
        int length = A.length;
        if(length == 0){
            return -1;
        }
        int count = 0;
        int cadidate = A[length / 2];
        for (int a : A) {
            if (a == cadidate) {
                count++;
            }
        }
        if (count < (length / 2) + 1) {
            return -1;
        }
        for (int i = 0; i < B.length; i++) {
            if (B[i] == cadidate) {
                return i;
            }
        }
        return 0;
    }

}

Analysis

▶

example
example test

✔

Code: 09:59:28 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

// 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) {
        // write your code in Java SE 8
        int[] B = A.clone();
        Arrays.sort(A);
        int length = A.length;
        if(length == 0){
            return -1;
        }
        int count = 0;
        int cadidate = A[length / 2];
        for (int a : A) {
            if (a == cadidate) {
                count++;
            }
        }
        if (count < (length / 2) + 1) {
            return -1;
        }
        for (int i = 0; i < B.length; i++) {
            if (B[i] == cadidate) {
                return i;
            }
        }
        return 0;
    }

}

Analysis

▶

example
example test

✔

Code: 09:59:32 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

// 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) {
        // write your code in Java SE 8
        int[] B = A.clone();
        Arrays.sort(A);
        int length = A.length;
        if(length == 0){
            return -1;
        }
        int count = 0;
        int cadidate = A[length / 2];
        for (int a : A) {
            if (a == cadidate) {
                count++;
            }
        }
        if (count < (length / 2) + 1) {
            return -1;
        }
        for (int i = 0; i < B.length; i++) {
            if (B[i] == cadidate) {
                return i;
            }
        }
        return 0;
    }

}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

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

▶

example
example test

✔

▶

small_nondominator
all different and all the same elements

✔

▶

small_half_positions
half elements the same, and half + 1 elements the same

✔

▶

small
small test

✔

▶

small_pyramid
decreasing and plateau, small

✔

▶

extreme_empty_and_single_item
empty and single element arrays

✔

▶

extreme_half1
array with exactly N/2 values 1, N even + [0,0,1,1,1]

✔

▶

extreme_half2
array with exactly floor(N/2) values 1, N odd + [0,0,1,1,1]

✔

▶

extreme_half3
array with exactly ceil(N/2) values 1 + [0,0,1,1,1]

✔

▶

medium_pyramid
decreasing and plateau, medium

✔

▶

large_pyramid
decreasing and plateau, large

✔

▶

medium_random
random test with dominator, N = 10,000

✔

▶

large_random
random test with dominator, N = 100,000

✔