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 2 minutes

Notes

not defined yet

Task timeline

Code: 03:56:11 UTC, java, verify, result: Passed

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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) {
        Stack<Integer> item = new Stack<>();
        for(int i=0; i<A.length; i++){
            if(item.isEmpty() || item.peek() == A[i]){
                item.push(A[i]);
            }else{
                item.pop();
            }
        }
        int cnt=0;
        int pos = -1;
        if(item.isEmpty()){
            return -1;
        }
        int num = item.pop();
        for(int i=0; i<A.length; i++){
            if(A[i] == num){
                if(pos == -1){
                    pos = i;
                }
                cnt++;
            }
        }
        if(A.length / 2 < cnt) {
            return pos;
        }else{
            return -1;
        }
        // write your code in Java SE 8
    }
}

Analysis

▶

example
example test

✔

Code: 03:56:17 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

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) {
        Stack<Integer> item = new Stack<>();
        for(int i=0; i<A.length; i++){
            if(item.isEmpty() || item.peek() == A[i]){
                item.push(A[i]);
            }else{
                item.pop();
            }
        }
        int cnt=0;
        int pos = -1;
        if(item.isEmpty()){
            return -1;
        }
        int num = item.pop();
        for(int i=0; i<A.length; i++){
            if(A[i] == num){
                if(pos == -1){
                    pos = i;
                }
                cnt++;
            }
        }
        if(A.length / 2 < cnt) {
            return pos;
        }else{
            return -1;
        }
        // write your code in Java SE 8
    }
}

Analysis

▶

example
example test

✔

Code: 03:56:21 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

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) {
        Stack<Integer> item = new Stack<>();
        for(int i=0; i<A.length; i++){
            if(item.isEmpty() || item.peek() == A[i]){
                item.push(A[i]);
            }else{
                item.pop();
            }
        }
        int cnt=0;
        int pos = -1;
        if(item.isEmpty()){
            return -1;
        }
        int num = item.pop();
        for(int i=0; i<A.length; i++){
            if(A[i] == num){
                if(pos == -1){
                    pos = i;
                }
                cnt++;
            }
        }
        if(A.length / 2 < cnt) {
            return pos;
        }else{
            return -1;
        }
        // write your code in Java SE 8
    }
}

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

✔