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

int solution(vector<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 C++

Time spent on task 1 minutes

Notes

not defined yet

Task timeline

Code: 09:13:49 UTC, cpp, 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

// you can use includes, for example:
#include <bits/stdc++.h>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    int len = A.size();
    if(len == 0){
        return -1;
    }else if(len == 1){
        return 0;
    }

    map<int, int> dominator_cnt; 
    for(int i=0; i<len; i++){
        // A[i] not in dominator_cnt
        if(dominator_cnt.find(A[i]) == dominator_cnt.end()){
            dominator_cnt[A[i]] = 1;
            // cout <<"not exist" << A[i] << endl;
            // cout <<"cnt" << dominator_cnt[A[i]] << endl;
        }else{
            dominator_cnt[A[i]]++;
            if(dominator_cnt[A[i]] > len/2){
                return i;
            }
            // cout <<"exist" << A[i] << endl;
            // cout <<"cnt" << dominator_cnt[A[i]] << endl;
        }
    }
    return -1;
}

Analysis

▶

example
example test

✔

Code: 09:13:54 UTC, cpp, 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

// you can use includes, for example:
#include <bits/stdc++.h>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    int len = A.size();
    if(len == 0){
        return -1;
    }else if(len == 1){
        return 0;
    }

    map<int, int> dominator_cnt; 
    for(int i=0; i<len; i++){
        // A[i] not in dominator_cnt
        if(dominator_cnt.find(A[i]) == dominator_cnt.end()){
            dominator_cnt[A[i]] = 1;
            // cout <<"not exist" << A[i] << endl;
            // cout <<"cnt" << dominator_cnt[A[i]] << endl;
        }else{
            dominator_cnt[A[i]]++;
            if(dominator_cnt[A[i]] > len/2){
                return i;
            }
            // cout <<"exist" << A[i] << endl;
            // cout <<"cnt" << dominator_cnt[A[i]] << endl;
        }
    }
    return -1;
}

Analysis

▶

example
example test

✔

Code: 09:13:59 UTC, cpp, 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

// you can use includes, for example:
#include <bits/stdc++.h>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    int len = A.size();
    if(len == 0){
        return -1;
    }else if(len == 1){
        return 0;
    }

    map<int, int> dominator_cnt; 
    for(int i=0; i<len; i++){
        // A[i] not in dominator_cnt
        if(dominator_cnt.find(A[i]) == dominator_cnt.end()){
            dominator_cnt[A[i]] = 1;
            // cout <<"not exist" << A[i] << endl;
            // cout <<"cnt" << dominator_cnt[A[i]] << endl;
        }else{
            dominator_cnt[A[i]]++;
            if(dominator_cnt[A[i]] > len/2){
                return i;
            }
            // cout <<"exist" << A[i] << endl;
            // cout <<"cnt" << dominator_cnt[A[i]] << endl;
        }
    }
    return -1;
}

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

✔