Test results - Codility

Tasks Details

easy

1. EquiLeader

Find the index S such that the leaders of the sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N - 1] are the same.

100%

A non-empty array A consisting of N integers is given.

The leader of this array is the value that occurs in more than half of the elements of A.

An equi leader is an index S such that 0 ≤ S < N − 1 and two sequences A[0], A[1], ..., A[S] and A[S + 1], A[S + 2], ..., A[N − 1] have leaders of the same value.

For example, given array A such that:

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

we can find two equi leaders:

0, because sequences: (4) and (3, 4, 4, 4, 2) have the same leader, whose value is 4.

2, because sequences: (4, 3, 4) and (4, 4, 2) have the same leader, whose value is 4.

The goal is to count the number of equi leaders.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the number of equi leaders.

For example, given:

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

the function should return 2, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000];

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

Solution

Programming language used C#

Total time used 2 minutes

Effective time used 2 minutes

Notes

not defined yet

Task timeline

Code: 06:01:57 UTC, cs, 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 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

using System;
// you can also use other imports, for example:
using System.Collections.Generic;
using System.Collections;
// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        var N = A.Length;
        int best = 0; 
        int count = 0;
        var hash = new Dictionary<int,int>();
        
        // search leader
        foreach (int element in A)
        {
            // count element at the same time
            if(hash.ContainsKey(element))
                hash[element]++;
            else
                hash.Add(element,1);
            
            if (count == 0)
            {
                best = element;
                count = 1;
            }
            else
            {
                // Vote current choice up or down
                count += (best == element) ? 1 : -1;
            }
        }
        
        int leaderCount = hash[best];
        // avoid searching if it's not a leader
        if ( leaderCount < N/2)
            return 0;
            
        // search equi
        int lcsf = 0;
        int equi = 0;
        for(int index =0; index < N; index++)
        {
            if(A[index] == best)
            {
                lcsf++;
            }
            if(lcsf > (index+1)/2 &&  leaderCount-lcsf > (N-index-1)/2)
            {
                equi++;
            }
        }
    
        return equi;
    }
}

Analysis

▶

example
example test

✔

Code: 06:02:03 UTC, cs, 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 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

using System;
// you can also use other imports, for example:
using System.Collections.Generic;
using System.Collections;
// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        var N = A.Length;
        int best = 0; 
        int count = 0;
        var hash = new Dictionary<int,int>();
        
        // search leader
        foreach (int element in A)
        {
            // count element at the same time
            if(hash.ContainsKey(element))
                hash[element]++;
            else
                hash.Add(element,1);
            
            if (count == 0)
            {
                best = element;
                count = 1;
            }
            else
            {
                // Vote current choice up or down
                count += (best == element) ? 1 : -1;
            }
        }
        
        int leaderCount = hash[best];
        // avoid searching if it's not a leader
        if ( leaderCount < N/2)
            return 0;
            
        // search equi
        int lcsf = 0;
        int equi = 0;
        for(int index =0; index < N; index++)
        {
            if(A[index] == best)
            {
                lcsf++;
            }
            if(lcsf > (index+1)/2 &&  leaderCount-lcsf > (N-index-1)/2)
            {
                equi++;
            }
        }
    
        return equi;
    }
}

Analysis

▶

example
example test

✔

Code: 06:02:05 UTC, cs, 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 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

using System;
// you can also use other imports, for example:
using System.Collections.Generic;
using System.Collections;
// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        var N = A.Length;
        int best = 0; 
        int count = 0;
        var hash = new Dictionary<int,int>();
        
        // search leader
        foreach (int element in A)
        {
            // count element at the same time
            if(hash.ContainsKey(element))
                hash[element]++;
            else
                hash.Add(element,1);
            
            if (count == 0)
            {
                best = element;
                count = 1;
            }
            else
            {
                // Vote current choice up or down
                count += (best == element) ? 1 : -1;
            }
        }
        
        int leaderCount = hash[best];
        // avoid searching if it's not a leader
        if ( leaderCount < N/2)
            return 0;
            
        // search equi
        int lcsf = 0;
        int equi = 0;
        for(int index =0; index < N; index++)
        {
            if(A[index] == best)
            {
                lcsf++;
            }
            if(lcsf > (index+1)/2 &&  leaderCount-lcsf > (N-index-1)/2)
            {
                equi++;
            }
        }
    
        return equi;
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example
example test

✔

▶

single
single element

✔

▶

double
two elements

✔

▶

simple
simple test

✔

▶

small_random
small random test with two values, length = ~100

✔

▶

small
random + 200 * [MIN_INT] + random ,length = ~300

✔

▶

large_random
large random test with two values, length = ~50,000

✔

▶

large
random(0,1) + 50000 * [0] + random(0, 1), length = ~100,000

✔

▶

large_range
1, 2, ..., N, length = ~100,000

✔

▶

extreme_large
all the same values

✔