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#

Time spent on task 15 minutes

Notes

not defined yet

Task timeline

Code: 08:51:33 UTC, cs, verify, result: Failed

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

using System;
// you can also use other imports, for example:
using System.Collections.Generic;

// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        int n = A.Length;
        int size =0;
        int value=0;
        Stack<int> s = new Stack<int>();
        
        for (int i=0; i<n; i++)
        {
            if(size ==0)
            {
                size +=1;
                s.Push(A[i]);
            }
            else
            {
                if (s.Peek() != A[i]) size -=1;
                else size +=1;
            }           
        }
        int candidate = -1;
        if (size>0) candidate = s.Peek();
        int count =0;
        int leader= -1;
        
        for (int i=0; i<n; i++)
        {
         if (A[i] == candidate) count +=1;
         if (count > n/2) leader = candidate;
        }
        
        int equiLeaders=0;
        int leaders=0;
        
        
        for (int i=0; i<n; i++)
        {
            if (A[i] == leader) leaders++;
            if (leaders > (i+1)/2 && leaders > (n-1-i)/2) equiLeaders++;
               
        }        
        
        return equiLeaders;
    }
}

Analysis

▶

example
example test

✘

WRONG ANSWER
got 4 expected 2

Code: 08:56:37 UTC, cs, verify, result: Failed

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

using System;
// you can also use other imports, for example:
using System.Collections.Generic;

// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        int n = A.Length;
        int size =0;
        int value=0;
        Stack<int> s = new Stack<int>();
        
        for (int i=0; i<n; i++)
        {
            if(size ==0)
            {
                size +=1;
                s.Push(A[i]);
            }
            else
            {
                if (s.Peek() != A[i]) size -=1;
                else size +=1;
            }           
        }
        int candidate = -1;
        if (size>0) candidate = s.Peek();
        int count =0;
        int leader= -1;
        
        for (int i=0; i<n; i++)
        {
         if (A[i] == candidate) count +=1;
         if (count > n/2) leader = candidate;
        }
        
        int equiLeaders=0;
        int leaders=0;
        
        
        for (int i=0; i<n; i++)
        {
            if (A[i] == leader) leaders++;
            if (leaders > (i+1)/2 && leaders > (n-1-i)/2) equiLeaders+=1;             
        }        
        
        return equiLeaders;
    }
}

Analysis

▶

example
example test

✘

WRONG ANSWER
got 4 expected 2

Code: 08:58:59 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

using System;
// you can also use other imports, for example:
using System.Collections.Generic;

// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        int n = A.Length;
        int size =0;
        int value=0;
        Stack<int> s = new Stack<int>();
        
        for (int i=0; i<n; i++)
        {
            if(size ==0)
            {
                size +=1;
                s.Push(A[i]);
            }
            else
            {
                if (s.Peek() != A[i]) size -=1;
                else size +=1;
            }           
        }
        int candidate = -1;
        if (size>0) candidate = s.Peek();
        int count =0;
        int leader= -1;
        
        for (int i=0; i<n; i++)
        {
         if (A[i] == candidate) count +=1;
         if (count > n/2) leader = candidate;
        }
        
        int equiLeaders=0;
        int leaders=0;
        
        
        for (int i=0; i<n; i++)
        {
            if (A[i] == leader) leaders++;
            if (leaders > (i+1)/2 && count -leaders > (n-1-i)/2) equiLeaders++;             
        }        
        
        return equiLeaders;
    }
}

Analysis

▶

example
example test

✔

Code: 09:03:50 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

using System;
// you can also use other imports, for example:
using System.Collections.Generic;

// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        int n = A.Length;
        int size =0;
        int value=0;
        Stack<int> s = new Stack<int>();
        
        for (int i=0; i<n; i++)
        {
            if(size ==0)
            {
                size +=1;
                s.Push(A[i]);
            }
            else
            {
                if (s.Peek() != A[i]) size -=1;
                else size +=1;
            }           
        }
        int candidate = -1;
        if (size>0) candidate = s.Peek();
        int count =0;
        int leader= -1;
        
        for (int i=0; i<n; i++)
        {
         if (A[i] == candidate) count +=1;
         if (count > n/2) leader = candidate;
        }
        
        int equiLeaders=0;
        int leaders=0;
        
        for (int i=0; i<n; i++)
        {
            if (A[i] == leader) leaders++;
            if (leaders > (i+1)/2 && count-leaders > (n-1-i)/2) equiLeaders++;             
        }        
        
        return equiLeaders;
    }
}

Analysis

▶

example
example test

✔

Code: 09:03:52 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

using System;
// you can also use other imports, for example:
using System.Collections.Generic;

// you can use Console.WriteLine for debugging purposes, e.g.
// Console.WriteLine("this is a debug message");

class Solution {
    public int solution(int[] A) {
        int n = A.Length;
        int size =0;
        int value=0;
        Stack<int> s = new Stack<int>();
        
        for (int i=0; i<n; i++)
        {
            if(size ==0)
            {
                size +=1;
                s.Push(A[i]);
            }
            else
            {
                if (s.Peek() != A[i]) size -=1;
                else size +=1;
            }           
        }
        int candidate = -1;
        if (size>0) candidate = s.Peek();
        int count =0;
        int leader= -1;
        
        for (int i=0; i<n; i++)
        {
         if (A[i] == candidate) count +=1;
         if (count > n/2) leader = candidate;
        }
        
        int equiLeaders=0;
        int leaders=0;
        
        for (int i=0; i<n; i++)
        {
            if (A[i] == leader) leaders++;
            if (leaders > (i+1)/2 && count-leaders > (n-1-i)/2) equiLeaders++;             
        }        
        
        return equiLeaders;
    }
}

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

✔