Codility Lesson 10

10.1 - Count Factors

A positive integer D is a factor of a positive integer N if there exists an integer M such that N = D * M.

For example, 6 is a factor of 24, because M = 4 satisfies the above condition (24 = 6 * 4).

Write a function:

class Solution 
{
  public int solution(int N); 
}

that, given a positive integer N, returns the number of its factors.

For example, given N = 24, the function should return 8, because 24 has 8 factors, namely 1, 2, 3, 4, 6, 8, 12, 24. There are no other factors of 24.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [1..2,147,483,647].

using System;

/**
 * 10.1 - Count Factors
 * Paulo Santos
 * 15.Dec.2022
 */
class Solution {
    public int solution(int N) {

        var res = 0;
        var root = (int)Math.Sqrt(N);
        if (Math.Pow(root, 2) != N) 
            root += 1;
        else 
            res += 1;
        
        for (var i = 1; i < root; i++)
            if ((N % i) == 0)
                res += 2;
        
        return res;
    }
}

---

10.2 - Min Perimeter Rectangle

An integer N is given, representing the area of some rectangle.

The area of a rectangle whose sides are of length A and B is A * B, and the perimeter is 2 * (A + B).

The goal is to find the minimal perimeter of any rectangle whose area equals N. The sides of this rectangle should be only integers.

For example, given integer N = 30, rectangles of area 30 are:

• (1, 30), with a perimeter of 62,
• (2, 15), with a perimeter of 34,
• (3, 10), with a perimeter of 26,
• (5, 6), with a perimeter of 22.

Write a function:

class Solution 
{
  public int solution(int N); 
}

that, given an integer N, returns the minimal perimeter of any rectangle whose area is exactly equal to N.

For example, given an integer N = 30, the function should return 22, as explained above.

Write an efficient algorithm for the following assumptions:

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

using System;

/**
 * 10.2 - Min Perimeter Rectangle
 * Paulo Santos
 * 15.Dec.2022
 */
class Solution {
    public int solution(int N) {

        /*
         * Math gives that the min
         * perimeter for a given area N
         * is Sqrt(N).
         */
        var root = (int)Math.Sqrt(N);
        if (Math.Pow(root, 2) == N) 
            return (4 * root);
        if ((root * (root + 1)) == N)            
            return (4 * root + 2);
        
        var min = int.MaxValue;
        for(var i = 1; i <= root; i++) 
            if ((i * (N / i)) == N) 
                min = Math.Min(min, 2 * (i + (N / i)));
            
        return min;
    }
}

---

10.3 - Flags

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

A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N − 1 and A[P − 1] < A[P] > A[P + 1].

For example, the following array A:

    A[0] = 1
    A[1] = 5
    A[2] = 3
    A[3] = 4
    A[4] = 3
    A[5] = 4
    A[6] = 1
    A[7] = 2
    A[8] = 3
    A[9] = 4
    A[10] = 6
    A[11] = 2

has exactly four peaks: elements 1, 3, 5 and 10.

You are going on a trip to a range of mountains whose relative heights are represented by array A, as shown in a figure below. You have to choose how many flags you should take with you. The goal is to set the maximum number of flags on the peaks, according to certain rules.

Flags can only be set on peaks. What's more, if you take K flags, then the distance between any two flags should be greater than or equal to K. The distance between indices P and Q is the absolute value |P − Q|.

For example, given the mountain range represented by array A, above, with N = 12, if you take:

• two flags, you can set them on peaks 1 and 5;
• three flags, you can set them on peaks 1, 5 and 10;
• four flags, you can set only three flags, on peaks 1, 5 and 10.

You can therefore set a maximum of three flags in this case.

Write a function:

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

that, given a non-empty array A of N integers, returns the maximum number of flags that can be set on the peaks of the array.

For example, the following array A:

    A[0] = 1
    A[1] = 5
    A[2] = 3
    A[3] = 4
    A[4] = 3
    A[5] = 4
    A[6] = 1
    A[7] = 2
    A[8] = 3
    A[9] = 4
    A[10] = 6
    A[11] = 2

the function should return 3, as explained above.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [1..400,000];
• each element of array A is an integer within the range [0..1,000,000,000].

using System;

/**
 * 10.3 - Flags
 * Paulo Santos
 * 15.Dec.2022
 */
class Solution {
    public int solution(int[] A) {

        /*
         * List all the peakes
         */
        var len = A.Length;
        var peaks = new int[len];
        var cnt = 0;
        for (var i = len - 2; i > 0; i--) {
            if ((A[i - 1] < A[i]) && (A[i] > A[i + 1])) {
                peaks[i] = i;
                cnt += 1;
            }
            else 
                peaks[i] = peaks[i + 1];
        }

        /*
         * Plant the flags
         */
        for (var flags = peaks.Length - 1; flags > 0; flags--) {
            var planted = 0;
            var i = 1;
            while ((i < len) && 
                   (peaks[i] != 0) && 
                   (planted < flags)) {
                i = peaks[i];
                planted += 1;
                i += flags;
            }
            if (flags == planted)
                return planted;
        }
        
        return 0;
    }
}

---

10.4 - Peaks

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

A peak is an array element which is larger than its neighbors. More precisely, it is an index P such that 0 < P < N − 1, A[P − 1] < A[P] and A[P] > A[P + 1].

For example, the following array A:

    A[0] = 1
    A[1] = 2
    A[2] = 3
    A[3] = 4
    A[4] = 3
    A[5] = 4
    A[6] = 1
    A[7] = 2
    A[8] = 3
    A[9] = 4
    A[10] = 6
    A[11] = 2

has exactly three peaks: 3, 5, 10.

We want to divide this array into blocks containing the same number of elements. More precisely, we want to choose a number K that will yield the following blocks:

• A[0], A[1], ..., A[K − 1],
• A[K], A[K + 1], ..., A[2K − 1],
• ...
• A[N − K], A[N − K + 1], ..., A[N − 1].

What's more, every block should contain at least one peak. Notice that extreme elements of the blocks (for example A[K − 1] or A[K]) can also be peaks, but only if they have both neighbors (including one in an adjacent blocks).

The goal is to find the maximum number of blocks into which the array A can be divided.

Array A can be divided into blocks as follows:

• one block (1, 2, 3, 4, 3, 4, 1, 2, 3, 4, 6, 2). This block contains three peaks.
• two blocks (1, 2, 3, 4, 3, 4) and (1, 2, 3, 4, 6, 2). Every block has a peak.
• three blocks (1, 2, 3, 4), (3, 4, 1, 2), (3, 4, 6, 2). Every block has a peak. Notice in particular that the first block (1, 2, 3, 4) has a peak at A[3], because A[2] < A[3] > A[4], even though A[4] is in the adjacent block.

However, array A cannot be divided into four blocks, (1, 2, 3), (4, 3, 4), (1, 2, 3) and (4, 6, 2), because the (1, 2, 3) blocks do not contain a peak. Notice in particular that the (4, 3, 4) block contains two peaks: A[3] and A[5].

The maximum number of blocks that array A can be divided into is three.

Write a function:

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

that, given a non-empty array A consisting of N integers, returns the maximum number of blocks into which A can be divided.

If A cannot be divided into some number of blocks, the function should return 0.

For example, given:

    A[0] = 1
    A[1] = 2
    A[2] = 3
    A[3] = 4
    A[4] = 3
    A[5] = 4
    A[6] = 1
    A[7] = 2
    A[8] = 3
    A[9] = 4
    A[10] = 6
    A[11] = 2

the function should return 3, 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 [0..1,000,000,000].

using System;
using System.Collections.Generic;

/**
 * 10.4 - Peaks
 * Paulo Santos
 * 15.Dec.2022
 */
class Solution {
    public int solution(int[] A) {

        /*
         * List all the peakes
         */
        var len = A.Length;
        var peaks = new List<int>();
        for (var i = 1 ; i < len - 1; i++) 
            if ((A[i - 1] < A[i]) && (A[i] > A[i + 1])) 
                peaks.Add(i);
        
        if (peaks.Count == 0)
            return 0;

        for (var size = peaks.Count; size >= 0; size--) {
            if ((A.Length % size) == 0) {
                var blockSize = A.Length / size;
                var found = new bool[size];
                var foundCnt = 0;
                foreach(var p in peaks) {
                    var blkNum = p / blockSize;
                    if (!found[blkNum]) {
                        found[blkNum] = true;
                        foundCnt += 1;
                    }
                }
                
                if (foundCnt == size) 
                    return size;
            }
        }
        return 0;
    }
}