Codility Lesson 03

Lesson 3

3.1 - Frog Jump

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

Write a function:

class Solution 
{ 
  public int solution(int X, int Y, int D); 
}

that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:

 X = 10
 Y = 85
 D = 30

the function should return 3, because the frog will be positioned as follows:

after the first jump, at position 10 + 30 = 40

after the second jump, at position 10 + 30 + 30 = 70

after the third jump, at position 10 + 30 + 30 + 30 = 100

Write an efficient algorithm for the following assumptions:

• X, Y and D are integers within the range [1..1,000,000,000];
• X ≤ Y.

using System;

/* Lesson 3.1 - Odd Occurrences in Array
 * Paulo Santos
 * 24.Nov.2022
 */
class Solution {
    public int solution(int X, int Y, int D) {
		
		/*
		 * Check the input
		 */
		if ((X < 1 || X > 1000000000) ||
			(Y < 1 || X > 1000000000) ||
			(D < 1 || X > 1000000000))
			throw new ArgumentException();
		
		/*
		 * Calculate the answer
		 */
        var ans = ((double)(Y - X) / D);
        if (ans != Math.Floor((double)ans))
            return (int)(Math.Floor(ans) + 1);

        return (int)ans;
    }
}

---

3.2 - Perm Missing Element

An array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.

Your goal is to find that missing element.

Write a function:

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

that, given an array A, returns the value of the missing element.

For example, given array A such that:

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

the function should return 4, as it is the missing element.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [0..100,000];
• the elements of A are all distinct;
• each element of array A is an integer within the range [1..(N + 1)].

using System;

/* Lesson 3.2 - Perm Missing Element
 * Paulo Santos
 * 24.Nov.2022
 */
class Solution {
    public int solution(int[] A) {

        /*
         * Check the input
         */
        if (A == null || A.Length == 0)
            return 1;

        Array.Sort(A);

        var ans = 0;
        foreach(var i in A) {
            if((++ans) != i)
                return (ans);
        }

        return (ans + 1);
    }
}

---

3.3 - Tape Equilibrium

A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.

Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].

The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])| In other words, it is the absolute difference between the sum of the first part and the sum of the second part.

For example, consider array A such that:

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

We can split this tape in four places:

P = 1, difference = |3 − 10| = 7

P = 2, difference = |4 − 9| = 5

P = 3, difference = |6 − 7| = 1

P = 4, difference = |10 − 3| = 7

Write a function:

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

that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.

For example, given:

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

the function should return 1, as explained above.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [2..100,000];
• each element of array A is an integer within the range [−1,000..1,000].

using System;
using System.Linq;

/* Lesson 3.3 - Tape Equilibrium
 * Paulo Santos
 * 24.Nov.2022
 */
class Solution {
    public int solution(int[] A) {

        /*
         * Check the input
         */
        if (A == null)
            throw new ArgumentNullException();

        var sum1 = A.Sum();
        var sum2 = 0;
        var ans = int.MaxValue;
        for(var i = 0; i < A.Length - 1; i++) {
            sum2 += A[i];
            sum1 -= A[i];
            var dif = Math.Abs(sum1 - sum2);
            ans = (dif < ans) ? dif : ans;
        }

        return ans;

    }
}