1. TapeEquilibrium - Codility solution

Source:

Task description

A non-empty zero-indexed 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] = 3We 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 zero-indexed 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] = 3the function should return 1, as explained above.
Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Solution

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

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

class Solution {
    public int solution(int[] A) {
        // write your code in C# 6.0 with .NET 4.5 (Mono)
        int count = A.Length;
        if(count>100000||count<=1){
            throw new Exception("Not a valid range");
        }
        int []temp= new int[count];
        int sum =0;
        for(int i=1;i<A.Length;i++){
            sum+=A[i];
        }
        int diff = A[0]-sum;
        if(diff<0){
            diff = -diff;
        }
        temp[0]= diff;
        int firstsum=A[0];
        for(int i=1; i<A.Length-1;i++){
            sum -= A[i];
            firstsum+=A[i];
            int diff2 = firstsum-sum;
            if(diff2<0){
                diff2=-diff2;    
            }
            temp[i] = diff2;
        }
        int min=temp[0];
        for(int i=1; i<temp.Length-1;i++){
                if(temp[i]<min){
                    min=temp[i];    
                }
        }
        
        return min;
    }    
}

1. CountDiv Compute number of integers divisible by k in range [a..b].

Source: https://codility.com/demo/take-sample-test/count_div/

Task description

Write a function:

class Solution { public int solution(int A, int B, int K); }

that, given three integers A, B and K, returns the number of integers within the range [A..B] that are divisible by K, i.e.:

{ i : A ≤ i ≤ B, i mod K = 0 }

For example, for A = 6, B = 11 and K = 2, your function should return 3, because there are three numbers divisible by 2 within the range [6..11], namely 6, 8 and 10.
Assume that:

• A and B are integers within the range [0..2,000,000,000];
• K is an integer within the range [1..2,000,000,000];
• A ≤ B.

Complexity:

• expected worst-case time complexity is O(1);
• expected worst-case space complexity is O(1).

Solution:

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 B, int K) { if(K==0) return 0; int left = A/K; int right = B/K; int modLeft = A%K; int modRight = B%K; if((modLeft==00 || modRight==0)) return (B-A)/K +1 ; return right-left; } }

Javascript: Get query string value by name

function getParameterByName(name) {
    name = name.replace(/[\[]/, "\\\[").replace(/[\]]/, "\\\]");
    var regex = new RegExp("[\\?&]" + name + "=([^&#]*)"),
        results = regex.exec(location.search);
    return results == null ? "" : decodeURIComponent(results[1].replace(/\+/g, " "));
}