Learning Assistance application privacy policy

Privacy Policy

Ahmed Attallah Abdelbaky built the Learning Assistance app as a Free app. This Application is provided by Ahmed Attallah Abdelbaky at no cost and is intended for use as is.
This page is used to inform visitors regarding my policies with the collection, use, and disclosure of Personal Information if anyone decided to use my Application.
If you choose to use my Application, then you agree to the collection and use of information in relation to this policy. The Personal Information that I collect is used for providing and improving the Application. I will not use or share your information with anyone except as described in this Privacy Policy.
The terms used in this Privacy Policy have the same meanings as in our Terms and Conditions, which is accessible at Learning Assistance unless otherwise defined in this Privacy Policy.
Information Collection and Use
The information that I request will be retained on your device and is not collected by me in any way.
The app does use third party services that may collect information used to identify you.
Link to privacy policy of third party service providers used by the app
Log Data
I want to inform you that whenever you use my Service, in a case of an error in the app I collect data and information (through third party products) on your phone called Log Data. This Log Data may include information such as your device Internet Protocol (“IP”) address, device name, operating system version, the configuration of the app when utilizing my Service, the time and date of your use of the Service, and other statistics.
Cookies
Cookies are files with a small amount of data that are commonly used as anonymous unique identifiers. These are sent to your browser from the websites that you visit and are stored on your device's internal memory.
This Service does not use these “cookies” explicitly. However, the app may use third party code and libraries that use “cookies” to collect information and improve their services. You have the option to either accept or refuse these cookies and know when a cookie is being sent to your device. If you choose to refuse our cookies, you may not be able to use some portions of this Service.
Service Providers
I may employ third-party companies and individuals due to the following reasons:
  • To facilitate our Service;
  • To provide the Service on our behalf;
  • To perform Service-related services; or
  • To assist us in analyzing how our Service is used.
I want to inform users of this Service that these third parties have access to your Personal Information. The reason is to perform the tasks assigned to them on our behalf. However, they are obligated not to disclose or use the information for any other purpose.
Security
I value your trust in providing us your Personal Information, thus we are striving to use commercially acceptable means of protecting it. But remember that no method of transmission over the internet, or method of electronic storage is 100% secure and reliable, and I cannot guarantee its absolute security.
Links to Other Sites
This Service may contain links to other sites. If you click on a third-party link, you will be directed to that site. Note that these external sites are not operated by me. Therefore, I strongly advise you to review the Privacy Policy of these websites. I have no control over and assume no responsibility for the content, privacy policies, or practices of any third-party sites or services.
Children’s Privacy
I do not knowingly collect personally identifiable information from children under 13. In the case I discover that a child under 13 has provided me with personal information, I immediately delete this from our servers. If you are a parent or guardian and you are aware that your child has provided us with personal information, please contact me so that I will be able to do necessary actions.
Changes to This Privacy Policy
I may update our Privacy Policy from time to time. Thus, you are advised to review this page periodically for any changes. I will notify you of any changes by posting the new Privacy Policy on this page. These changes are effective immediately after they are posted on this page.
Contact Us
If you have any questions or suggestions about my Privacy Policy, do not hesitate to contact me at dev.ahmad3ttallah@yahoo.com.

OddOccurrencesInArray

A non-empty zero-indexed array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

For example, in array A such that:

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

the elements at indexes 0 and 2 have value 9,the elements at indexes 1 and 3 have value 3,the elements at indexes 4 and 6 have value 9,the element at index 5 has value 7 and is unpaired.

Write a function:
class Solution { public int solution(int[] A); }

that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.

For example, given array A such that:

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

the function should return 7, as explained in the example above.

 ----------- Solution

1- If you know that XOR function you should be aware of the following
A xor A ==> 0 
So every pair of array items will cancel each other, which means that if you did xor for all array items you will get the odd item.

public int solution(int[] A) {
    int notFound = 0;
    for(int i=0; i < A.length; i++)
        notFound ^= A[i];
    return notFound;
}







BinaryGap

BinaryGap داخل رقم صحيح موجب N هو أكبر عدد من الأصفار محاطين بالرقم واحد في الطرفين داخل التمثيل الثنائي للرقم N .
على سبيل المثال الرقم 9 ، التمثيل الثنائي له هو 1001 يحتوي على BinaryGap بالطول 2 . و الرقم 529 الممثل ثنائياً بـ 1000010001 يحتوي اثنين من الـ BinaryGap واحد بالطول 4 و آخر بالطول 3 . و الرقم 20 الممثل بـ 10100 يحتوي BinaryGap واحدة بالطول 1 . و الرقم 15 الممثل بـ 1111 لا يحتوي أياً من BinaryGap .

أكتب دالة :

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

و التي تقبل رقم صحيح موجب N تعطي أكبر BinaryGap داخل التمثيل الثنائي لهذا الرقم . يجب أن تعطي الدالة النتيجة 0 لو أن الرقم N لا يحتوي BinaryGap .

على سبيل المثال:
الرقم N = 1041 إذا تم إدخاله لى الدالة يجب أن تعطي 5 ، لأن التمثيل الثنائي للرقم 1041 هو 10000010001 و أكبر طول لـ BinaryGap داخله هو 5 .

---------------- الحل

١- لكي تحصل على التمثيل الثنائي لرقم ما بالجافا:
Integer.toBinaryString(N);

٢- و باستخدام الـ regex التالي:
(?<=1)(.*?)(?=1)
يمكنك أن تحصل على كل حالات الأصفار في هذا التمثيل الثنائي و منه يمكنك حساب أكبرهم .

الحل كالتالي :

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Solution {
    public static int solution(int N) {
        int max = 0;
        String x= Integer.toBinaryString(N);
        Matcher m = Pattern.compile("(?<=1)(.*?)(?=1)").matcher(x);
        while(m.find()) {
            if(m.group(1).length() > max)
                max = m.group(1).length();
        }
        return max;
    }
}

BinaryGap

A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.

For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps.

Write a function:

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

that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn't contain a binary gap.

For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5.

---------------- Solution

1- For getting the binary code of number using java you can use:
Integer.toBinaryString(N);
 
2- Using regex pattern 
(?<=1)(.*?)(?=1)
you can get all instances between 1s so you can check length of all of them to get the largest one.

The implementation is as below:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Solution {
    public static int solution(int N) {
        int max = 0;
        String x= Integer.toBinaryString(N);
        Matcher m = Pattern.compile("(?<=1)(.*?)(?=1)").matcher(x);
        while(m.find()) {
            if(m.group(1).length() > max)
                max = m.group(1).length();
        }
        return max;
    }
}



شاعر العرب

شاعر العرب تطبيق يوفر آلاف القصائد المعروضة بشكل جيد جداً و يتيح لك الاحتفاظ بما زرته من قصائد على جهازك لتتصفحة بحرية بدون الاتصال بالإنترنت.
يتم تحميل ما تريد بشكل مبدئي لمرة واحدة تكون فيها متصلاً بالإنترنت ثم تتصفحه بعد ذلك بحرية.
يتيح لك التطبيق القدرة على مشاركة أبيات الشعر مع أصدقائك و على مواقع التواصل الاجتماعي.
يمكنك أن تسجل أبيات بصوتك إذا كنت تملك القدرة على إلقاء الشعر و رفعه على الإنترنت "SoundCloud" و مشاركة هذا المقطع الصوتي مع أصدقائك و على مواقع التواصل الاجتماعي.
يمكنك التطبيق من تحديد أبيات شعرية في "المفضلة" لمراجعتها مباشرة مرة أخرى.

هذا الإصدار يحوي مجموعة من الشعراء القدامى و جاري إضافة المزيد من الشعراء.
هذا الإصدار يعد طفرة كبيرة حيث يحوي أكثر من مليون و ربع المليون بيت من الشعر و ما يربو على ألفي شاعر.
هذا الإصدار قيد التجربة و التطوير.
لازال هناك الكثير من التطوير و يمكنكم إرسال مقترحاتكم حتى يتم إضافة المزيد من التحسينات.

https://play.google.com/store/apps/details?id=net.hopesun.sha3eral3arabandroid