Tag: 1000

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I'll give you 1000 wiki backlinks mix profile and articles for $ 2

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  • Go to the Android Play Store and To install Triviata on your phone. (IOS coming soon)
  • Under Android: Register after installation and activate your bounty code: Go to the user profile in the mobile app> tap your avatar five times> enter the premium code into the popup that you received from your reward manager.
  • If you did not receive the bonus code, you can use the following: 2J2XB,
  • Verify that the bounty program is enabled under your username in the app.
  • Rate the app now. Do not comment on bounties, tokens or airdrops!(No reward)
  • Follow us on Twitter: https://twitter.com/verasitytech. https://twitter.com/rjmark_verasity. https://twitter.com/Triviata1
  • Tasks: Open and play at least one game of trivia each day for 90 days. From the main screen, click the Earn Now button to complete as many quotes as you can on the Quote Wall in 90 days. Other tasks include: installing other games, reaching levels in games, surveys, and viewing video ads. If you earn 25 million or more crystals after 90 days, you will receive an additional reward equal to three times your crystals! For information on how much you can earn, see "How to earn more than threefold."
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Java – Divide a series of numbers into groups of 10, 100, 1000, 10000 and so on

Specify a beginning and end of a range as a string, for example:

String start = "1230000";
String end   = "1285999";

Generate the shortest list of prefixes that span the entire range.
Thus, for each number in the specified range, there is a list item that can be used as a prefix. For the example above, the list would include:

 123,124,125,126,127,1280,1281,1282,1283,1284,1285

explanation

The length of the strings may vary (6 – 12) start.length() and end.length() are always the same.
The above range can be subdivided manually into the subregions:

    //1230000 - 1239999
    //1240000 - 1249999
    //1250000 - 1259999
    //1260000 - 1269999
    //1270000 - 1279999
    //1280000 - 1289999 this range is not completly included so make the ranges smaller with a factor 10
    //1280000 - 1280999
    //1281000 - 1281999
    //1282000 - 1282999
    //1283000 - 1283999
    //1284000 - 1284999
    //1285000 - 1285999

By storing the common substring of each start and end number of the subregions, it can be easily confirmed that any number in the range starts with 123,124,125,126,127,1280,1281,1282,1283,1284 or 1285.

Here's a very naive but working solution for printing the prefixes:

public static void main(String() args) {
    String start = "1230000";
    String end   = "1285999";

    while(start.charAt(start.length()-1) == '0' && end.charAt(start.length()-1) == '9'){
        start = start.substring(0,start.length()-1);
        end = end.substring(0,end.length()-1);
    }

    long st = Integer.parseInt(start);
    long en = Integer.parseInt(end);
    for(long i = st; i <= en; ){
         if(i % 1000 == 0 && i + 999 <= en){
            System.out.println(i/1000); 
            i = i + 1000;
        }
        else if(i % 100 == 0 && i + 99 <= en){
            System.out.println(i/100); 
            i = i + 100;
        }
        else if(i % 10 == 0 && i + 9 <= en){
            System.out.println(i/10); 
            i = i + 10;
        }
        else {
            System.out.println(i); 
            i = i + 1;
        }
    }
}

For some not so nice areas like the ones above, I have to expand my for loop:

public static void main(String() args) {
    String start = "1279995";
    String end   = "1285999";

    while(start.charAt(start.length()-1) == '0' && end.charAt(start.length()-1) == '9'){
        start = start.substring(0,start.length()-1);
        end = end.substring(0,end.length()-1);
    }

    long st = Integer.parseInt(start);
    long en = Integer.parseInt(end);
    for(long i = st; i <= en; ){
        if(i % 1000000 == 0 && i + 999999 <= en){
            System.out.println(i/1000000); 
            i = i + 1000000;
        }
        else if(i % 100000 == 0 && i + 99999 <= en){
            System.out.println(i/100000); 
            i = i + 100000;
        }
        else if(i % 10000 == 0 && i + 9999 <= en){
            System.out.println(i/10000); 
            i = i + 10000;
        }
        else if(i % 1000 == 0 && i + 999 <= en){
            System.out.println(i/1000); 
            i = i + 1000;
        }
        else if(i % 100 == 0 && i + 99 <= en){
            System.out.println(i/100); 
            i = i + 100;
        }
        else if(i % 10 == 0 && i + 9 <= en){
            System.out.println(i/10); 
            i = i + 10;
        }
        else {
            System.out.println(i); 
            i = i + 1;
        }
    }
}

Which I did not find very nice. Any ideas on how to optimize my approach or if someone has a different approach (I'm open to everything, regex, streams, recursion ...)?

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