We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Greedy
  4. Sherlock and The Beast
  5. Discussions

Sherlock and The Beast

Sort by

recency

|

529 Discussions

|

  • danielcristianv
     + 0 comments

    This is my solution. Count 5 (the number of '5' digits) starts completely divisible by 3. Count 3 (the number of '3' digits) is basically the remainder, after substracting count 5 from n. We need to remove multiples of 3 from count5 and add them to count3 until count3 will become divisible by 5. It won't take long. Think about it, any number divided by 5 will have the following remainders 0, 1, 2, 3, 4. If it's 0 you're already there. If it's 1, you need to add 3 three times and consider modulo 5 group with addition: 1 + 3 = 4; 4 + 3 = 2; 2 + 3 = 0; If it's 2, you only need one addition to 3: 2 + 3 = 0 For 3: 3 + 3 = 1; 1 + 3 = 4; 4 + 3 = 2; 2 + 3 = 1 For 4: 4 + 3 = 2; 2 + 3 = 0;

    void decentNumber(int n) { int count5 = (n / 3) * 3; int count3 = n - count5;

    while(count3 % 5)
{
    count3 += 3;
}

count5 = n - count3;
if(count5 < 0)
{
    cout << "-1" << endl;
    return;
}
for(int i = 0; i < count5; i++)
{
    cout << "5";
}
for(int i = 0; i < count3; i++)
{
    cout << "3";
}

cout << endl;

    }

  • ricardo1512
     + 0 comments
    def decentNumber(n):
    fives3 = 0
    threes5 = 0

    for k in range(n//5+1):
        threes5 = k
        if (n - 5*k) % 3 == 0:
            fives3 = (n - 5*k)//3
            break
    
    if fives3*3 + threes5*5 != n:
        print(-1)
        return

    string = '5'*3*fives3 + '3'*5*threes5
    print(string)
    return

  • edgalvazquez
     + 0 comments

    In C#

    public static void decentNumber(int n)
{
    int max5 = 0;
    int max3 = 0;
    // Any number with less than 3 digits
    // isn't Descent 
    if(n < 3)
    {
        Console.WriteLine("-1");
        return;
    }
    
    // If the number can be modded by 3, then
    // we can fill it with fives which will give
    // us the highest value number
    if(n % 3 == 0)
    {
        Console.WriteLine(new string('5', n));
        return;
    }
        
    //Check how many groups of 5's we can put in
    max5 = (n / 3) - 1;
    // Check the number of 3's we can put after them
    max3 = n - (max5 * 3);
    // If the number of 3's doesn't mod by 5 then
    // start removing groups of 5
    if(max3 % 5 != 0)
    {
        // set max3 to zero incase there isn't a number
        // of 5's and 3's that will work
        max3 = 0;
        // While we still have groups of 5's, keep 
        // removing them
        while(max5 > 0)
        {
            // Remove a group of 5's
            max5--;
            // Get the number of remaining spaces
            max3 = n - (max5 * 3);
            // If those can be modded by 5, then
            // we found a good group combo and can
            // exit the loop
            if(max3 % 5 == 0)
                break;
            // Else set max3 back to 0
            max3 = 0;
        }
    }
    
    // If we ended up with no groups of 5's and 
    // no groups 3's, then no good combination was found
    if(max5 <= 0 && max3 <= 0)
    {
        Console.WriteLine("-1");
        return;
    }
    
    // Print out the group combinations
    Console.Write(new string('5', max5 * 3));
    Console.WriteLine(new string('3', max3));
}

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can have the implementation link here : https://youtu.be/R01NQ4Zt2qA

    void decentNumber(int n) {
    int three = ((3 - (n % 3)) % 3 ) * 5;
    if(three > n) cout << -1;
    else cout << string(n - three, '5') << string(three, '3');
    cout << endl;
}

  • shreyavishwalin1
     + 0 comments
    def decentNumber(n):
    for fives in range(n, -1, -1):
        if fives % 3 == 0 and (n - fives) % 5 == 0:
            print('5' * fives + '3' * (n - fives))
            return
    print(-1)