Counting the number of "Nines", "Threes" and "Ones" in an integer betw-CodePudding

I came across a problem and I was unable to find a solution for it over the net. Maybe I wasn't entering in the right words for the accurate search. However, the problem goes like this:

Find how many of Nines, Threes and Ones it would take when added together to reach the given integer. When also multiplied by 1, 3 and 9.

In other words, say n = 25, we will then have 2 (Nines), 2 (Threes) and 1 (Ones). When we multiply them

2 * 9 = 18

2 * 3 = 6

1 * 1 = 1

total = 18 6 1 = 25 = n.

I was able to find a solution; however I was wondering if there's a smaller version of the same.

Here's my code.

def one_3_9(num):
    temp = num
    threes = []
    ones = []
    nines = []
    val_ = 0

    while temp != 0:
        if 0 < num <=8:
           if num >= 3:
               val_ = num // 3
               threes.append(val_)
               temp = temp - (3 * val_)
               ones.append(temp)
               temp = temp - temp
           elif num < 3:
               ones.append(num)
               temp = temp - temp

        if 26 >= num >= 9:
            val_ = num // 9
            nines.append(val_)
            temp = temp - (9 * val_)
            ones.append(temp)
            temp = temp - temp
            if temp >= 3:
                num = temp
                val_ = num // 3
                threes.append(val_)
                temp = temp - (3 * val_)
                ones.append(temp)
                temp = temp - temp
       else:
           print("Number Should be more than 0 and less than 26.")
           break
   print("threes: ", threes)
   print("Ones : ", ones)
   print("nines: ", nines)

I completely AGREE the answer is long-winded and perhaps so is my explanation. However, I really hope I make sense with this example.

Thank you.

CodePudding user response：

I came across a problem and I was unable to find a solution for it over the net. Maybe I wasn't entering in the right words for the accurate search. However, the problem goes like this:

Find how many of Nines, Threes and Ones it would take when added together to reach the given integer. When also multiplied by 1, 3 and 9.

In other words, say n = 25, we will then have 2 (Nines), 2 (Threes) and 1 (Ones). When we multiply them

2 * 9 = 18

2 * 3 = 6

1 * 1 = 1

total = 18 6 1 = 25 = n.

I was able to find a solution; however I was wondering if there's a smaller version of the same.

Here's my code.

def one_3_9(num):
    temp = num
    threes = []
    ones = []
    nines = []
    val_ = 0

    while temp != 0:
        if 0 < num <=8:
           if num >= 3:
               val_ = num // 3
               threes.append(val_)
               temp = temp - (3 * val_)
               ones.append(temp)
               temp = temp - temp
           elif num < 3:
               ones.append(num)
               temp = temp - temp

        if 26 >= num >= 9:
            val_ = num // 9
            nines.append(val_)
            temp = temp - (9 * val_)
            ones.append(temp)
            temp = temp - temp
            if temp >= 3:
                num = temp
                val_ = num // 3
                threes.append(val_)
                temp = temp - (3 * val_)
                ones.append(temp)
                temp = temp - temp
       else:
           print("Number Should be more than 0 and less than 26.")
           break
   print("threes: ", threes)
   print("Ones : ", ones)
   print("nines: ", nines)

I completely AGREE the answer is long-winded and perhaps so is my explanation. However, I really hope I make sense with this example.

Thank you.

CodePudding user response：

What you're describing is calculating the three digits of your number in base 3, assuming it's less than 27.

Here is some simpler code:

def one_3_9(num):
    nines = num // 9
    threes = (num%9) // 3
    ones = num % 3
    print("Nines:", nines)
    print("Threes:", threes)
    print("Ones:", ones)

CodePudding user response：

Loop through the values in descending order, use modulus division to find how many times it can go into the value and subtract that result from your running value:

def goes_into(n, vals):
    result = {}
    for i in sorted(vals, reverse=True):
        into = n // i
        result[i] = into
        n -= i * into
    if n % i:
        raise ValueError("n is not attainable by vals")
    return result

>>> goes_into(25, [1, 3, 9])
{9: 2, 3: 2, 1: 1}

CodePudding user response：

One of the approaches:

def calculate(num):
    nines = num//9
    threes = (num-(9*nines))//3
    ones = num - (9*nines) - (3*threes)
    print (f'{num} requires: 9s: {nines}, 3s: {threes}, 1s: {ones}')
    
calculate(49)
calculate(25)
calculate(37)

Output:

49 requires: 9s: 5, 3s: 1, 1s: 1
25 requires: 9s: 2, 3s: 2, 1s: 1
37 requires: 9s: 4, 3s: 0, 1s: 1