Python Challenge - Number Permutation

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Does anybody know how to do this? I’ve been trying, but I can’t.

from itertools import product as p def makeNumber(z): # Write your code here products = [] solutions = 0 for i in range(1,6): products.append(x for x in p(range(1,10), repeat=i)) for x in products: for y in x: if sum(y) == z: print(y) solutions += 1 return '\nThere are {} solutions'.format(solutions) print(makeNumber(3))

@aryabadri8979868753 found a solution, posted below

from itertools import product, chain def makeNumber(z): res = [] for i in range(1, 6): res.append([x for x in product((y for y in range(1, 10)), repeat=i) if sum(x) == z]) return len(list(chain.from_iterable(res))) print(makeNumber(45))

I did a brute force approach instead of recursion this time:
I iterated through all possible series of 5 numbers from 0 - 9 (with repetition), and counted only the relevant ones (meaning the ones where the numbers added up to the given number z, where there were no zeros that weren’t at the beginning).
I used an integer (meaning its digits) as each permutation, so that I could iterate through all the permutations.

my code
def checkPermutation(n, sum_of_digits = None, max_exp_of_10 = 4):
  # get sum of digits of n, check if equals sum_of_digits
  # can have leading 0s, but no 0s after a non-zero
  previous_was_nonzero = False
  total = 0  # will store sum of digits so far

  checker = 10 ** max_exp_of_10

  while checker >= 1:
    digit = n // checker
    if digit == 0:
      if previous_was_nonzero:
        # if zero comes after nonzero, return False
        return False
      total += (digit % 10)
      previous_was_nonzero = True
    n = n % checker
    checker = checker // 10
  if sum_of_digits is None:
    return total
  return (total == sum_of_digits)

def makeNumber(z):
    if (z < 1) or (z > 45):
      return 0
    count = 0
    for n in range(1, 100000):
      if checkPermutation(n, z):
        count += 1
    return count


def makeNumber(z):

    lst = [[]] # temp lst to create tuple in, inner list.
    cnt = 0 # store the count of permitations

    for idx in range(5): # Limit to 5 digit in tuple
      lst = [tuple(i,) + (x,) for i in lst for x in range(1, z+1) if x <= 9]
      # create permutations in a lst

      for i in lst: # loop trough list and put a condition to get the count of the sum that equals z
        if sum(i) == z:
          cnt += 1
    return cnt


def D(z, n): if(z == 0): return 1 elif(z < 0 or n == 0): return 0 else: total = 0 for i in range(1,10): total += D(z-i, n-1) return total def makeNumber(z): # Write your code here if (z == 0): return 0 return D(z, 5) for i in range(45): print(i, makeNumber(i))
from itertools import product def makeNumber(z): #defining initial array, digit array and sorting array a = [] l = range(1, 10) s = [] #getting all possible permutations for i in range(1, 6): s.append(product(l, repeat=i)) #searching for elements in initial array with sum equal to z for element in s: for i in element: if sum(i) == z: a.append(i) #returning number of elements found return len(a) print(makeNumber(4))

A small summary of my code:

  1. Importing product function from itertools (to not find the permutations manually).
  2. Initializing our permutations counter on 0 and creating a list for saving them.
  3. Iterating throught 1 to 5 (amount of possible numbers in our permutations).
  4. Saving on our list “products” the result of all the possible permutations after calling the function products (imported).
  5. Checking if any of these permutations match our parameter z. In this case, we update the counter.
from itertools import product def makeNumber(z): count = 0 products = list() for it in range(1,6): products = product(range(1,10), repeat=it) for perm in products: if sum(perm) == z: count += 1 return count print(makeNumber(23))
import itertools def makeNumber(z): return len([perm for _ in range(1, 5 + 1) for perm in itertools.product(range(1, 10), repeat=_) if sum(perm) == z]) print(makeNumber(45))
def perm(z, dig, base): # If number is 0 or remaining digits are 0 if z < 1 or dig < 1: return 0 # If number is lower than base, count current number as a permutation total = 1 if z < base else 0 # Calculate the number of permutations for each number at base for i in range(1,base): total += perm(z-i, dig-1, base) return total def makeNumber(z): return perm(z,5,10) # number, digits, base print(makeNumber(21))