Check if all sub-numbers have distinct Digit product in Python

Suppose we have a number n, we have to check whether all sub-numbers of this number have unique digit product or not. As we know, n digit number has n*(n+1)/2 sub-numbers. For example, the sub-numbers of 135 are 1, 3, 5, 13, 35, 135. And the digit product of a number is product of its digits.

So, if the input is like n = 235, then the output will be True as sub numbers are [2, 3, 5, 23, 35, 235], digit products are [2, 3, 5, 6, 15, 30]

To solve this, we will follow these steps −

Define a function dig_prod() . This will take digits

• product := 1
• for each d in digits, do
  • product := product * d
• return product
• From the main method do the following:
• num_str := num as string
• length := size of num_str
• digits := a list of size length, and initially all values are null
• prod_set := a new empty set
• for i in range 0 to length, do
  • digits[i] := num_str[i] as integer
• for i in range 0 to length - 1, do
  • for j in range i to length - 1, do
    • item := dig_prod(digits[from index i to j])
    • if item is in prod_set, then
      • return False
    • otherwise,
      • insert item into prod_set
• return True

Let us see the following implementation to get better understanding −

Example

 Live Demo

def dig_prod(digits):
   product = 1
   for d in digits:
      product *= d
   return product
def solve(num):
   num_str = str(num)
   length = len(num_str)
   digits = [None] * length
   prod_set = set()
   for i in range(0, length):
      digits[i] = int(num_str[i])
   for i in range(0, length):
      for j in range(i, length):
         item = dig_prod(digits[i:j+1])
         if item in prod_set:
            return False
         else:
            prod_set.add(item)
   return True
n = 235
print(solve(n))

Input

235

Output

True

Arnab Chakraborty

Updated on: 30-Dec-2020

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