Daily Leetcode 1689 - 27/06/2022

Problem Summary

Today’s problem is 1689: Partitioning Into Minimum Number Of Deci-Binary Numbers. Our goal is to write n, an integer as the minimum number of deci-binary numbbers, e.g 32 = 10+11+11. There are multiple ways to write 32 as the sum of deci-binary numbers but we’re looking at the smallest.

My thought process

This one is tricky, but has a very short solution. I simply tabulated values, noticing that single digit numbers are their own answer. Then, if I took a number, say 17, I’d need 1+1+1+1+1+1+11 which is 7 terms. Then, I looked at 39 = 11+11+11+(1+1+1+1+1+1) which is 9 terms. There is a pattern of the result being the largest digit in n. The idea is that we’re going to need k terms to build up that specific digit, and is the limiting factor.

The code

class Solution:
def minPartitions(self, n: str) -> int:
for 0 <= n <= 9:
return n
12 -> 11 + 1
13 -> 11 + 1 +1
17 -> 11+ 1+...+1 (7)
25 -> 10+10+1+1+1+1+1 -> 11 + 11 + 1 + 1 + 1
We need k for the largest digit
return int(max(n))

A rather short solution, but not obvious without a bit of writing.

Pitfalls, and where I stumbled

It took a while to realise the pattern, and the independence of digits to the left/right of the largest digit. I also had considered a recursive solution but this is overkill, and would be too slow. There are similar such number related problems where you’ll need to look for a pattern to simplify your logic to $O(n)$.