LC091 - Decode Ways

Problem

A message containing letters from A-Z can be encoded into numbers using the following mapping:

• 'A' -> "1"

• 'B' -> "2"

• ...

• 'Z' -> "26"

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:

• "AAJF" with the grouping (1 1 10 6)

• "KJF" with the grouping (11 10 6)

Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".

Given a string s containing only digits, return the number of ways to decode it.

The test cases are generated so that the answer fits in a 32-bit integer.

Example

Input: s = "12"

Output: 2

Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).

Case

• 1 -> 1, 10-19

• 2 -> 2, 20-16

• 3-9 -> 3-9

Solution

Cached Recursive

Count based on how many ways the characters can be interpreted. Use cache decorator @cache for caching. Time complexity is $O(n)$, and $O(n)$ space complexity.

def numDecodings(self, s: str) -> int:
	# cache results
	@cache
	def recurse(i):
		# reach end
		if i >= len(s)
			return 1
		# start on zero
		elif s[i] == "0":
			return 0
		# double digit handling
		if i < len(s) - 1 and int(s[i:i+2]) <= 26:
			return recurse(i+1) + recurse(i+2)
		else:
			return recurse(i+1)
	return recurse(0)

Dynamic Programming

Notice that how many ways a string can be decoded is dependent on how many ways its constituent substrings can be decoded. Time and space complexity are still both $O(n)$.

def numDecodings(self, s: str) -> int:
	pre = 0
	cur = 1

	for i in range(len(s)):
		nxt = 0
		# if cur not 0, continue
		if s[i] != "0":
			nxt = cur
		# if cur and pre <= 26
		if i > 0 and s[i-1] != "0" and (s[i-1] == "1" or s[i-1] == "2" and s[i] <= "6"):
			nxt += pre
			
		pre = cur
		cur = nxt
		
	return cur