91. Decode Ways

Medium

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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 1:

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

Example 2:

Input: s = "226"
Output: 3
Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Example 3:

Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").

Constraints:

• 1 <= s.length <= 100

• s contains only digits and may contain leading zero(s).

Solution

First, I build a tree of possible decodings I can do from a random string.

• The number of leaves in the tree essentially is the number of ways the string can be decoded.

• We are going to build our tree with DFS from our original string, trying to decode either as:

  • A single digit (and call dfs again with remaining string)

  • Both single digit and double digit

We can see redundant sub-trees so this calls up for dynamic programming.

My solution :

class Solution:
    def numDecodings(self, s: str) -> int:
        # Similar to staircase problem
        arr = [0]*(len(s)+1)
        arr[0] = 1
        arr[1] = 0 if s[0] == '0' else 1
        for index in range(2, len(s)+1):
            # one step
            if 0 < int(s[index-1:index]) <=9:
                arr[index] += arr[index-1]
            if 10 <= int(s[index-2:index]) <= 26:
                arr[index] += arr[index-2]
        return arr[len(s)]
            

Other Solution 👍

class Solution:
    def numDecodings(self, s:str) -> int:
        if len(s) == 0 or s is None:
            return 0

        @lru_cache(maxsize=None)
        def dfs(string):
            if len(string)>0:
                if string[0] == '0':
                    return 0
            if string == "" or len(string) == 1:
                return 1
            if int(string[0:2]) <= 26:
                first = dfs(string[1:])
                second = dfs(string[2:])
                return first+second
            else:
                return dfs(string[1:])

        result_sum = dfs(s)

        return result_sum