91 Decode Ways - Medium

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 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).

Problem Analysis:

• use dynamic programming and memorization
• base case the whole string itself is 1 way of decoding
• the number of solution of i is composed of the number of solution found in i+1
• if the current char at index i is "1", the number of solution can be found in i+1 and i+2
• similarly, if the current char i is "2" and the char at i+1 is less than or equal to 6, the number of solution can be found in i+1 and i+2

Solutions:

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        # base case is 1 solution
        dp = {len(s):1} # length : # of soln

        def dfs (i):
            if i in dp:
                return dp[i]
            if s[i] == "0":
                return 0
            
            res = dfs(i+1)

            if (i+1 < len(s) and (s[i] == "1" or
                s[i] == "2" and s[i+1] in "0123456")):

                res += dfs(i+2)
            dp[i] = res
            return res
        return dfs(0)

# Testcase
s= "226"

dp =
{3: 1} 
{2: 1, 3: 1} 
{1: 2, 2: 1, 3: 1} 
{0: 3, 1: 2, 2: 1, 3: 1}

output = 3

# solution 2

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        # base case is 1 solution
        dp = {len(s):1} # length : # of soln

        for i in range(len(s)-1, -1, -1):
            if s[i] == "0":
                dp[i] = 0 
            else:
                dp[i] = dp[i+1]
            
            if (i + 1 < len(s) and (s[i] == "1" or
                s[i] == "2" and s[i+1] in "0123456")):
                dp[i] += dp[i+2]

        return dp[0]

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