91. Decode Ways

String, DP, Review · 23 Sep 2023

Problem Statement

link: https://leetcode.com/problems/decode-ways/ https://leetcode.cn/problems/decode-ways/

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

Input: s = "226" Output: 3

Input: s = "06" Output: 0

Solution Approach

The problem is approached using dynamic programming, where we maintain a state for the number of ways to decode up to the current index.

Algorithm

Initialization: Use three pointers (dp0, dp1, dp2) to represent possible decoding ways up to the last two, last, and current characters.
Iterative Decoding: For each character: Check if it’s a valid single character decoding (not ‘0’) and update dp2. Check if it combines with the previous character for a valid two-character decoding and update dp2.
Transition: Shift the pointers for the next iteration, setting dp2 to 0. By the end, dp2 contains the total number of ways to decode the input string.

Implement

    class Solution:
      def numDecodings(self, s: str) -> int:
          n = len(s)
          dp0, dp1, dp2 = 0, 1, 0
          for i in range(1, n + 1):
              dp2 = 0
              if s[i - 1] != '0':
                  dp2 += dp1
              if i > 1 and s[i - 2] != '0' and int(s[i - 2: i]) <= 26:  
                  dp2 += dp0
              dp0, dp1 = dp1, dp2
          return dp2