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").
1 <= s.length <= 100scontains only digits and may contain leading zero(s).
class Solution:
def numDecodings(self, s: str) -> int:
if not s or s[0] == '0':
return 0
# Length of the string
n = len(s)
# dp[i] represents the number of ways to decode s[:i]
dp = [0] * (n + 1)
dp[0], dp[1] = 1, 1
for i in range(2, n + 1):
# Single character decoding (must not be '0')
if s[i-1] != '0':
dp[i] += dp[i-1]
# Double character decoding (must be between 10 to 26)
two_digit = int(s[i-2:i]) # Convert the two characters to an integer
if 10 <= two_digit <= 26:
dp[i] += dp[i-2]
return dp[n]The time complexity of this solution is O(n), where n is the length of the string s. This is because we traverse the string once and perform constant-time operations for each character.
The space complexity is O(n) due to the use of the dp array to store the number of decoding ways for each substring up to i.