Skip to main content

Decode Ways II

 Decode Ways II





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".

In addition to the mapping above, an encoded message may contain the '*' character, which can represent any digit from '1' to '9' ('0' is excluded). For example, the encoded message "1*" may represent any of the encoded messages "11", "12", "13", "14", "15", "16", "17", "18", or "19". Decoding "1*" is equivalent to decoding any of the encoded messages it can represent.

Given a string s containing digits and the '*' character, return the number of ways to decode it.

Since the answer may be very large, return it modulo 109 + 7.

 

Example 1:

Input: s = "*"
Output: 9
Explanation: The encoded message can represent any of the encoded messages "1", "2", "3", "4", "5", "6", "7", "8", or "9".
Each of these can be decoded to the strings "A", "B", "C", "D", "E", "F", "G", "H", and "I" respectively.
Hence, there are a total of 9 ways to decode "*".
Example 2:

Input: s = "1*"
Output: 18
Explanation: The encoded message can represent any of the encoded messages "11", "12", "13", "14", "15", "16", "17", "18", or "19".
Each of these encoded messages have 2 ways to be decoded (e.g. "11" can be decoded to "AA" or "K").
Hence, there are a total of 9 * 2 = 18 ways to decode "1*".
Example 3:

Input: s = "2*"
Output: 15
Explanation: The encoded message can represent any of the encoded messages "21", "22", "23", "24", "25", "26", "27", "28", or "29".
"21", "22", "23", "24", "25", and "26" have 2 ways of being decoded, but "27", "28", and "29" only have 1 way.
Hence, there are a total of (6 * 2) + (3 * 1) = 12 + 3 = 15 ways to decode "2*".
 

Constraints:

1 <= s.length <= 105
s[i] is a digit or '*'.


Solution :

class Solution {
public:
    int numDecodings(string s) {
        long e0 = 1, e1 = 0, e2 = 0, f0, f1, f2;
        for ( char c : s ) {
            if ( '*' == c ) {
                f0 = 9 * e0 + 9 * e1 + 6 * e2;
                f1 = f2 = e0;
            } else {
                f0 = int(c > '0') * e0 + e1 + int(c < '7') * e2;
                f1 = '1' == c ? e0 : 0;
                f2 = '2' == c ? e0 : 0;
            }
            e0 = f0 % 1000000007;
            e1 = f1;
            e2 = f2;
        }
        return int(e0);
    }
};

Labels:

Comments

Post a Comment

Popular posts from this blog

Leetcode 424. Longest Repeating Character Replacement. Python (Sliding Window)

  424 .  Longest Repeating Character Replacement You are given a string  s  and an integer  k . You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most  k  times. Return  the length of the longest substring containing the same letter you can get after performing the above operations .   Example 1: Input: s = "ABAB", k = 2 Output: 4 Explanation: Replace the two 'A's with two 'B's or vice versa. Example 2: Input: s = "AABABBA", k = 1 Output: 4 Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA". The substring "BBBB" has the longest repeating letters, which is 4.   Constraints: 1 <= s.length <= 10 5 s  consists of only uppercase English letters. 0 <= k <= s.length Solution :  class Solution: def characterReplacement(self, s: str, k: int) -> int: hm = {} ans = 0
Post a Comment
Read more

Leetcode 371. Sum of Two Integers. C++ / Java

371 .  Sum of Two Integers   Given two integers  a  and  b , return  the sum of the two integers without using the operators   +   and   - .   Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5   Constraints: -1000 <= a, b <= 1000 Solution :  C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java :  class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :
Post a Comment
Read more

Leetcode 242. Valid Anagram. Python (3 Ways)

  242 .  Valid Anagram Given two strings  s  and  t , return  true   if   t   is an anagram of   s , and   false   otherwise . An  Anagram  is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.   Example 1: Input: s = "anagram", t = "nagaram" Output: true Example 2: Input: s = "rat", t = "car" Output: false   Constraints: 1 <= s.length, t.length <= 5 * 10 4 s  and  t  consist of lowercase English letters.   Follow up:  What if the inputs contain Unicode characters? How would you adapt your solution to such a case? class Solution: def isAnagram(self, s: str, t: str) -> bool: return sorted(s) == sorted(t) # return Counter(s) == Counter(t) # cs, ct = {}, {} # if len(s) != len(t): # return False # for i in range(len(s)): # cs[s[i]] = 1 + cs.get(s[i], 0) #
Post a Comment
Read more