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Formatted question description: https://leetcode.ca/all/639.html

# 639. Decode Ways II (Hard)

A message containing letters from A-Z is being encoded to numbers using the following mapping way:

'A' -> 1
'B' -> 2
...
'Z' -> 26


Beyond that, now the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9.

Given the encoded message containing digits and the character '*', return the total number of ways to decode it.

Also, since the answer may be very large, you should return the output mod 109 + 7.

Example 1:

Input: "*"
Output: 9
Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".


Example 2:

Input: "1*"
Output: 9 + 9 = 18


Note:

1. The length of the input string will fit in range [1, 105].
2. The input string will only contain the character '*' and digits '0' - '9'.

Related Topics:
Dynamic Programming

Similar Questions:

## Solution 1. DP

// OJ: https://leetcode.com/problems/decode-ways-ii/
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/decode-ways-ii/solution/
class Solution {
void add(long &a, long b) { a = (a + b) % ((int)1e9+7); }
public:
int numDecodings(string s) {
long prev1 = s[0] == '*' ? 9 : (s[0] != '0'), prev2 = 1;
for (int i = 1; i < s.size(); ++i) {
long cur = 0;
if (s[i] == '*') {
cur = 9 * prev1;
if (s[i - 1] == '1') add(cur, 9 * prev2);
else if (s[i - 1] == '2') add(cur, 6 * prev2);
else if (s[i - 1] == '*') add(cur, 15 * prev2);
} else {
cur = s[i] != '0' ? prev1 : 0;
if (s[i - 1] == '1' || (s[i - 1] == '2' && s[i] <= '6')) add(cur, prev2);
else if (s[i - 1] == '*') add(cur, (s[i] <= '6' ? 2 : 1) * prev2);
}
prev2 = prev1;
prev1 = cur;
}
return prev1;
}
};


Java

• class Solution {
public int numDecodings(String s) {
if (s == null || s.length() == 0)
return 0;
final int MODULO = 1000000007;
int length = s.length();
long[] dp = new long[length + 1];
dp[length] = 1;
dp[length - 1] = s.charAt(length - 1) == '0' ? 0 : s.charAt(length - 1) == '*' ? 9 : 1;
for (int i = length - 2; i >= 0; i--) {
if (s.charAt(i) == '0')
dp[i] = 0;
else {
char c1 = s.charAt(i), c2 = s.charAt(i + 1);
if (c1 == '*' && c2 == '*') {
int twos = 0;
for (int j = 1; j <= 9; j++) {
for (int k = 1; k <= 9; k++) {
int num = j * 10 + k;
if (num <= 26)
twos++;
}
}
dp[i] = 9 * dp[i + 1] + twos * dp[i + 2];
} else if (c1 == '*') {
int twos = 0;
for (int num = 10 + c2 - '0'; num < 100; num += 10) {
if (num <= 26)
twos++;
}
dp[i] = 9 * dp[i + 1] + twos * dp[i + 2];
} else if (c2 == '*') {
int twos = 0;
int low = (c1 - '0') * 10 + 1, high = (c1 - '0') * 10 + 9;
for (int num = low; num <= high; num++) {
if (num <= 26)
twos++;
}
dp[i] = dp[i + 1] + twos * dp[i + 2];
} else {
int num = Integer.parseInt(s.substring(i, i + 2));
dp[i] = num <= 26 ? dp[i + 1] + dp[i + 2] : dp[i + 1];
}
}
dp[i] %= MODULO;
}
return (int) dp[0];
}
}

• // OJ: https://leetcode.com/problems/decode-ways-ii/
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/decode-ways-ii/solution/
class Solution {
void add(long &a, long b) { a = (a + b) % ((int)1e9+7); }
public:
int numDecodings(string s) {
long prev1 = s[0] == '*' ? 9 : (s[0] != '0'), prev2 = 1;
for (int i = 1; i < s.size(); ++i) {
long cur = 0;
if (s[i] == '*') {
cur = 9 * prev1;
if (s[i - 1] == '1') add(cur, 9 * prev2);
else if (s[i - 1] == '2') add(cur, 6 * prev2);
else if (s[i - 1] == '*') add(cur, 15 * prev2);
} else {
cur = s[i] != '0' ? prev1 : 0;
if (s[i - 1] == '1' || (s[i - 1] == '2' && s[i] <= '6')) add(cur, prev2);
else if (s[i - 1] == '*') add(cur, (s[i] <= '6' ? 2 : 1) * prev2);
}
prev2 = prev1;
prev1 = cur;
}
return prev1;
}
};

• '''
>>> 1e9
1000000000.0
>>> int(1e9 + 7)
1000000007

1e9 is a method of writing the number 1,000,000,000 in scientific notation, and is short for 1*(10^9)'''
class Solution:
def numDecodings(self, s: str) -> int:
mod = int(1e9 + 7)
n = len(s)

# dp[i - 2], dp[i - 1], dp[i]
a, b, c = 0, 1, 0
for i in range(1, n + 1):
# 1 digit
if s[i - 1] == "*":
c = 9 * b % mod
elif s[i - 1] != "0":
c = b
else:
c = 0

# 2 digits
if i > 1:
if s[i - 2] == "*" and s[i - 1] == "*":
c = (c + 15 * a) % mod
elif s[i - 2] == "*":
if s[i - 1] > "6":
c = (c + a) % mod
else:
c = (c + 2 * a) % mod
elif s[i - 1] == "*":
if s[i - 2] == "1":
c = (c + 9 * a) % mod
elif s[i - 2] == "2":
c = (c + 6 * a) % mod
elif (
s[i - 2] != "0"
and (ord(s[i - 2]) - ord("0")) * 10 + ord(s[i - 1]) - ord("0") <= 26
):
c = (c + a) % mod

a, b = b, c

return c

############

# ref: https://leetcode.com/problems/decode-ways-ii/discuss/105262/Python-6-lines-DP-solution

one = {str(i): 1 for i in range(1, 10)}
one.update({'*': 9, '0': 0})

two = {str(i): 1 for i in range(10, 27)}
two.update({'*' + str(i): 2 if i <= 6 else 1 for i in range(10)})
two.update({'1*': 9, '2*': 6, '**': 15})

class Solution:
def numDecodings(self, s):
dp = (1, one.get(s[:1], 0))

for i in range(1, len(s)):
dp = (dp[1], (one.get(s[i]) * dp[1] + two.get(s[i - 1: i + 1], 0) * dp[0]) % 1000000007)

return dp[-1]