# Solution For Decode Ways

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 answer is guaranteed to fit in a 32-bit integer.

Example 1:

Input: s = “12”
Output: 2
Explanation: “12” can be decoded as “AB” (1 2) or “L” (12).

Example 2:

Input: s = “226”
Output: 3
Explanation: “226” can be decoded as “BZ” (2 26), “VF” (22 6), or “BBF” (2 2 6).

Example 3:

Input: s = “0”
Output: 0
Explanation: There is no character that maps to a number starting with 0.
The only valid mappings with 0 are ‘J’ -> “10” and ‘T’ -> “20”, neither of which start with 0.
Hence, there are no valid ways to decode this since all digits need to be mapped.

Example 4:

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).

Solution:

Approach:
We can solve this problem using dynamic programming. We can define dp[i] as the number of ways to decode the substring from 0 to i in the given string s. We can utilize the previously calculated dp values to calculate the current dp value. We can check two cases while calculating the current dp value:

Case 1: If the current digit is not zero, we can include it as a single digit and add the number of ways to decode the substring from (i-1) to the current index.

Case 2: If the current digit, along with the previous digit, forms a valid two-digit number, we can include it as a two-digit number and add the number of ways to decode the substring from (i-2) to the current index.

We can return dp[n-1], where n is the length of the given string s, as the desired answer.

Algorithm:

1. Initialize dp as 1, since there is one way to decode a string of length 1.
2. If the first digit is zero, return 0, since the first digit cannot be mapped to any letter.
3. Loop from i=1 to n-1, where n is the length of the given string s.
4. If s[i] is not zero, add dp[i-1] to dp[i], since we can include s[i] as a single digit.
5. If s[i-1] and s[i] form a valid two-digit number, add dp[i-2] to dp[i], since we can include s[i-1] and s[i] as a two-digit number.
6. Return dp[n-1], where n is the length of the given string s, as the desired answer.

Time Complexity: O(n), where n is the length of the given string s.
Space Complexity: O(n), where n is the length of the given string s.

Code:

class Solution {
public:
int numDecodings(string s) {
int n = s.size();
vector dp(n, 0);
dp = 1;
if(s == ‘0’) return 0;
for(int i = 1; i < n; i++) {
if(s[i] != ‘0’) dp[i] += dp[i-1];
int two_digit_number = (s[i-1] – ‘0’)*10 + (s[i] – ‘0’);
if(two_digit_number >= 10 && two_digit_number <= 26) {
if(i >= 2) dp[i] += dp[i-2];
else dp[i]++;
}
}
return dp[n-1];
}
};

## Step by Step Implementation For Decode Ways

```There are two ways to decode a string.

1. If the first character is a letter, then we can decode the string by reading the letter and the following number.
2. If the first character is a number, then we can decode the string by reading the number and the following letter.

For example, given the string "1234", we can decode it as "abcd" or "lmno".

We can use a dynamic programming approach to solve this problem. We can create an array dp where dp[i] represents the number of ways to decode the string s[0..i-1].

We can initialize dp = 1, since there is only one way to decode an empty string.

Then, we can iterate over the string s, and for each character s[i], we can update dp[i] as follows:

If s[i] is a letter, then we can decode the string by reading the letter and the following number. This means that we can update dp[i] as dp[i] = dp[i-1].

If s[i] is a number, then we can decode the string by reading the number and the following letter. This means that we can update dp[i] as dp[i] = dp[i-2], if s[i-1] is a valid letter.

We can continue this process until we reach the end of the string. The final answer is dp[s.length() - 1].

class Solution {
public int numDecodings(String s) {
if (s == null || s.length() == 0) {
return 0;
}
int n = s.length();
int[] dp = new int[n+1];
dp = 1;
for (int i = 1; i <= n; i++) {
char c = s.charAt(i-1);
if (c >= '1' && c <= '9') {
dp[i] += dp[i-1];
}
if (i >= 2) {
int num = (s.charAt(i-2) - '0') * 10 + (c - '0');
if (num >= 10 && num <= 26) {
dp[i] += dp[i-2];
}
}
}
return dp[n];
}
}```
```There are two ways to decode a message encoded as a sequence of digits:

1. Treat the digit as a normal character and move to the next digit.
2. Treat the two digits as a single character and move to the next two digits.

Given a sequence of digits, return the total number of ways to decode the message.

def numDecodings(s):
# Base case
if s == "":
return 0

# First digit
if s == "0":
return 0
else:
ways = 1

# Second digit
if len(s) >= 2:
if s == "0":
if s > "2":
return 0
else:
ways = 1
elif s == "1" or (s == "2" and s <= "6"):
ways = 2
else:
ways = 1

# Third digit and beyond
for i in range(2, len(s)):
if s[i] == "0":
if s[i-1] == "0" or s[i-1] > "2":
return 0
else:
ways = ways
elif s[i-1] == "1" or (s[i-1] == "2" and s[i] <= "6"):
ways = ways + ways
else:
ways = ways

return ways```
```There are two ways to decode a message:

1. Use the mapping of letters to numbers (e.g., 'a' -> 1, 'b' -> 2, ..., 'z' -> 26) to determine the number for each letter.
2. Use the number of ways to decode a message that is one letter shorter.

If the first letter of the message is '0', then it can't be decoded. Otherwise, the number of ways to decode the message is the sum of the number of ways to decode the message if the first letter is removed and the number of ways to decode the message if the first two letters are removed.

function decodeWays(message) {
// Base case:
if (message.length === 0) {
return 1;
}
// If the first letter is '0', then it can't be decoded.
if (message === '0') {
return 0;
}
// Otherwise, the number of ways to decode the message is the
// sum of the number of ways to decode the message if the first
// letter is removed and the number of ways to decode the message
// if the first two letters are removed.
return decodeWays(message.substring(1)) +
decodeWays(message.substring(2));
}```
```There are two ways to decode a message:

1. Use the letters of the alphabet to decode a message.
2. Use numbers to decode a message.

If the first character of the message is a letter, then that letter is the first letter of the decoded message. If the first character of the message is a number, then the number is the first number of the decoded message.

For example, if the message is "1234", then the decoded message is "abcd".

If the message is "12345", then the decoded message is "abcde".

If the message is "123456", then the decoded message is "abcdef".```
```There are two ways to decode a string:

1) Recursive

2) Dynamic Programming

Recursive:

We can decode a string by breaking it down into smaller substrings and then decode each substring recursively. For example, “12” can be decoded as “1” + “2” or “12”. Thus, the number of ways to decode “12” is equal to the sum of the number of ways to decode “1” and the number of ways to decode “2”.

Dynamic Programming:

We can also use dynamic programming to solve this problem. We can create an array dp where dp[i] represents the number of ways to decode the substring s[0…i-1].

If s[i-1] is a valid character (i.e., it is not '0'), we can decode s[0…i-1] as s[0…i-2] + s[i-1].

If s[i-2] and s[i-1] form a valid character code (i.e., s[i-2] is not '0' and the two characters form a valid character code), we can decode s[0…i-1] as s[0…i-3] + s[i-2] + s[i-1].

Thus, we have the following recursive relation:

dp[i] = dp[i-1] + dp[i-2] (if s[i-1] is not '0')

dp[i] = dp[i-2] (if s[i-1] and s[i-2] form a valid character code)

We can initialize dp = 1 and dp = 1 since there is only one way to decode an empty string and there is only one way to decode a string of length 1.

Thus, the number of ways to decode a string of length n is given by dp[n].```
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