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 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").

Constraints:

1 <= s.length <= 100
s contains only digits and may contain leading zero(s).

Solution

/**
 * @param {string} s
 * @return {number}
 */
var numDecodings = function(s) {
    const dp = [1, s[0] === "0" ? 0 : 1];
    for (let i = 1; i < s.length; i++) {
        let count = 0
        if (s[i] !== "0") { // in range [1, 9]
            count += dp[i];
        }
        if (s[i - 1] === "1" || s[i - 1] === "2" && s[i] <= "6") { // in range [10, 26]
            count += dp[i - 1];
        }
        dp.push(count);
    }
    return dp[s.length];
};

We will implement a DP solution. Let dp[i] be the number of ways to decode s[0:i].

For the base case, if i = 0, then dp[0] = 1 since the only way to decode the empty string is (); if i = 1, then there's 1 way to decode a single character provided it's not "0", otherwise there 0 ways to decode it.
For the recursive case, when s[i] is introduced, new groupings exist if s[i] is in range ["1", "9"] and/or if s[i - 1:i + 1] is in range ["10", "26"]. The new number of groupings (starting at 0) is incremented by:
- The total number of groupings not including s[i] (ie. dp[i]) for the first case.
- The total number of groupings not including s[i - 1:i + 1] (ie. dp[i - 1]) for the second case.

Problem​

Solution​

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