91. Decode Ways

The Problem

Link to original problem on Leetcode.

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.

Examples

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).

My Solution

This is a dynamic programming problem. It needs to be solved by breaking the problem down into sub-problems. We can do it either recursively or iteratively.

Recursion with Memoization

To solve this recursively, we will call the function again and again with smaller sub-sections of the original string until we trigger a base case. Here, we'll assume that an empty string means we've found a correct decoding and to return 1. If at any point the string starts with "0", we know it can't be properly mapped and return 0. These smaller sub-strings will be memoized in out memo variable, so that we don't waste resources recomputing things we've seen before.

// We modify the function to accept a Map named memo. We'll// use this to cache previously seen strings that we've// decoded. It's initialized to an empty Map.function numDecodings(s: string, memo: Map<string, number> = new Map()): number {	// Base case: if the string is empty, we've found 1 way.	if (s === '') return 1;	// If the string has a leading zero, it cannot have any	// successful mapping, so return 0.	if (s.slice(0, 1) === '0') return 0;	// Don't recompute if we've seen this string before. Just	// return what's in our map.	if (memo.has(s)) return memo.get(s);	// We'll initialize variables to analyze the pieces of the	// string that would be either a single digit to letter	// decoding, or a two digit number to letter decoding.	let singleDigit = 0, doubleDigit = 0;	// For single digit decodings, we need to know that the number	// is between 1 and 9 inclusive. If so, we recursively call	// the function on the rest of the string after this single	// digit. If there was only one letter in the string left,	// we'd be calling with an empty string and get our base case	// returned.	if (1 <= +s.slice(0, 1) && 9 >= +s.slice(0, 1)) {		singleDigit = numDecodings(s.slice(1), memo);	}	// For double digit decodings, our two letters of the string	// must parse as an int between 10 and 26 inclusive. If it	// does, we call the function again with the rest of	// the string after those two letters.	if (10 <= +s.slice(0, 2) && 26 >= +s.slice(0, 2)) {		doubleDigit = numDecodings(s.slice(2), memo);	}	// To prevent needless recalculation of repeating sequences in	// the string, save the evaluations to our Map.	memo.set(s, singleDigit + doubleDigit);	// Finally, return the memoized value of the string.	return memo.get(s);};

Solving Iteratively with Dynamic Programming

The iterative solution is, in my opinion, easier to understand. We create an array dp to hold a running count of valid decodings. We will iterate through the string s, examining the integer values of the previous character and the pair of previous two characters. (We initialized dp[0] and dp[1] to make sure we don't go out of range when we do this.)

If the currently examined sub-string matches the conditions for either the single- or double-digit decoding, we add to dp[i] the previous value of dp for either single- or double-digits. So if we're looking at single-digits, and +s.slice(i - 1, i) meets the criteria, then dp[i] will have the value of dp[i - 1] added to it. It's important that we're adding to an existing value, not setting it equal to the value, because we're going to be modifying any given dp[i] multiple times as we check for both valid single- and double-digit decodings.

As we go through the string, values of dp[i] ought to increase as i increased and we find ever more valid decodings. Finally, dp[s.length] will have the total number of valid decodings for the string s.

function numDecodings(s: string): number {	if (s === '') return 0;	// Initialize dp as an array with length equal to 1 plus the	// length of the string, all values set to zero.	const dp: number[] = Array.from({length: s.length + 1}, () => 0);	// We initialize our dp cache's first two values. dp[0] is 1,	// and this is what we'll add to dp[2] if our first double-	// digit check is valid. dp[1] is either 0 or 1, depending on	// the first character of s. Since leading zeroes aren't	// valid, then we only initialize dp[1] = 1 if there is no	// leading zero. This is the value that will be added to dp[2]	// if the first single-digit decoding validates.	dp[0] = 1;	if (s.charAt(0) !== '0') dp[1] = 1;	for (let i = 2; i <= s.length; i++) {		// For deciding dp[i], we look back at the previous 1 and 2		// characters of s to see if they are valid encodings for		// a letter.		const singleDigit = +s.slice(i - 1, i);		const doubleDigit = +s.slice(i - 2, i);		// If the single-digit encoding is valid, dp[i] should have		// the value of dp[i - 1] added to it.		if (1 <= singleDigit && 9 >= singleDigit) {			dp[i] += dp[i - 1];		}		// Now we also add dp[i - 2] to dp[i] if the double-digit		// encoding is valid.		if (10 <= doubleDigit && 26 >= doubleDigit) {			dp[i] += dp[i - 2];		}	}	// Finally, the last value of dp has the total number of valid	// encodings.	return dp[s.length];};