Decode Ways: Unraveling Encoded Messages with Dynamic Programming

The "Decode Ways" problem is a dynamic programming challenge that revolves around decoding a string of digits into alphabets, akin to the way messages were encoded in the past.

Problem Statement

Given a string s containing only digits, return the number of ways to decode it into letters using the mapping: '1' -> 'A', '2' -> 'B', ..., '26' -> 'Z'.

Example

• Input: s = "12" Output: 2 Explanation: It could be decoded as "AB" (1 2) or "L" (12).

• Input: s = "226" Output: 3 Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Dynamic Programming Solution

Breaking Down the Solution

• Handle Leading Zero: If the string starts with '0', return 0, as it can't be decoded.
• Initialize dp Array: Create a dp array to store the number of ways to decode up to each character in the string.
• Iterate and Update dp: For each character, check if it forms a valid one-digit or two-digit number and update dp accordingly.
• Return Decoding Count: The last element in the dp array gives the total number of ways to decode the entire string.

Conclusion

The Decode Ways problem is an interesting application of dynamic programming in string processing. It showcases how to approach problems where solutions depend on the number of ways previous subproblems have been solved, illustrating the flexibility and utility of dynamic programming in a wide range of scenarios.

Rust Solution

Rust's s: &str is turned into raw bytes via s.as_bytes(), sidestepping UTF-8 boundary checks so bytes[0] == b'0' is a plain byte comparison against a byte literal. The DP table is built with vec![0i32; m + 1], pinning the element type to i32 up front rather than leaving it untyped as the JS array does. Subtracting b'0' yields a byte that would overflow, so each digit is widened with as i32 before the two-digit arithmetic. The loop walks the inclusive range 2..=m, and the trailing dp[m] with no semicolon is the return value — no return keyword needed.

Go Solution

Go converts the string with []byte(s) and indexes that, so bytes[0] == '0' compares a byte against the untyped rune constant '0'. The table is a slice allocated with make([]int, n+1), and each digit is pulled out with an explicit int(bytes[i-1] - '0') conversion, since Go won't mix the byte and int arithmetic implicitly. The classic three-clause for i := 2; i <= n; i++ drives the loop, and the function returns dp[n] directly.