91. Decode Ways

Problem Statement

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).

Clarification Questions

  1. Leading zeros: Does a string starting with ‘0’ have any valid decodings? (No, ‘0’ doesn’t map to anything and leading zeros are invalid for 2-digit mappings)
  2. Invalid characters: Can the string contain non-digit characters? (Constraint says only digits)
  3. Result size: Does the result fit in a standard integer? (Yes, fits in 32-bit integer)

Interview Deduction Process

  1. Subproblem identification: To find the ways to decode a string of length n, we can look at the ways to decode the string up to n-1 (if the last digit is valid) and the string up to n-2 (if the last two digits are valid).
  2. State definition: Let dp[i] be the number of ways to decode the prefix of the string of length i+1.
  3. Transitions:
    • Single digit: If s[i] != '0', we can append s[i] to any decoding of s[0...i-1]. So dp[i] += dp[i-1].
    • Two digits: If the substring s[i-1...i] represents a number between 10 and 26, we can append this 2-digit number to any decoding of s[0...i-2]. So dp[i] += dp[i-2].
  4. Space optimization: Notice that dp[i] only depends on dp[i-1] and dp[i-2]. We can reduce space complexity from $O(n)$ to $O(1)$ by using two variables.

Solution Approach

This is a classic 1D dynamic programming problem, similar to the Fibonacci sequence or the Climbing Stairs problem, but with added validity checks for the digits.

Solution 1: Standard 1D DP

class Solution:
def numDecodings(self, s):
    if(not s  or  s[0] == '0') return 0
    n = s.length()
    list[int> dp(n, 0)
    dp[0] = 1
    for(i = 1 i < n i += 1) :
    # Single char
    if s[i] != '0':
        dp[i] += dp[i - 1]
    # Two chars
    two_digits = (s[i - 1] - '0')  10 + (s[i] - '0')
    if s[i - 1] != '0'  and  two_digits <= 26:
        (dp[i - 2] if                 dp[i] += (i >= 2  else 1))
return dp[n - 1]

Solution 2: Space Optimized DP

class Solution:
def numDecodings(self, s):
    if(not s  or  s[0] == '0') return 0
    n = s.length()
    prev1 = 1, prev2 = 1
    for(i = 1 i < n i += 1) :
    curr = 0
    # Single char
    if s[i] != '0':
        curr += prev1
    # Two chars
    two_digits = (s[i - 1] - '0')  10 + (s[i] - '0')
    if s[i - 1] != '0'  and  two_digits <= 26:
        (prev2 if                 curr += (i >= 2  else 1))
    prev2 = prev1
    prev1 = curr
return prev1

Complexity Analysis

  • Time Complexity: $O(n)$, where $n$ is the length of the string. We iterate through the string once.
  • Space Complexity:
    • Solution 1: $O(n)$ for the DP array.
    • Solution 2: $O(1)$ as we only use a few integer variables.

Template Reference

See Dynamic Programming Templates: 1D DP for more similar patterns.