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).

Common Approaches

Typical techniques for this pattern:

Approach Time Space Notes
1D DP (this problem) O(n) O(n) or O(1) Linear recurrence
2D DP O(nm) O(nm) or O(n) Grid or two-sequence problems
State machine DP O(n) O(1) Buy/sell, hold/not-hold states
Memoization (top-down) Same as DP O(n) Recursive + cache

Thinking Process

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 {
public:
    int numDecodings(string s) {
        if(s.empty() || s[0] == '0') return 0;

        int n = s.length();
        vector<int> dp(n, 0);
        dp[0] = 1;
        for(int i = 1; i < n; i++) {
            // Single char
            if(s[i] != '0') {
                dp[i] += dp[i - 1];
            }
            // Two chars
            int two_digits = (s[i - 1] - '0') * 10 + (s[i] - '0');
            if(s[i - 1] != '0' && two_digits <= 26) {
                dp[i] += (i >= 2 ? dp[i - 2] : 1);
            }
        }
        return dp[n - 1];
    }
};

Solution 2: Space Optimized DP

class Solution {
public:
    int numDecodings(string s) {
        if(s.empty() || s[0] == '0') return 0;

        int n = s.length();
        int prev1 = 1, prev2 = 1;
        for(int i = 1; i < n; i++) {
            int curr = 0;
            // Single char
            if(s[i] != '0') {
                curr += prev1;
            }
            // Two chars
            int two_digits = (s[i - 1] - '0') * 10 + (s[i] - '0');
            if(s[i - 1] != '0' && two_digits <= 26) {
                curr += (i >= 2 ? prev2 : 1);
            }
            prev2 = prev1;
            prev1 = curr;
        }
        return prev1;
    }
};

Complexity

  • 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.
1D DP recurrence dp[i] 0 1 2 ? dp[i] from smaller indices / subproblems

Common Mistakes

  • Skipping edge cases (empty input, single element, boundaries).
  • Off-by-one errors in loops and index ranges.
  • Forgetting to handle the case when no valid answer exists.

Key Takeaways

  • 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.

References

Template Reference

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