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Decode Ways

Medium Dynamic Programming 1D DP Original on LeetCode

Decode Ways is a medium dynamic programming problem solved with the 1d dp pattern. The best approach, optimal (two variables), runs in O(n) time and O(1) space. Below are 3 approaches in Java, from recursion up.

Problem

A message of letters was encoded as digits with A = 1, B = 2, ..., Z = 26. Given the digit string, return the number of ways to decode it.

Examples

Example 1

Input
s = "226"
Output
3
Why
2 2 6, 22 6, 2 26.

Example 2

Input
s = "06"
Output
0
Why
A code cannot start with 0.

Constraints

  • 1 <= s.length <= 100; digits only.
  • A = 1, ..., Z = 26.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
RecursionO(2ⁿ)O(n)
TabulationO(n)O(n)
Optimal (two variables)O(n)O(1)

1Recursion

TimeO(2ⁿ)
SpaceO(n)

From position i, take one digit (if not 0) or two digits (if 10–26) and recurse.

  1. If i == n return 1; if s[i] == '0' return 0.
  2. ways(i + 1) + (valid pair ? ways(i + 2) : 0).
Java
class Solution {
    public int numDecodings(String s) {
        return ways(s, 0);
    }

    private int ways(String s, int i) {
        if (i == s.length()) return 1;
        if (s.charAt(i) == '0') return 0;
        int r = ways(s, i + 1);
        if (i + 1 < s.length() && Integer.parseInt(s.substring(i, i + 2)) <= 26) r += ways(s, i + 2);
        return r;
    }
}

2Tabulation

TimeO(n)
SpaceO(n)

dp[0] = 1 (empty prefix). For each i, add dp[i - 1] when the last digit alone is valid, and dp[i - 2] when the last two digits form 10–26.

  1. one = s[i - 1]; two = s[i - 2..i - 1] as a number.
  2. if one != '0': dp[i] += dp[i - 1]; if 10 <= two <= 26: dp[i] += dp[i - 2].
Java
class Solution {
    public int numDecodings(String s) {
        int n = s.length();
        int[] dp = new int[n + 1];
        dp[0] = 1;
        for (int i = 1; i <= n; i++) {
            if (s.charAt(i - 1) != '0') dp[i] += dp[i - 1];
            if (i >= 2) {
                int two = (s.charAt(i - 2) - '0') * 10 + (s.charAt(i - 1) - '0');
                if (two >= 10 && two <= 26) dp[i] += dp[i - 2];
            }
        }
        return dp[n];
    }
}

3Optimal (two variables)

TimeO(n)
SpaceO(1)

dp[i] needs only dp[i - 1] and dp[i - 2].

  1. prev2 = 1 (dp[0]), prev1 = dp[1]; roll forward.
Java
class Solution {
    public int numDecodings(String s) {
        int prev2 = 1, prev1 = s.charAt(0) == '0' ? 0 : 1;
        for (int i = 2; i <= s.length(); i++) {
            int cur = 0;
            if (s.charAt(i - 1) != '0') cur += prev1;
            int two = (s.charAt(i - 2) - '0') * 10 + (s.charAt(i - 1) - '0');
            if (two >= 10 && two <= 26) cur += prev2;
            prev2 = prev1;
            prev1 = cur;
        }
        return prev1;
    }
}

Edge cases to test

  • Leading zero
  • Zeros that must pair with the previous digit (10, 20)
  • Invalid pairs such as 30

Hints

Hint 1

dp[i] = ways to decode the first i characters. Add dp[i - 1] if s[i - 1] is 1–9, and dp[i - 2] if s[i - 2..i - 1] is 10–26.

FAQ

What is the best time complexity for Decode Ways?

Optimal (two variables) runs in O(n) time and O(1) extra space.

Which pattern does Decode Ways use?

It is a dynamic programming problem that uses the 1d dp pattern. Other problems with the same pattern: Fibonacci Number, Climbing Stairs, Min Cost Climbing Stairs (2 jumps).

Is there a brute force solution for Decode Ways?

Yes. Recursion takes O(2ⁿ) time and O(n) space. From position i, take one digit (if not 0) or two digits (if 10–26) and recurse.

Which edge cases should I test for Decode Ways?

Leading zero; Zeros that must pair with the previous digit (10, 20); Invalid pairs such as 30.