How do you solve Decode Ways on LeetCode?

Decode Ways (LeetCode #91) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Leading zeros in the string are invalid and should be handled early.

What is the brute force solution for Decode Ways?

The brute force approach for Decode Ways is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible ways to decode the string by recursively checking each character and its possible combinations. This method is straightforward but inefficient as it explores all potential paths. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Decode Ways?

Decode Ways uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Decode Ways?

Always clarify edge cases with the interviewer, such as empty strings or strings starting with '0'. Explain your thought process as you develop your solution, especially when transitioning from brute force to dynamic programming. Practice writing both recursive and iterative solutions to solidify your understanding of the problem.

What are common mistakes when solving Decode Ways?

Not handling leading zeros correctly, which can lead to invalid decodings. Forgetting to check the bounds when accessing characters in the string.

#91

Decode Ways

Medium

StringDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses dynamic programming to build up the number of ways to decode the string iteratively. This avoids the overhead of recursion and reduces redundant calculations by storing intermediate results.

⚙️

Algorithm

4 steps

1Step 1: Create a DP array of size n+1, where n is the length of the string, initialized with 0.
2Step 2: Set DP[0] = 1 (an empty string has one way to be decoded).
3Step 3: Iterate through the string, updating the DP array based on valid single and double digit decodings.
4Step 4: Return DP[n] as the result.

solution.py8 lines

1def numDecodings(s):
2    if not s or s[0] == '0': return 0
3    dp = [0] * (len(s) + 1)
4    dp[0], dp[1] = 1, 1
5    for i in range(2, len(s) + 1):
6        if s[i-1] != '0': dp[i] += dp[i-1]
7        if 10 <= int(s[i-2:i]) <= 26: dp[i] += dp[i-2]
8    return dp[-1]

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the string once, and the space complexity is O(n) due to the DP array.

1Leading zeros in the string are invalid and should be handled early.
2The combinations of characters can be treated as a tree structure where each node represents a decision to take one or two characters.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.