Solution Approach
Solution Overview
There are 4 main approaches to solve this problem:
- Recursive Approach (Brute Force) - Complex approach
- Memoization (Top-Down DP) - Complex approach
- Tabulation (Bottom-Up DP) - Complex approach
- Optimized Space Complexity - Complex approach
Approach 1: Recursive Approach (Brute Force)
Let's start with a recursive approach to understand the problem better. We can break down the problem into smaller subproblems.
Thinking Process: At each position in the string, we have at most two choices:
- Decode a single digit (if it's not '0')
- Decode two digits together (if they form a valid number between 10 and 26)
We can use recursion to explore all possible ways to decode the string. For each position, we try both options (if valid) and sum up the total number of ways.
Intuition: This is a classic decision tree problem where at each step we have at most two choices. The total number of ways is the sum of ways from all valid choices.
Algorithm:
- Start from the beginning of the string.
- At each position i, we have two options:
- a. Take a single digit if it's not '0'.
- b. Take two digits if they form a valid number between 10 and 26.
- Recursively count the ways for the remaining substring for each valid option.
- The base case is when we reach the end of the string, which counts as 1 way.
- Return the sum of ways from all valid choices.
Time Complexity:
In the worst case, we might have two choices at each position, leading to an exponential number of recursive calls.
Space Complexity:
The space complexity is O(n) due to the recursion stack, where n is the length of the input string.
Approach 2: Memoization (Top-Down DP)
The recursive approach works but is inefficient due to overlapping subproblems. We can optimize it using memoization.
Thinking Process: We notice that we're solving the same subproblems multiple times in the recursive approach. For example, if we have the string "12345", the ways to decode "345" might be calculated multiple times.
To avoid this redundancy, we can use memoization to store the results of subproblems we've already solved. This way, when we encounter the same subproblem again, we can simply look up the result instead of recalculating it.
Intuition: Memoization transforms our exponential solution into a linear one by eliminating redundant calculations.
Algorithm:
- Create a memo array/map to store results of subproblems.
- Define a recursive function numDecodings(s, i) that returns the number of ways to decode the substring starting at index i.
- Before computing, check if the result for index i is already in the memo. If yes, return it.
- At each position i, we have two options:
- a. Take a single digit if it's not '0'.
- b. Take two digits if they form a valid number between 10 and 26.
- Recursively count the ways for the remaining substring for each valid option.
- Store the result in the memo before returning.
- The base case is when we reach the end of the string, which counts as 1 way.
- Return numDecodings(s, 0) as the final answer.
Time Complexity:
With memoization, each position is processed at most once, resulting in a linear time complexity.
Space Complexity:
We use O(n) space for the memoization array and the recursion stack.
Approach 3: Tabulation (Bottom-Up DP)
We can further optimize our solution by using a bottom-up dynamic programming approach, which eliminates the need for recursion.
Thinking Process: Instead of starting from the beginning and working our way to the end (top-down), we can start from the end and work our way to the beginning (bottom-up).
Let's define dp[i] as the number of ways to decode the substring s[i...n-1]. The base case is dp[n] = 1, representing an empty string which has 1 way to decode (doing nothing).
Intuition: By building our solution from smaller subproblems to larger ones, we can avoid recursion and its associated overhead.
Algorithm:
- Create a DP array of size n+1, where n is the length of the string.
- Initialize dp[n] = 1 (base case: empty string has 1 way to decode).
- Iterate from the end of the string to the beginning (i = n-1 to 0):
- a. If the current digit is not '0', add dp[i+1] to dp[i].
- b. If the current digit and the next digit form a valid number (10-26), add dp[i+2] to dp[i].
- Return dp[0] as the final answer.
Time Complexity:
We process each position once, resulting in a linear time complexity.
Space Complexity:
We use an array of size n+1 to store the DP values.
Approach 4: Optimized Space Complexity
We can further optimize the space complexity of our solution by observing that we only need the last two DP values at any point.
Thinking Process: When calculating dp[i], we only need the values of dp[i+1] and dp[i+2]. This means we don't need to store the entire DP array; we can just keep track of these two values.
Intuition: By using just two variables instead of an array, we can reduce the space complexity from O(n) to O(1).
Algorithm:
- Initialize two variables: dp1 = 1 (representing dp[n]) and dp2 = 0 (a placeholder).
- Iterate from the end of the string to the beginning (i = n-1 to 0):
- a. Initialize current = 0.
- b. If the current digit is not '0', add dp1 to current.
- c. If the current digit and the next digit form a valid number (10-26), add dp2 to current.
- d. Update dp2 = dp1 and dp1 = current for the next iteration.
- Return dp1 as the final answer.
Time Complexity:
We still process each position once, resulting in a linear time complexity.
Space Complexity:
We only use a constant amount of extra space regardless of the input size.
Pseudocode
1234567891011121314151617181920function numDecodings(s): // Base case: empty string has 1 way to decode if s is empty: return 1 // If the string starts with '0', it cannot be decoded if s[0] == '0': return 0 // Initialize result ways = 0 // Option 1: Take a single digit ways += numDecodings(s[1:]) // Option 2: Take two digits if valid if length of s >= 2 and (s[0] == '1' or (s[0] == '2' and s[1] <= '6')): ways += numDecodings(s[2:]) return waysString Problems
Learning Objective
Understand different approaches to solve the secret message decoder problem and analyze their efficiency.
Solution Approaches
Approach 1: Recursive Approach (Brute Force)
Let's start with a recursive approach to understand the problem better. We can break down the problem into smaller subproblems.
Thinking Process: At each position in the string, we have at most two choices:
- Decode a single digit (if it's not '0')
- Decode two digits together (if they form a valid number between 10 and 26)
We can use recursion to explore all possible ways to decode the string. For each position, we try both options (if valid) and sum up the total number of ways.
Intuition: This is a classic decision tree problem where at each step we have at most two choices. The total number of ways is the sum of ways from all valid choices.
Algorithm:
- Start from the beginning of the string.
- At each position i, we have two options:
- a. Take a single digit if it's not '0'.
- b. Take two digits if they form a valid number between 10 and 26.
- Recursively count the ways for the remaining substring for each valid option.
- The base case is when we reach the end of the string, which counts as 1 way.
- Return the sum of ways from all valid choices.
Approach 2: Memoization (Top-Down DP)
The recursive approach works but is inefficient due to overlapping subproblems. We can optimize it using memoization.
Thinking Process: We notice that we're solving the same subproblems multiple times in the recursive approach. For example, if we have the string "12345", the ways to decode "345" might be calculated multiple times.
To avoid this redundancy, we can use memoization to store the results of subproblems we've already solved. This way, when we encounter the same subproblem again, we can simply look up the result instead of recalculating it.
Intuition: Memoization transforms our exponential solution into a linear one by eliminating redundant calculations.
Algorithm:
- Create a memo array/map to store results of subproblems.
- Define a recursive function numDecodings(s, i) that returns the number of ways to decode the substring starting at index i.
- Before computing, check if the result for index i is already in the memo. If yes, return it.
- At each position i, we have two options:
- a. Take a single digit if it's not '0'.
- b. Take two digits if they form a valid number between 10 and 26.
- Recursively count the ways for the remaining substring for each valid option.
- Store the result in the memo before returning.
- The base case is when we reach the end of the string, which counts as 1 way.
- Return numDecodings(s, 0) as the final answer.
Approach 3: Tabulation (Bottom-Up DP)
We can further optimize our solution by using a bottom-up dynamic programming approach, which eliminates the need for recursion.
Thinking Process: Instead of starting from the beginning and working our way to the end (top-down), we can start from the end and work our way to the beginning (bottom-up).
Let's define dp[i] as the number of ways to decode the substring s[i...n-1]. The base case is dp[n] = 1, representing an empty string which has 1 way to decode (doing nothing).
Intuition: By building our solution from smaller subproblems to larger ones, we can avoid recursion and its associated overhead.
Algorithm:
- Create a DP array of size n+1, where n is the length of the string.
- Initialize dp[n] = 1 (base case: empty string has 1 way to decode).
- Iterate from the end of the string to the beginning (i = n-1 to 0):
- a. If the current digit is not '0', add dp[i+1] to dp[i].
- b. If the current digit and the next digit form a valid number (10-26), add dp[i+2] to dp[i].
- Return dp[0] as the final answer.
Approach 4: Optimized Space Complexity
We can further optimize the space complexity of our solution by observing that we only need the last two DP values at any point.
Thinking Process: When calculating dp[i], we only need the values of dp[i+1] and dp[i+2]. This means we don't need to store the entire DP array; we can just keep track of these two values.
Intuition: By using just two variables instead of an array, we can reduce the space complexity from O(n) to O(1).
Algorithm:
- Initialize two variables: dp1 = 1 (representing dp[n]) and dp2 = 0 (a placeholder).
- Iterate from the end of the string to the beginning (i = n-1 to 0):
- a. Initialize current = 0.
- b. If the current digit is not '0', add dp1 to current.
- c. If the current digit and the next digit form a valid number (10-26), add dp2 to current.
- d. Update dp2 = dp1 and dp1 = current for the next iteration.
- Return dp1 as the final answer.
Complexity Analysis
Recursive Approach (Brute Force) Approach
Time Complexity
In the worst case, we might have two choices at each position, leading to an exponential number of recursive calls.
Space Complexity
The space complexity is O(n) due to the recursion stack, where n is the length of the input string.
Memoization (Top-Down DP) Approach
Time Complexity
With memoization, each position is processed at most once, resulting in a linear time complexity.
Space Complexity
We use O(n) space for the memoization array and the recursion stack.
Tabulation (Bottom-Up DP) Approach
Time Complexity
We process each position once, resulting in a linear time complexity.
Space Complexity
We use an array of size n+1 to store the DP values.
Optimized Space Complexity Approach
Time Complexity
We still process each position once, resulting in a linear time complexity.
Space Complexity
We only use a constant amount of extra space regardless of the input size.
Pseudocode
Recursive Approach (Brute Force) Approach
1234567891011121314151617181920function numDecodings(s): // Base case: empty string has 1 way to decode if s is empty: return 1 // If the string starts with '0', it cannot be decoded if s[0] == '0': return 0 // Initialize result ways = 0 // Option 1: Take a single digit ways += numDecodings(s[1:]) // Option 2: Take two digits if valid if length of s >= 2 and (s[0] == '1' or (s[0] == '2' and s[1] <= '6')): ways += numDecodings(s[2:]) return waysMemoization (Top-Down DP) Approach
12345678910111213141516171819202122232425262728function numDecodings(s): memo = new Map() function decode(i): // Base case: reached the end of the string if i == length of s: return 1 // If already computed, return from memo if i in memo: return memo[i] // If current digit is '0', it cannot be decoded on its own if s[i] == '0': return 0 // Option 1: Take a single digit ways = decode(i + 1) // Option 2: Take two digits if valid if i + 1 < length of s and (s[i] == '1' or (s[i] == '2' and s[i+1] <= '6')): ways += decode(i + 2) // Store result in memo memo[i] = ways return ways return decode(0)Tabulation (Bottom-Up DP) Approach
123456789101112131415161718192021function numDecodings(s): n = length of s dp = new array of size n+1 // Base case: empty string has 1 way to decode dp[n] = 1 // Fill dp array from right to left for i from n-1 down to 0: // If current digit is '0', it cannot be decoded on its own if s[i] == '0': dp[i] = 0 else: // Option 1: Take a single digit dp[i] = dp[i+1] // Option 2: Take two digits if valid if i+1 < n and (s[i] == '1' or (s[i] == '2' and s[i+1] <= '6')): dp[i] += dp[i+2] return dp[0]