Decode Ways

▶

Steps

setup

Check for empty string or leading zero

The algorithm first checks if the input string is empty or starts with '0'. If so, it returns 0 immediately because no valid decodings exist.

💡 This step prevents invalid inputs from proceeding, ensuring the algorithm only processes valid strings.

Line:

if not s or s[0] == '0':
    return 0

💡 Input validity is crucial before starting the decoding count.

setup

Initialize dp base cases

Initialize two variables prev2 and prev1 to 1, representing dp[0] and dp[1]. dp[0] = 1 means there is one way to decode an empty string, and dp[1] = 1 means one way to decode the first character '1'.

💡 Setting these base cases is essential for building the solution iteratively.

Line:

prev2, prev1 = 1, 1

💡 Base cases anchor the dynamic programming solution and allow subsequent states to be computed.

fill_cells

Start iteration at index 1

Begin iterating from index 1 (second character '2'). Initialize current ways (curr) to 0 before checking valid decodings.

💡 Each iteration computes the number of ways to decode the substring up to the current index.

Line:

for i in range(1, n):
    curr = 0

💡 We prepare to calculate dp[i+1] based on previous dp states.

compare

Check single digit validity at index 1

Check if the current character '2' is not '0'. Since it is '2', add prev1 (dp[1]) to curr. This accounts for decoding '2' as 'B'.

💡 Single digit decoding is valid if the character is not zero.

Line:

if s[i] != '0':
    curr += prev1

💡 Single digit decoding contributes to the total ways at this index.

compare

Check two-digit validity at index 1

Extract the two-digit number formed by s[0:2] = '12'. Since 12 is between 10 and 26, add prev2 (dp[0]) to curr. This accounts for decoding '12' as 'L'.

💡 Two-digit decoding is valid if the number is between 10 and 26 inclusive.

Line:

two_digit = int(s[i-1:i+1])
if 10 <= two_digit <= 26:
    curr += prev2

💡 Two-digit decoding adds additional ways to decode the substring.

fill_cells

Update dp variables after index 1

Set prev2 to prev1 (1) and prev1 to curr (2). This shifts the window forward for the next iteration, representing dp[i-1] and dp[i].

💡 Updating these variables allows the algorithm to use only constant space.

Line:

prev2, prev1 = prev1, curr

💡 The dp state moves forward, ready for the next character.

traverse

Iteration complete, no more characters

The iteration ends because the string length is 2. The final number of decode ways is stored in prev1, which is 2.

💡 The algorithm returns the total decode ways after processing all characters.

Line:

return prev1

💡 The final dp state holds the answer to the problem.

reconstruct

Interpret final result

The algorithm outputs 2, meaning there are two ways to decode '12': 'AB' and 'L'.

💡 Understanding the output confirms the correctness of the decoding logic.

Line:

print(numDecodings('12'))  # Output: 2

💡 The count matches the expected decode ways for the input.

prune

Summary of dp states

The dp states show dp[0]=1 (empty string), dp[1]=1 (decode '1'), dp[2]=2 (decode '12'). This confirms the incremental build-up of decode ways.

💡 Reviewing dp states helps solidify understanding of the dynamic programming approach.

💡 Each dp state represents the count of decode ways up to that index.

finalize

End of visualization

The visualization ends here, having shown all key steps of the algorithm for input '12'. The final answer is 2 decode ways.

💡 This completes the step-by-step understanding of the decode ways algorithm.

💡 You have seen how the algorithm processes each character and updates decode counts.

def numDecodings(s: str) -> int:
    n = len(s)
    if not s or s[0] == '0':  # STEP 1
        return 0
    prev2, prev1 = 1, 1  # STEP 2
    for i in range(1, n):  # STEP 3
        curr = 0  # STEP 3
        if s[i] != '0':  # STEP 4
            curr += prev1
        two_digit = int(s[i-1:i+1])  # STEP 5
        if 10 <= two_digit <= 26:
            curr += prev2
        prev2, prev1 = prev1, curr  # STEP 6
    return prev1  # STEP 7

if __name__ == '__main__':
    print(numDecodings('12'))  # STEP 8

📊

Decode Ways - Watch the Algorithm Execute, Step by Step

Watching this step-by-step helps you understand how the algorithm efficiently counts decode ways without building a full DP table, by only keeping track of the last two states.

Item 0 - wt:0 val:0

i\w	0	1	2
i=0	1	?	?

Start

Item 0 - wt:0 val:0

i\w	0	1	2
i=0	1	1	?

dp[0]=1

Item 1 - wt:0 val:0

i\w	0	1	2
i=0	1	1	?

Processing index 1

Item 1 - wt:2 val:1

i\w	0	1	2
i=0	1	1	?

Single digit valid

Item 1 - wt:12 val:2

i\w	0	1	2
i=0	1	1	?

Two-digit valid

Item 1 - wt:12 val:2

i\w	0	1	2
i=0	1	1	2

dp[1]=1

Item 1 - wt:12 val:2

i\w	0	1	2
i=0	1	1	2

Answer

Item 1 - wt:12 val:2

i\w	0	1	2
i=0	1	1	2

Final Answer = 2

Item 1 - wt:12 val:2

i\w	0	1	2
i=0	1	1	2

dp[0]=1

Item 1 - wt:12 val:2

i\w	0	1	2
i=0	1	1	2

Final Answer

Key Takeaways

✓ The algorithm uses only two variables to track decode ways, making it space efficient.

This insight is hard to see from code alone because the dp array is implicit, but the visualization shows how states shift.

✓ Single and two-digit checks combine to incrementally build the total decode count.

Seeing these checks step-by-step clarifies how the algorithm handles different decoding cases.

✓ The final dp state holds the answer, confirming the correctness of the approach.

The visualization highlights the answer cell, making it clear where the result comes from.

Practice

1. What is the time complexity of the optimal dynamic programming solution for the House Robber II problem using the space-optimized approach shown below?

def rob_linear(nums, start, end):
    prev, curr = 0, 0
    for i in range(start, end + 1):
        prev, curr = curr, max(curr, prev + nums[i])
    return curr

def rob(nums):
    n = len(nums)
    if n == 1:
        return nums[0]
    return max(rob_linear(nums, 0, n - 2), rob_linear(nums, 1, n - 1))

medium

A. O(n) with constant extra space

B. O(n²) due to nested loops

C. O(2^n) because of recursion without memoization

D. O(n) because each house is processed twice in two linear scans

Solution

1. Step 1: Identify loops in rob_linear

   rob_linear runs a single loop from start to end, which is O(n) in worst case.

2. Step 2: Count calls in rob

   rob calls rob_linear twice, each over roughly n/2 elements, so total O(n) time.

3. Step 3: Consider space usage

   The DP uses only constant extra space (variables prev and curr) during computation.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Two linear scans over n elements with constant space -> O(n) time and O(1) space [OK]

Hint: Two linear scans sum to O(n) time with constant space [OK]

Common Mistakes:

• Confusing recursion with DP
• Assuming nested loops cause O(n²)
• Ignoring that calls are sequential, not nested

2. The following code attempts to solve the House Robber II problem but contains a subtle bug. Identify the line that causes incorrect results for some inputs.

def rob_linear(nums, start, end):
    prev, curr = 0, 0
    for i in range(start, end):  # BUG: should be end + 1
        prev, curr = curr, max(curr, prev + nums[i])
    return curr

def rob(nums):
    n = len(nums)
    if n == 1:
        return nums[0]
    return max(rob_linear(nums, 0, n - 2), rob_linear(nums, 1, n - 1))

medium

A. Line with 'for i in range(start, end):'

B. Line with 'prev, curr = 0, 0'

C. Line with 'return max(rob_linear(nums, 0, n - 2), rob_linear(nums, 1, n - 1))'

D. Line with 'if n == 1:'

Solution

1. Step 1: Check loop range in rob_linear

   The loop uses range(start, end), which excludes the 'end' index, missing the last house in the subarray.

2. Step 2: Impact on correctness

   Because the last house in the range is skipped, the DP does not consider all houses, leading to incorrect results.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Changing to range(start, end + 1) fixes the bug [OK]

Hint: Loop must include 'end' index in range [OK]

Common Mistakes:

• Off-by-one in loop ranges
• Confusing inclusive vs exclusive ranges
• Incorrect base cases

3. Consider the following buggy code for minimum cost climbing stairs. Which line contains the subtle bug that causes incorrect results on some inputs?

medium

A. Line 4: Loop range starting at 2

B. Line 7: Returning prev1 instead of min(prev1, prev2)

C. Line 6: Updating prev2 and prev1

D. Line 3: Initialization of prev2 and prev1

Solution

1. Step 1: Understand final return logic

   The top can be reached from either last step or second last step, so minimum of prev1 and prev2 must be returned.

2. Step 2: Identify incorrect return

   Returning only prev1 ignores the possibility that prev2 might be cheaper, causing wrong results.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Return min(prev1, prev2) to cover both possible last steps [OK]

Hint: Return min of last two costs, not just last [OK]

Common Mistakes:

• Forgetting min at return
• Misinitializing base cases
• Incorrect loop boundaries

4. If the problem is changed to allow unlimited transactions instead of at most two, which modification to the DP approach is correct?

hard

A. Keep the same DP with k=2 but run the loop more times

B. Use a simpler DP tracking only holding and not holding states without transaction count

C. Use the same code but reset profit and maxDiff arrays after every two transactions

D. Use brute force to check all possible transaction pairs

Solution

1. Step 1: Understand unlimited transactions impact

   With unlimited transactions, tracking transaction count is unnecessary; only holding or not holding states matter.

2. Step 2: Simplify DP states

   Use a DP that tracks max profit holding stock or not holding stock each day, allowing multiple buys and sells without limit.

3. Step 3: Reject other options

   Keeping k=2 or resetting arrays is incorrect; brute force is inefficient and unnecessary.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Unlimited transactions reduce problem to simpler state machine [OK]

Hint: Unlimited transactions remove need to track transaction count [OK]

Common Mistakes:

• Trying to reuse k=2 DP for unlimited
• Resetting DP arrays incorrectly

5. Suppose the cooldown constraint is removed, allowing buying immediately after selling. Which modification to the DP state machine yields the correct optimal profit?

hard

A. Remove the 'cool' state and update 'hold' as max(hold, free - prices[i]) with free = max(free, free)

B. Use a greedy approach buying at every local minimum and selling at every local maximum

C. Keep 'cool' state but allow buying on cooldown day by updating hold = max(hold, cool - prices[i])

D. Remove 'cool' state and update hold = max(hold, free - prices[i]) with free = max(free, free)

Solution

1. Step 1: Understand cooldown removal effect

   Without cooldown, no need to track cooldown state; buying can happen immediately after selling.

2. Step 2: Adjust DP states

   Remove 'cool' state; hold updated as max(hold, free - prices[i]); free updated as max(free, free) (effectively unchanged).

3. Step 3: Why other options fail

   Keeping 'cool' or using greedy breaks correctness or adds unnecessary complexity.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Removing cooldown state matches classic Best Time II problem [OK]

Hint: No cooldown means no cooldown state needed [OK]

Common Mistakes:

• Trying to keep cooldown state unnecessarily
• Using greedy which fails on some inputs