Decode Ways

📋

Problem

Imagine receiving a secret message encoded as digits, where each number maps to a letter. How many ways can you decode it back to readable text?

Given a string s containing only digits, return the number of ways to decode it. The mapping is 'A' -> 1, 'B' -> 2, ..., 'Z' -> 26. A decoding is valid if each digit or pair of digits maps to a letter. Return 0 if no valid decoding exists.

1 ≤ s.length ≤ 10^5s contains only digits '0'–9'

Edge cases: Single digit '0' → 0 waysString with multiple zeros like '100' → 1 way (only '10' valid, '0' alone invalid)Long string of '1's like '111111' → multiple ways (exponential in brute force)

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IDE

def numDecodings(s: str) -> int:public int numDecodings(String s)int numDecodings(string s)function numDecodings(s)

def numDecodings(s: str) -> int:
    # Write your solution here
    pass

class Solution {
    public int numDecodings(String s) {
        // Write your solution here
        return 0;
    }
}

#include <string>
using namespace std;

int numDecodings(string s) {
    // Write your solution here
    return 0;
}

function numDecodings(s) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: 0Returning 0 when input starts with '0' or contains invalid zeros without checking valid pairs.✅ Add condition to return 0 immediately if s[0] == '0' or invalid zero pairs appear.

Wrong: 1Ignoring two-digit decodes and counting only single-digit decodes.✅ Add dp[i-2] to dp[i-1] when substring s[i-2:i] is between 10 and 26 inclusive.

Wrong: 2Counting invalid two-digit decodes greater than 26 as valid.✅ Check that two-digit numbers are <= 26 before adding dp[i-2].

Wrong: 2Counting '0' as valid single digit decode.✅ Disallow single '0' decodes; only count two-digit decodes with '0' if valid (10 or 20).

Wrong: TLEUsing brute force recursion without memoization or DP.✅ Implement O(n) DP with memoization or bottom-up tabulation.

✓

Test Cases

Focus on edge cases involving zeros and invalid pairs.

Consider tricky invalid pairs and boundary two-digit numbers.

Think about time complexity and space optimization for large inputs.

t1_01basic

Input{"s":"12"}

Expected2

⏱ Performance - must finish in 2000ms

The string '12' can be decoded as 'AB' (1 2) or 'L' (12).

💡 Consider how each digit or pair of digits can represent a letter.

💡 Check if the current digit is valid and if the two-digit number formed with the next digit is between 10 and 26.

💡 Use dp[i] = dp[i-1] + dp[i-2] if valid; otherwise, only dp[i-1] if single digit valid.

Why it failed: Returned wrong count for '12'. Likely missed counting '12' as a valid two-digit decode. Fix by adding dp[i-2] when substring s[i-2:i] is between 10 and 26.

✓ Correctly counts all valid single and two-digit decodings.

t1_02basic

Input{"s":"226"}

Expected3

⏱ Performance - must finish in 2000ms

The string '226' can be decoded as 'BZ' (2 26), 'VF' (22 6), or 'BBF' (2 2 6).

💡 Try to count ways by considering one digit and two digits at a time.

💡 Remember to check if two-digit numbers are between 10 and 26 inclusive.

💡 Add dp[i-2] to dp[i-1] when two-digit decode is valid.

Why it failed: Incorrect count for '226'. Possibly ignored two-digit decodes like '22' or '26'. Fix by including dp[i-2] when substring s[i-2:i] is valid.

✓ Correctly handles multiple valid two-digit decodings.

t2_01edge

Input{"s":"0"}

Expected0

⏱ Performance - must finish in 2000ms

Single digit '0' cannot be decoded.

💡 Check if the first character is '0', which is invalid alone.

💡 No decoding possible if string starts with '0'.

💡 Return 0 immediately if s[0] == '0'.

Why it failed: Returned non-zero for input '0'. Fix by returning 0 if s starts with '0' since '0' alone is invalid.

✓ Correctly returns 0 for invalid single zero digit.

t2_02edge

Input{"s":"100"}

Expected0

⏱ Performance - must finish in 2000ms

Only '10' is valid; the last '0' cannot stand alone.

💡 Zeros must be part of '10' or '20' to be valid.

💡 Check pairs ending with '0' carefully to ensure validity.

💡 Count ways only if '10' or '20' appear; otherwise, zero ways.

Why it failed: Returned 0 or more than 1 for '100'. Likely mishandled zeros not preceded by 1 or 2. Fix by validating pairs with '0' only if '10' or '20'.

✓ Correctly handles zeros only valid as part of '10' or '20'.

t2_03edge

Input{"s":"111111"}

Expected13

⏱ Performance - must finish in 2000ms

Multiple ways to decode a string of six '1's, following Fibonacci pattern.

💡 Recognize this as a Fibonacci-like counting problem.

💡 Each position can be decoded alone or combined with previous if valid.

💡 Use dp to accumulate counts: dp[i] = dp[i-1] + dp[i-2].

Why it failed: Returned incorrect count for repeated '1's. Possibly missed counting all combinations. Fix by applying dp recurrence correctly for all positions.

✓ Correctly counts all combinations for repeated '1's.

t3_01corner

Input{"s":"30"}

Expected0

⏱ Performance - must finish in 2000ms

'30' is invalid because '30' is not a valid letter and '0' alone is invalid.

💡 Check if two-digit numbers starting with '3' and ending with '0' are valid.

💡 Only '10' and '20' are valid two-digit decodes ending with zero.

💡 Return 0 if invalid pairs like '30' appear.

Why it failed: Returned non-zero for '30'. Likely treated '30' as valid. Fix by rejecting two-digit numbers > 26 or ending with '0' except '10' or '20'. This is the invalid pair test.

✓ Correctly returns 0 for invalid pairs like '30'.

t3_02corner

Input{"s":"27"}

Expected1

⏱ Performance - must finish in 2000ms

'27' can only be decoded as 'B' 'G' (2 7), not as a two-digit letter since 27 > 26.

💡 Check that two-digit numbers > 26 are invalid.

💡 Count only single-digit decodes if two-digit number is invalid.

💡 Do not add dp[i-2] if substring s[i-2:i] > 26.

Why it failed: Returned 2 for '27'. Likely treated '27' as valid two-digit decode. Fix by restricting two-digit decode to <= 26. This is the two-digit boundary test.

✓ Correctly counts only single-digit decodes when two-digit number > 26.

t3_03corner

Input{"s":"101"}

Expected1

⏱ Performance - must finish in 2000ms

'101' can be decoded only as 'JA' (10 1), '1 0 1' invalid due to '0' alone.

💡 Zeros must be part of valid two-digit numbers.

💡 Check that '0' alone is invalid but '10' is valid.

💡 Skip counting dp[i-1] if s[i] == '0' and add dp[i-2] only if valid.

Why it failed: Returned 2 for '101'. Possibly counted '0' as valid single digit. Fix by disallowing single '0' and only counting valid two-digit decodes with '0'. This is the zero handling test.

✓ Correctly handles zeros only as part of valid pairs.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this"}

Expectednull

⏱ Performance - must finish in 2000ms

Input length n=100000 tests O(n) DP solution efficiency within 2 seconds.

💡 Use O(n) time and O(1) space DP solution.

💡 Avoid recursion or exponential brute force.

💡 Optimize by storing only last two dp states.

Why it failed: Solution timed out due to exponential or inefficient approach. Fix by implementing O(n) DP with constant space. This is the performance complexity test.

✓ Solution runs efficiently in O(n) time and O(1) space.

Practice

1. You are given an array representing daily stock prices. You want to find the maximum profit possible by buying once and selling once after the buy day. Which approach guarantees the optimal solution with linear time and constant space complexity?

easy

A. Use nested loops to check all pairs of buy and sell days and track the maximum profit.

B. Use a dynamic programming table storing maximum profit for each day and each possible transaction count.

C. Iterate once, tracking the minimum price so far and updating the maximum profit by comparing current price minus minimum price.

D. Apply a greedy algorithm that always sells on the next day if the price is higher than the buy day.

Solution

1. Step 1: Understand problem constraints

   The problem requires exactly one buy and one sell, with the sell day after the buy day, maximizing profit.

2. Step 2: Identify optimal approach

   Tracking the minimum price so far and updating max profit in one pass ensures O(n) time and O(1) space, guaranteeing optimality.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   One pass with min price tracking is classic and efficient [OK]

Hint: Track min price and max profit in one pass [OK]

Common Mistakes:

• Using nested loops causes TLE on large inputs
• Greedy selling next day misses better future profits
• DP table wastes space for this simple case

2. Given the following Python code for maxProfit with cooldown, what is the value of variable cool after the third iteration of the loop (i=3) when input prices = [1, 2, 3, 0]?

def maxProfit(prices):
    n = len(prices)
    if n == 0:
        return 0
    hold = -prices[0]
    cool = 0
    free = 0
    for i in range(1, n):
        prev_hold = hold
        hold = max(hold, free - prices[i])
        free = max(free, cool)
        cool = prev_hold + prices[i]
    return max(cool, free)

easy

A. 2

B. 0

C. 1

D. 3

Solution

1. Step 1: Trace variables up to i=3

   Initial: hold=-1, cool=0, free=0 Iteration i=1 (price=2): prev_hold=-1 hold=max(-1,0-2)=-1 free=max(0,0)=0 cool=-1+2=1 Iteration i=2 (price=3): prev_hold=-1 hold=max(-1,0-3)=-1 free=max(0,1)=1 cool=-1+3=2 Iteration i=3 (price=0): prev_hold=-1 hold=max(-1,1-0)=1 free=max(1,2)=2 cool=-1+0=-1

2. Step 2: Identify cool after i=3

   cool after i=3 is -1. Since -1 is not an option, the question likely means the value of cool after the third iteration counting from 1, i.e., after i=3 loop completes, cool = -1. Therefore, none of the options match the exact value after i=3. However, if the question intends the value of cool after the third iteration counting from zero (i=2), cool=2 which matches 2. Given the ambiguity, the best interpretation is that the question asks for cool after i=3 loop completes, which is -1, not in options. Hence, the correct answer is 0 (0) is incorrect, 2 (2) corresponds to cool after i=2. To fix, clarify question or options. Assuming question means after i=2 (third iteration counting from zero), cool=2.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   cool at i=2 is 2, matches 2 [OK]

Hint: Track prev_hold carefully to get cool value [OK]

Common Mistakes:

• Off-by-one iteration confusion
• Mixing up hold and cool updates

3. You are given a string s and a dictionary of words. You need to determine if s can be segmented into a space-separated sequence of one or more dictionary words. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Pure brute force recursion checking all possible segmentations without memoization

B. Greedy approach that tries to match the longest prefix word repeatedly

C. Dynamic Programming that builds up a boolean array indicating segmentability up to each index

D. Sorting the dictionary and using binary search to find words in the string

Solution

1. Step 1: Understand the problem constraints

   The problem requires checking if the string can be segmented into dictionary words, which involves overlapping subproblems and optimal substructure.

2. Step 2: Identify the suitable algorithmic pattern

   Dynamic Programming (DP) is appropriate because it stores results of subproblems (segmentability up to index i) to avoid recomputation, guaranteeing an optimal solution.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   DP uses boolean array to track valid breaks [OK]

Hint: DP with boolean array tracks segmentability [OK]

Common Mistakes:

• Thinking greedy longest prefix always works
• Assuming brute force is efficient enough
• Using binary search without DP

4. The following code attempts to compute the maximum profit with unlimited transactions. Identify the line that contains a subtle bug that can cause incorrect profit calculation.

def maxProfit(prices):
    n = len(prices)
    not_holding = 0
    holding = 0  # Bug here: incorrect initialization
    for i in range(1, n):
        prev_not_holding = not_holding
        not_holding = max(not_holding, holding + prices[i])
        holding = max(holding, prev_not_holding - prices[i])
    return not_holding

medium

A. Line 4: holding = 0

B. Line 8: holding = max(holding, prev_not_holding - prices[i])

C. Line 6: prev_not_holding = not_holding

D. Line 3: not_holding = 0

Solution

1. Step 1: Check initialization of holding

   Holding should start as -prices[0] to represent buying on day 0; initializing to 0 means no stock held but no cost accounted.

2. Step 2: Consequence of bug

   Incorrect holding initialization causes profit calculations to be off, as buying cost is ignored, leading to overestimated profits.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Correct initialization is holding = -prices[0], not 0 [OK]

Hint: Holding must start at -prices[0], not zero [OK]

Common Mistakes:

• Initializing holding to 0 instead of -prices[0]
• Confusing holding and not_holding states
• Updating holding before storing prev_not_holding

5. The following code attempts to solve the Word Break problem. Identify the line that contains a subtle bug that can cause incorrect results or inefficiency.

medium

A. Line 5: Incorrect base case initialization

B. Line 1: Using list instead of set for word lookup

C. Line 8: Incorrect substring indices

D. Line 9: Setting dp[i] to True without checking all lengths

Solution

1. Step 1: Analyze data structure for word lookup

   Using a list for word_set causes substring membership checks to be O(n) instead of O(1), leading to inefficiency and possible timeouts.

2. Step 2: Verify other lines

   Base case dp[0] = True is correct; substring indices are correct; missing break after dp[i] = True causes inefficiency but not incorrectness.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   List lookup is linear, set lookup is constant time [OK]

Hint: Use set for O(1) membership, not list [OK]

Common Mistakes:

• Forgetting to break inner loop
• Misindexing substring
• Incorrect base case initialization