Generate All Permutations Using Bitmask DP

📋

Generate All Permutations Using Bitmask DP

Edge cases:

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IDE

def generate_permutations(n: int) -> list:public List<List<Integer>> generatePermutations(int n)vector<vector<int>> generatePermutations(int n)function generatePermutations(n)

def generate_permutations(n):
    # Write your solution here
    pass

class Solution {
    public List<List<Integer>> generatePermutations(int n) {
        // Write your solution here
        return new ArrayList<>();
    }
}

#include <vector>
using namespace std;

vector<vector<int>> generatePermutations(int n) {
    // Write your solution here
    return {};
}

function generatePermutations(n) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: [[0,1,2],[0,2,1],[1,2,0],[2,0,1],[2,1,0]]Missing permutation [1,0,2] due to greedy approach picking next element in ascending order only.✅ At each recursion step, iterate over all unused elements instead of picking the smallest only.

Wrong: [[0,0,1],[0,1,1],[1,0,0]]Elements reused multiple times due to missing or incorrect bitmask checks.✅ Use bitmask to mark elements as visited and skip those already included.

Wrong: [[0,1,2],[0,2,1],[1,0,2],[1,2,0],[2,0,1]]Off-by-one error in bitmask indexing causing last permutation missing.✅ Use (1 << i) consistently for element i's bit and check bounds carefully.

Wrong: Non-empty output for n=0Missing base case for n=0 returning empty list.✅ Add base case: if n == 0 return [].

Wrong: Timeout or crash on n=10Inefficient recursion without memoization or pruning.✅ Implement memoization with bitmask keys to avoid recomputation.

✓

Test Cases

Focus on handling smallest and boundary inputs correctly with proper base cases.

Watch out for common algorithmic traps like greedy selection or element reuse.

Prepare for performance challenges by optimizing recursion and using memoization.

t1_01basic

Input3

Expected[[0,1,2],[0,2,1],[1,0,2],[1,2,0],[2,0,1],[2,1,0]]

⏱ Performance - must finish in 2000ms

All permutations of [0,1,2] are generated using bitmask to track which elements are used.

💡 Think about how to represent visited elements using bits.

💡 Use recursion with a bitmask to mark which indices are used.

💡 At each step, try adding an element if its bit is not set in the mask.

Why it failed: Output missing some permutations or duplicates due to incorrect bitmask usage. Fix by ensuring each element is included exactly once by checking if its bit is unset before recursion.

✓ Correctly generates all permutations using bitmask tracking.

t1_02basic

Input2

Expected[[0,1],[1,0]]

⏱ Performance - must finish in 2000ms

All permutations of [0,1] generated correctly with bitmask tracking visited elements.

💡 For n=2, there are 2! = 2 permutations to generate.

💡 Check that your bitmask updates correctly after choosing an element.

💡 Ensure recursion stops when the permutation length equals n.

Why it failed: Incorrect output due to not stopping recursion at length n or incorrect bitmask updates. Fix by adding base case for length n and updating mask properly.

✓ Correctly handles smaller input with full permutation generation.

t2_01edge

Input0

Expected[]

⏱ Performance - must finish in 2000ms

No elements means no permutations; output is empty list.

💡 What should the output be when n=0?

💡 Check if your base case handles empty input correctly.

💡 Return empty list immediately if n=0.

Why it failed: Output non-empty or error for n=0 due to missing base case. Fix by returning [] when n=0 before recursion.

✓ Correctly returns empty list for n=0.

t2_02edge

Input1

Expected[[0]]

⏱ Performance - must finish in 2000ms

Single element has exactly one permutation: [0].

💡 For n=1, only one permutation exists.

💡 Ensure recursion handles single element correctly.

💡 Check bitmask updates and base case for length 1.

Why it failed: Fails to return [[0]] due to incorrect base case or bitmask logic. Fix by returning path when length equals n.

✓ Correctly returns single permutation for n=1.

t2_03edge

Input2

Expected[[0,1],[1,0]]

⏱ Performance - must finish in 2000ms

Exact boundary test for n=2 to verify correct permutation generation.

💡 Check that all permutations of length 2 are generated.

💡 Verify bitmask correctly tracks visited elements.

💡 Ensure recursion stops at length n.

Why it failed: Missing permutations or duplicates due to incorrect bitmask or recursion base case. Fix by proper bitmask checks and base case.

✓ Correctly generates all permutations for n=2.

t2_04edge

Input10

Expectednull

⏱ Performance - must finish in 2000ms

n=10 is the upper boundary; output size is 10! which is large but valid for performance test.

💡 Consider time complexity of generating all permutations.

💡 Bitmask DP reduces overhead but output size is factorial.

💡 Expect long runtime; optimize recursion and pruning.

Why it failed: Solution times out due to factorial output size and inefficient recursion. Fix by using memoization or iterative bitmask DP.

✓ Solution handles upper boundary input efficiently.

t3_01corner

Input3

Expected[[0,1,2],[0,2,1],[1,0,2],[1,2,0],[2,0,1],[2,1,0]]

⏱ Performance - must finish in 2000ms

Test to catch greedy approach that picks elements in ascending order only, missing permutations.

💡 Greedy approach may only generate permutations in sorted order.

💡 Check if your recursion tries all unused elements at each step.

💡 Fix by iterating over all elements and recursing on each unused one.

Why it failed: Output missing permutations due to greedy selection of next element only. Fix by exploring all unused elements at each recursion level.

✓ All permutations generated, no greedy trap.

t3_02corner

Input4

Expected[[0,1,2,3],[0,1,3,2],[0,2,1,3],[0,2,3,1],[0,3,1,2],[0,3,2,1],[1,0,2,3],[1,0,3,2],[1,2,0,3],[1,2,3,0],[1,3,0,2],[1,3,2,0],[2,0,1,3],[2,0,3,1],[2,1,0,3],[2,1,3,0],[2,3,0,1],[2,3,1,0],[3,0,1,2],[3,0,2,1],[3,1,0,2],[3,1,2,0],[3,2,0,1],[3,2,1,0]]

⏱ Performance - must finish in 2000ms

Test to catch confusion between 0/1 knapsack and permutation generation: reusing elements multiple times.

💡 Permutation requires each element used exactly once.

💡 Check that bitmask prevents reuse of elements.

💡 Fix by ensuring bitmask marks elements as visited and excludes them in recursion.

Why it failed: Output contains duplicates or repeated elements due to missing bitmask checks. Fix by enforcing bitmask to track visited elements strictly.

✓ No repeated elements; correct permutation generation.

t3_03corner

Input3

Expected[[0,1,2],[0,2,1],[1,0,2],[1,2,0],[2,0,1],[2,1,0]]

⏱ Performance - must finish in 2000ms

Test to catch off-by-one errors in bitmask indexing causing missing or duplicated permutations.

💡 Bitmask bits correspond to element indices from 0 to n-1.

💡 Check indexing when setting and checking bits.

💡 Fix by using (1 << i) to represent element i's bit consistently.

Why it failed: Incorrect bitmask indexing causes missing permutations or duplicates. Fix by correctly using (1 << i) for element i.

✓ Bitmask indexing correct; all permutations present.

t4_01performance

Input{"_description":"n=10 at constraint boundary - executor generates this input"}

Expectednull

⏱ Performance - must finish in 2000ms

n=10 requires generating 10! permutations; O(n * n!) complexity must complete within 2 seconds.

💡 Generating all permutations is factorial time; optimize recursion and memoization.

💡 Use bitmask DP to avoid recomputing states.

💡 Avoid unnecessary copying of arrays in recursion.

Why it failed: TLE due to exponential factorial complexity and inefficient recursion. Fix by using memoization and efficient bitmask operations.

✓ Solution runs within time limit using bitmask DP.

Practice

1. Consider the following code snippet that finds the single number in an array using XOR. What is the value of result after processing the third element in the input [2, 3, 2]?

easy

A. 3

B. 1

C. 0

D. 2

Solution

1. Step 1: Trace result after first element (2)

   result = 0 · 2 = 2

2. Step 2: Trace result after second element (3)

   result = 2 · 3 = 1

3. Step 3: Trace result after third element (2)

   result = 1 · 2 = 3

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Final XOR result is 3, the unique element [OK]

Hint: XOR accumulates unique element progressively [OK]

Common Mistakes:

• Off-by-one in loop iteration
• Confusing XOR with addition
• Misreading variable updates

2. Identify the bug in the following code that attempts to count the number of 1 bits in a signed 32-bit integer:

def hammingWeight(n: int) -> int:
    count = 0
    while n != 0:
        count += n & 1
        n >>= 1
    return count

medium

A. The loop condition 'while n != 0' causes infinite loop for negative numbers due to sign extension.

B. The bitwise AND operation 'n & 1' is incorrect; it should be logical AND 'n && 1'.

C. The right shift operator 'n >>= 1' should be replaced with left shift to count bits correctly.

D. The initial count variable should be initialized to 1 instead of 0.

Solution

1. Step 1: Understand signed right shift behavior

   In Python, right shifting a negative number propagates the sign bit, so n never becomes zero.

2. Step 2: Analyze loop condition

   Because n != 0 remains true for negative numbers, the loop never terminates, causing infinite loop.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Signed right shift on negative numbers causes infinite loop [OK]

Hint: Signed right shift on negatives causes infinite loop [OK]

Common Mistakes:

• Confusing bitwise AND with logical AND
• Wrong shift direction
• Incorrect count initialization

3. The following code attempts to find the single number in an array using XOR. Identify the line containing the subtle bug that causes incorrect results.

medium

A. Line 2: Initializing result to nums[0] instead of 0

B. Line 4: Returning result without sorting

C. Line 3: Looping over nums including the first element again

D. Line 1: Function definition

Solution

1. Step 1: Analyze initialization of result

   Result initialized to nums[0], then XORed again with nums[0] in loop, cancelling it out incorrectly.

2. Step 2: Understand correct initialization

   Result should start at 0 so XOR accumulates all elements exactly once.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Initializing to zero avoids double XOR of first element [OK]

Hint: XOR accumulator must start at zero [OK]

Common Mistakes:

• Initializing with first element causes double XOR
• Loop includes first element again
• Misplaced return statement

4. Suppose the input array can contain duplicate numbers and the missing number is still from 0 to n. Which modification to the XOR approach correctly finds the missing number?

hard

A. XOR approach fails; instead use a frequency count array to find the missing number

B. Use XOR as usual; duplicates cancel out so the missing number is found correctly

C. Sort the array and scan for the first missing number ignoring duplicates

D. Use sum approach but subtract duplicates only once to avoid errors

Solution

1. Step 1: Understand XOR behavior with duplicates

   Duplicates XORed twice cancel out, but duplicates in input break the assumption of unique elements, causing incorrect missing number.

2. Step 2: Identify correct approach for duplicates

   Frequency counting tracks occurrences accurately, allowing detection of missing number despite duplicates.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   XOR fails with duplicates; frequency count handles duplicates correctly [OK]

Hint: XOR assumes unique elements; duplicates break cancellation [OK]

Common Mistakes:

• Assuming XOR works with duplicates
• Trying to fix sum approach without frequency
• Ignoring duplicates impact on XOR

5. Suppose you want to generate a Gray code sequence where each number can be repeated multiple times (i.e., reuse allowed). Which modification to the optimal bitwise formula approach correctly handles this variant?

hard

A. Use brute force backtracking to generate sequences with repeated numbers allowed.

B. Modify the formula to i ^ (i >> 1) but allow duplicates by not enforcing uniqueness in the output list.

C. Use the same XOR formula but generate only n numbers instead of 2^n.

D. No modification needed; the original formula already allows repeated numbers.

Solution

1. Step 1: Understand the reuse constraint

   Allowing repeated numbers breaks the Gray code uniqueness property, so the original formula which generates unique numbers won't work.

2. Step 2: Identify suitable approach

   Brute force backtracking can explore sequences with repeated numbers and enforce the one-bit difference condition without uniqueness constraints.

3. Step 3: Why other options fail

   Original formula always generates unique numbers; simply allowing duplicates in output breaks Gray code definition; generating fewer numbers misses full coverage.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Backtracking is flexible enough to handle reuse constraints [OK]

Hint: Original formula assumes unique numbers; reuse needs backtracking [OK]

Common Mistakes:

• Assuming original formula can be tweaked trivially for reuse
• Thinking fewer numbers still form valid Gray code
• Ignoring uniqueness property of Gray code