Permutation Sequence (Kth)

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Steps

setup

Initialize factorial array

Create a factorial array of size n with all elements initialized to 1. This array will store factorial values from 0! to (n-1)!.

💡 Precomputing factorials allows quick calculation of how many permutations start with each number at each position.

Line:

factorial = [1] * n

💡 Factorials represent the number of permutations possible for the remaining positions.

setup

Compute factorial values

Compute factorial values for indices 1 and 2 by multiplying the previous factorial by the current index.

💡 This step fills the factorial array with correct values to use for index calculations later.

Line:

for i in range(1, n):
    factorial[i] = factorial[i - 1] * i

💡 factorial[1] = 1, factorial[2] = 2, representing 1! and 2! respectively.

setup

Initialize numbers list

Create a list of numbers from 1 to n to represent the available digits for the permutation.

💡 This list will shrink as numbers are chosen for the permutation.

Line:

numbers = list(range(1, n + 1))

💡 The numbers list starts as [1, 2, 3], representing all digits to choose from.

setup

Adjust k to zero-based index

Subtract 1 from k to convert it from 1-based to zero-based indexing for easier calculations.

💡 Zero-based indexing aligns with array indices and factorial calculations.

Line:

k -= 1  # zero-based index

💡 k changes from 3 to 2, meaning we want the 2nd index permutation in zero-based terms.

traverse

Calculate index for first position

Calculate idx by dividing k by factorial[2] (2!). idx = 2 // 2 = 1. This selects the second number in the list for the first position.

💡 This determines which number to pick first based on how many permutations each number leads.

Line:

idx = k // factorial[i]

💡 Choosing index 1 means the first digit is '2' (numbers[1]).

traverse

Update k after first selection

Update k to k modulo factorial[2], k = 2 % 2 = 0. This resets k for the next position's calculation.

💡 Updating k narrows down the position within the smaller permutation block.

Line:

k %= factorial[i]

💡 k=0 means the next selection is the first in the remaining permutations.

insert

Append first chosen number and remove it

Append '2' (numbers[1]) to the result and remove it from the numbers list, leaving [1, 3].

💡 Removing the chosen number prevents reuse and shrinks the problem size.

Line:

result.append(str(numbers[idx]))
numbers.pop(idx)

💡 The first digit of the permutation is fixed as '2'.

traverse

Calculate index for second position

Calculate idx = k // factorial[1] = 0 // 1 = 0. Select the first number in the updated list [1,3].

💡 This picks the next digit based on the updated k and factorial values.

Line:

idx = k // factorial[i]

💡 Choosing index 0 means the second digit is '1'.

traverse

Update k after second selection

Update k = k % factorial[1] = 0 % 1 = 0, preparing for the last digit selection.

💡 k remains zero, indicating the first permutation in the last block.

Line:

k %= factorial[i]

💡 k=0 means the next digit is the first in the remaining list.

insert

Append second chosen number and remove it

Append '1' (numbers[0]) to the result and remove it from the list, leaving [3].

💡 The permutation is building with the second digit fixed.

Line:

result.append(str(numbers[idx]))
numbers.pop(idx)

💡 The permutation so far is '21'.

traverse

Calculate index for last position

Calculate idx = k // factorial[0] = 0 // 1 = 0. Select the only remaining number '3'.

💡 The last digit is chosen from the remaining single number.

Line:

idx = k // factorial[i]

💡 The last digit completes the permutation.

insert

Append last chosen number and remove it

Append '3' to the result and remove it from the numbers list, which becomes empty.

💡 The permutation is now complete with all digits chosen.

Line:

result.append(str(numbers[idx]))
numbers.pop(idx)

💡 The final permutation string is '213'.

reconstruct

Return final permutation string

Join the result list into a string and return it as the kth permutation.

💡 This step outputs the final answer after all selections.

Line:

return ''.join(result)

💡 The algorithm efficiently computed the 3rd permutation without enumerating all permutations.

def getPermutation(n, k):
    factorial = [1] * n  # STEP 1
    for i in range(1, n):  # STEP 2
        factorial[i] = factorial[i - 1] * i
    numbers = list(range(1, n + 1))  # STEP 3
    k -= 1  # zero-based index adjustment STEP 4
    result = []
    for i in range(n - 1, -1, -1):  # STEP 5-12
        idx = k // factorial[i]  # STEP 5,8,11
        k %= factorial[i]  # STEP 6,9,12
        result.append(str(numbers[idx]))  # STEP 7,10,12
        numbers.pop(idx)
    return ''.join(result)  # STEP 13

if __name__ == '__main__':
    print(getPermutation(3, 3))  # Output: '213'

📊

Permutation Sequence (Kth) - Watch the Algorithm Execute, Step by Step

Watching this step-by-step reveals how the factorial number system helps directly compute the kth permutation without generating all permutations.

setup

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Result: ""

fill_row

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Result: ""

setup

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Result: ""

setup

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Result: ""

compare

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Result: ""

move_left

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Result: ""

delete

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Result: "2"

compare

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Result: "2"

move_left

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Result: "2"

delete

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Result: "21"

compare

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Result: "21"

delete

Result: "213"

record

Result: "213"

Key Takeaways

✓ The factorial number system partitions permutations into blocks, allowing direct calculation of the kth permutation without generating all permutations.

This insight is hard to see from code alone because the math behind factorial indexing is abstract without visualization.

✓ At each step, the algorithm selects a digit by dividing k by the factorial of remaining positions, then updates k to the remainder, progressively narrowing down the permutation.

Seeing k and idx update step-by-step clarifies how the algorithm zooms into the correct permutation.

✓ Removing chosen numbers from the list ensures no repeats and reduces the problem size, which is visually clear when watching the array shrink.

This concrete example shows how the numbers list changes, which is less obvious when reading code.

Practice

1. You are given a string containing only digits. Your task is to split it into exactly four parts such that each part represents a valid IP address segment (0 to 255, no leading zeros except for '0'). Which algorithmic approach guarantees finding all valid IP addresses efficiently?

easy

A. Sorting the digits and then partitioning into four segments

B. Greedy approach that picks the shortest valid segment at each step

C. Dynamic programming to count the number of valid segmentations

D. Backtracking with pruning to explore all valid segmentations

Solution

1. Step 1: Understand the problem constraints

   The problem requires enumerating all valid IP addresses, which involves exploring multiple segmentations of the string.

2. Step 2: Evaluate algorithm suitability

   Greedy approaches fail because they may miss valid segmentations; DP counting does not generate all solutions; sorting digits destroys original order. Backtracking with pruning systematically explores all valid partitions.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Backtracking explores all partitions with pruning to avoid invalid segments [OK]

Hint: Backtracking explores all segmentations with pruning [OK]

Common Mistakes:

• Assuming greedy can find all valid IPs
• Using DP counting without generating solutions

2. Consider the following code snippet for palindrome partitioning. Which line contains a subtle bug that causes incorrect or duplicate partitions?

def partition(s):
    def is_palindrome(left, right):
        while left < right:
            if s[left] != s[right]:
                return False
            left += 1
            right -= 1
        return True

    def backtrack(start, path):
        if start == len(s):
            result.append(path)  # Line X
            return
        for end in range(start, len(s)):
            if is_palindrome(start, end):
                path.append(s[start:end+1])
                backtrack(end+1, path)
                path.pop()

    result = []
    backtrack(0, [])
    return result

medium

A. Line 6: palindrome check misses single character substrings

B. Line X: appending path without copying causes shared references

C. Line 12: for loop range should be from start+1 to len(s)+1

D. Line 15: missing path.pop() after backtrack call

Solution

1. Step 1: Identify how results are stored

   Appending path directly stores a reference, so later modifications affect all stored results.

2. Step 2: Correct approach

   Use result.append(path[:]) to append a copy, preserving each partition snapshot.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Shared references cause incorrect final partitions [OK]

Hint: Always append a copy of path to results [OK]

Common Mistakes:

• Forgetting to copy path before appending
• Incorrect palindrome check boundaries
• Forgetting to pop after recursion

3. What is the time complexity of the optimal backtracking solution for the Word Search problem, where N is the number of cells in the board and L is the length of the word?

medium

A. O(N * 4^L) because each cell explores 4 directions for each character

B. O(N * 3^L) because after the first character, each step explores up to 3 directions

C. O(N * L) because each cell is visited once per character

D. O(N^2) because the board is searched multiple times for each character

Solution

1. Step 1: Identify branching factor per recursion

   From each cell, first character has 4 directions, subsequent steps have at most 3 (excluding the cell we came from).

2. Step 2: Calculate total complexity

   For each of N cells, explore up to 3^L paths, so total time is O(N * 3^L).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   3^L accounts for pruning revisits, not 4^L [OK]

Hint: Remember first step has 4 directions, next steps 3 due to no revisiting [OK]

Common Mistakes:

• Assuming 4^L for all steps
• Confusing L with N in complexity

4. Suppose the N-Queens problem is modified so that queens can be placed multiple times in the same column (i.e., column constraint is relaxed), but diagonal attacks still apply. Which modification to the bitmask backtracking algorithm correctly adapts to this variant?

hard

A. Remove the column bitmask tracking and only track diagonal attacks in recursion

B. Keep column bitmask but allow multiple queens by resetting it each row

C. Use a greedy approach placing queens in any column without pruning

D. Track columns as before but ignore diagonal bitmasks

Solution

1. Step 1: Understand relaxed constraints

   Since columns can have multiple queens, column bitmask is no longer needed to prune placements.

2. Step 2: Adapt pruning to only diagonal attacks

   Keep diag1 and diag2 bitmasks to prune diagonal conflicts, but remove column mask from available_positions calculation.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Removing column mask allows multiple queens per column while still pruning diagonals [OK]

Hint: Relaxed constraints remove column pruning, keep diagonal pruning [OK]

Common Mistakes:

• Resetting column mask each row incorrectly
• Ignoring diagonal pruning
• Using greedy without pruning

5. Suppose the problem is modified so that IP address segments can be reused multiple times (i.e., segments can overlap and be repeated). Which modification to the backtracking algorithm correctly handles this variant without generating duplicates or invalid IPs?

hard

A. Remove the pruning condition on remaining characters and segments to allow overlapping segments

B. Use a memoization cache keyed by (start, segments formed) to avoid recomputing paths

C. Allow segments to be reused by resetting start index to previous segment start after each addition

D. Change the algorithm to generate all permutations of segments without pruning

Solution

1. Step 1: Understand reuse implies overlapping segments

   Allowing reuse means the same substring can be used multiple times, potentially causing repeated computations.

2. Step 2: Apply memoization to avoid duplicates

   Memoization keyed by (start, segments formed) prevents recomputing the same subproblems and avoids duplicates.

3. Step 3: Evaluate other options

   Removing pruning causes inefficiency; resetting start index breaks segment order; generating all permutations ignores validity and pruning.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Memoization efficiently handles reuse without duplicates [OK]

Hint: Memoization avoids recomputation when segments can be reused [OK]

Common Mistakes:

• Removing pruning causes timeouts
• Resetting start index breaks segment order