How do you solve Maximum K to Sort a Permutation on LeetCode?

Maximum K to Sort a Permutation (LeetCode #3644) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: The AND operation restricts which elements can be swapped.

What is the time complexity of Maximum K to Sort a Permutation?

The optimal solution for Maximum K to Sort a Permutation runs in O(n log n) time and uses O(n) space. Sorting takes O(n log n) and we traverse the array once, leading to O(n log n) overall.

What is the brute force solution for Maximum K to Sort a Permutation?

The brute force approach for Maximum K to Sort a Permutation is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check all possible values of k from 0 to the maximum element in nums. For each k, we see if we can sort the array using allowed swaps based on the AND condition. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum K to Sort a Permutation?

Maximum K to Sort a Permutation uses the following concepts: Array, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Sorting.

What are the interview tips for Maximum K to Sort a Permutation?

Explain your thought process clearly. Use examples to illustrate your approach. Discuss time and space complexity.

What are common mistakes when solving Maximum K to Sort a Permutation?

Forgetting to check all elements in their correct positions. Not considering the effect of AND on possible swaps.

#3644

Maximum K to Sort a Permutation

Medium

ArrayBit ManipulationBit ManipulationSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The maximum k is the bitwise AND of all elements that are not in their correct position. This ensures that we can swap elements to sort the array.

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Algorithm

3 steps

1Step 1: Create a variable to hold the AND result, initialized to all bits set (e.g., ~0).
2Step 2: Iterate through nums and compute the AND for elements not in their sorted position.
3Step 3: Return the computed AND value.

solution.py7 lines

1def maxK(nums):
2    sorted_nums = sorted(nums)
3    k = ~0
4    for i in range(len(nums)):
5        if nums[i] != sorted_nums[i]:
6            k &= nums[i]
7    return k

ℹ

Complexity note: Sorting takes O(n log n) and we traverse the array once, leading to O(n log n) overall.

1The AND operation restricts which elements can be swapped.
2Elements must be in their sorted positions for k to be maximized.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.