How do you solve Minimum Incompatibility on LeetCode?

Minimum Incompatibility (LeetCode #1681) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(n) space complexity. Key insight: The problem requires careful management of subsets to avoid duplicates.

What is the brute force solution for Minimum Incompatibility?

The brute force approach for Minimum Incompatibility is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of subsets and calculating their incompatibility. While straightforward, this method is inefficient due to its exponential time complexity. The optimal Optimal Solution approach improves this to O(n * 2^n).

What are the interview tips for Minimum Incompatibility?

Always clarify the constraints and edge cases before diving into coding. Discuss your thought process and approach with the interviewer. Be prepared to optimize your solution after the initial brute force implementation.

What are common mistakes when solving Minimum Incompatibility?

Failing to check for duplicates in subsets. Not considering the case where k does not divide the length of nums.

#1681

Minimum Incompatibility

Hard

ArrayHash TableDynamic ProgrammingBit ManipulationBitmaskBacktrackingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n)
Space	O(1)	O(n)

💡

Intuition

Time O(n * 2^n)Space O(n)

The optimal solution uses backtracking with pruning to efficiently explore valid subsets. By keeping track of used elements and leveraging bit manipulation, we can significantly reduce the search space.

⚙️

Algorithm

3 steps

1Step 1: Sort the input array to facilitate easier subset creation.
2Step 2: Use a recursive function to try to build subsets while tracking the current incompatibility.
3Step 3: Use a bitmask to represent which elements have been used, allowing for efficient checks and updates.

solution.py35 lines

1from itertools import combinations
2
3def minIncompatibility(nums, k):
4    n = len(nums)
5    if n % k != 0:
6        return -1
7    nums.sort()
8    subset_size = n // k
9    min_incompatibility = float('inf')
10
11    def backtrack(start, used_mask, current_sum, count):
12        nonlocal min_incompatibility
13        if count == k:
14            min_incompatibility = min(min_incompatibility, current_sum)
15            return
16        if current_sum >= min_incompatibility:
17            return
18        for i in range(start, n):
19            if used_mask & (1 << i) == 0:
20                new_mask = used_mask
21                new_max = nums[i]
22                new_min = nums[i]
23                new_count = 1
24                for j in range(i + 1, n):
25                    if used_mask & (1 << j) == 0:
26                        new_mask |= (1 << j)
27                        new_max = max(new_max, nums[j])
28                        new_min = min(new_min, nums[j])
29                        new_count += 1
30                        if new_count == subset_size:
31                            backtrack(j + 1, new_mask, current_sum + (new_max - new_min), count + 1)
32                            break
33
34    backtrack(0, 0, 0, 0)
35    return min_incompatibility if min_incompatibility != float('inf') else -1

ℹ

Complexity note: The time complexity is O(n * 2^n) due to the recursive exploration of subsets and the use of bitmasks to track used elements, which allows for efficient checks.

1The problem requires careful management of subsets to avoid duplicates.
2Sorting the input array helps in efficiently calculating incompatibilities.

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