Watch 8 video solutions for Minimum Incompatibility, a hard level problem involving Array, Dynamic Programming, Bit Manipulation. This walkthrough by Hua Hua has 1,762 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an integer array nums and an integer k. You are asked to distribute this array into k subsets of equal size such that there are no two equal elements in the same subset.
A subset's incompatibility is the difference between the maximum and minimum elements in that array.
Return the minimum possible sum of incompatibilities of the k subsets after distributing the array optimally, or return -1 if it is not possible.
A subset is a group integers that appear in the array with no particular order.
Example 1:
Input: nums = [1,2,1,4], k = 2 Output: 4 Explanation: The optimal distribution of subsets is [1,2] and [1,4]. The incompatibility is (2-1) + (4-1) = 4. Note that [1,1] and [2,4] would result in a smaller sum, but the first subset contains 2 equal elements.
Example 2:
Input: nums = [6,3,8,1,3,1,2,2], k = 4 Output: 6 Explanation: The optimal distribution of subsets is [1,2], [2,3], [6,8], and [1,3]. The incompatibility is (2-1) + (3-2) + (8-6) + (3-1) = 6.
Example 3:
Input: nums = [5,3,3,6,3,3], k = 3 Output: -1 Explanation: It is impossible to distribute nums into 3 subsets where no two elements are equal in the same subset.
Constraints:
1 <= k <= nums.length <= 16nums.length is divisible by k1 <= nums[i] <= nums.lengthProblem Overview: You are given an array nums and an integer k. Split the array into k subsets of equal size. The incompatibility of a subset is max(subset) - min(subset). Each subset must contain unique values. The goal is to minimize the total incompatibility across all subsets.
Approach 1: Backtracking with Pruning (Exponential Time)
This approach builds the k groups recursively. Sort nums first so duplicates are easy to detect. During recursion, try placing each unused number into the current subset while ensuring no duplicate value exists in that subset. Track the current subset's min and max to compute incompatibility when the subset reaches size n/k. Pruning drastically reduces the search space: skip duplicate placements, stop when partial cost already exceeds the best answer, and avoid building identical subsets. Time complexity is roughly O(n!) in the worst case due to permutations, with O(n) auxiliary space for recursion and tracking used elements.
This method is straightforward to implement and works well when pruning eliminates large portions of the search tree. It demonstrates strong reasoning about constraints and subset construction.
Approach 2: Bitmask Dynamic Programming (Optimal for n ≤ 16)
The optimal solution uses bitmask representation and dynamic programming. Represent the used elements with a mask of length n. Each state dp[mask] stores the minimum incompatibility for selecting the elements represented by that mask. Precompute all valid subsets of size groupSize = n/k that contain unique numbers. For each valid subset, calculate its incompatibility max - min.
During DP transitions, iterate through masks whose bit count is divisible by groupSize. For the remaining unused elements, try adding a precomputed valid subset (another mask). Update dp[newMask] = min(dp[newMask], dp[mask] + subsetCost). Bit operations make membership checks and merges efficient. The overall complexity is about O(3^n) time with O(2^n) space, which works because n ≤ 16. This technique heavily relies on efficient subset enumeration and bit manipulation on the array indices.
Recommended for interviews: Interviewers typically expect the bitmask DP approach. A brute-force or backtracking explanation shows you understand subset generation, but the DP solution demonstrates mastery of state compression and optimization techniques commonly used in hard combinatorial problems.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Backtracking with Pruning | O(n!) worst case | O(n) | When exploring subset construction logic or when pruning drastically reduces the search space |
| Bitmask Dynamic Programming | O(3^n) | O(2^n) | Optimal for n ≤ 16; preferred interview solution using state compression |