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.
This approach involves recursively trying to distribute elements into k subsets. With each recursive step, we attempt to place each element into a subset, ensuring no two elements in a subset are the same. To optimize, we use pruning techniques to cut off unnecessary recursive calls when we hit an invalid state, such as trying to place duplicate elements into a subset.
This Python solution uses backtracking to distribute the elements of nums into k subsets of equal size, ensuring that no two equal elements reside in the same subset.
The algorithm first sorts the list to facilitate quick checks for duplicate values that cannot be accommodated. The backtracking function attempts to distribute elements among subsets and calculates incompatibility when a valid distribution is found.
Python
JavaScript
Time Complexity: O(k^n/k) due to combinatorial subset partitioning.
Space Complexity: O(n) for storing states and recursion stack.
This approach uses bitmasks to represent subsets and dynamic programming to calculate minimum incompatibilities efficiently. Each state in the DP table is a bitmask representing which elements have been used, and the value corresponds to the minimum incompatibility sum achievable with those used elements.
This C++ solution utilizes bitmask dynamic programming to find the minimum sum of incompatibilities. The bitmask is used to track the state of filled subsets, and dynamic programming transitions are used to fill subsets optimally.
Key Steps:
Time Complexity: O(3^n) due to subset enumeration with bitmask optimization.
Space Complexity: O(2^n) to store DP table states.
Let's assume that the size of each subset after partitioning is m, so m=\frac{n}{k}, where n is the length of the array.
We can enumerate all subsets i, where i \in [0, 2^n), if the binary representation of subset i has m ones, and the elements in subset i are not repeated, then we can calculate the incompatibility of subset i, denoted as g[i], i.e., g[i]=max_{j \in i} {nums[j]} - min_{j \in i} {nums[j]}.
Next, we can use dynamic programming to solve.
We define f[i] as the minimum sum of incompatibilities when the current partitioned subset state is i. Initially, f[0]=0, which means no elements are partitioned into the subset, and the rest f[i]=+infty.
For state i, we find all undivided and non-repeated elements, represented by a state mask. If the number of elements in state mask is greater than or equal to m, then we enumerate all subsets j of mask, and satisfy j \subset mask, then f[i \cup j]=min {f[i \cup j], f[i]+g[j]}.
Finally, if f[2^n-1]=+infty, it means that it cannot be partitioned into k subsets, return -1, otherwise return f[2^n-1].
The time complexity is O(3^n), and the space complexity is O(2^n). Here, n is the length of the array.
| Approach | Complexity |
|---|---|
| Backtracking with Pruning | Time Complexity: O(k^n/k) due to combinatorial subset partitioning. Space Complexity: O(n) for storing states and recursion stack. |
| Bitmask and Dynamic Programming | Time Complexity: O(3^n) due to subset enumeration with bitmask optimization. Space Complexity: O(2^n) to store DP table states. |
| Preprocessing + State Compression + Dynamic Programming | — |
| 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 |
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