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.lengthThis 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.
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:
Java
Time Complexity: O(3^n) due to subset enumeration with bitmask optimization.
Space Complexity: O(2^n) to store DP table states.
| 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. |
Find Minimum in Rotated Sorted Array - Binary Search - Leetcode 153 - Python • NeetCode • 356,602 views views
Watch 9 more video solutions →Practice Minimum Incompatibility with our built-in code editor and test cases.
Practice on FleetCode