1681. Minimum Incompatibility
Description
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 <= 16
nums.length
is divisible byk
1 <= nums[i] <= nums.length
Solution
minimum-incompatibility.py
class Solution:
def minimumIncompatibility(self, nums: List[int], k: int) -> int:
scores = collections.defaultdict(int)
n = len(nums)
per = n // k
candidates = []
for mask in range(1 << n):
count = 0
for j in range(n):
if mask & (1 << j) > 0:
count += 1
if count == per:
s = set()
for j in range(n):
if mask & (1 << j) > 0:
s.add(nums[j])
if len(s) != per: continue
candidates.append(mask)
scores[mask] = max(s) - min(s)
@cache
def go(mask):
if mask == 0: return 0
res = float('inf')
for candidate in candidates:
if (mask & candidate) == candidate:
res = min(res, go(mask ^ candidate) + scores[candidate])
return res
ans = go((1 << n) - 1)
return -1 if ans == float('inf') else ans