# Solving Leetcode Interviews in Seconds with AI: Minimum Incompatibility


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1681" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 by k 1 <= nums[i] <= nums.length  

	# Explanation
	Here's a solution to the problem, along with explanations:

*   **Core Idea:** Use dynamic programming with bit masking to explore all possible subset combinations. The state represents which elements from the original array have already been placed into subsets.
*   **Subset Generation:** For each state, iterate through the remaining elements and form valid subsets (no duplicates) of size `n/k`.
*   **Optimization:** Minimize the sum of incompatibilities by choosing the best combination of subsets at each DP state.

*   **Runtime Complexity:** O(2<sup>n</sup> * n<sup>2</sup>), where n is the length of `nums`.
    **Storage Complexity:** O(2<sup>n</sup>)

	
	# Code
	```python
	from functools import lru_cache

def min_incompatibility(nums, k):
    n = len(nums)
    subset_size = n // k
    counts = {}
    for num in nums:
        counts[num] = counts.get(num, 0) + 1
        if counts[num] > k:
            return -1

    nums.sort()

    @lru_cache(None)
    def dp(mask):
        if mask == (1 << n) - 1:
            return 0

        min_incompatibility = float('inf')

        available_indices = []
        for i in range(n):
            if not (mask & (1 << i)):
                available_indices.append(i)

        if not available_indices:
            return min_incompatibility

        first_index = available_indices[0]

        def get_incompatibility(subset_indices):
            subset_elements = [nums[i] for i in subset_indices]
            return max(subset_elements) - min(subset_elements)

        def is_valid_subset(subset_indices):
            subset_elements = [nums[i] for i in subset_indices]
            return len(set(subset_elements)) == len(subset_elements)

        def generate_subsets(start_index, current_subset, current_mask):
            nonlocal min_incompatibility

            if len(current_subset) == subset_size:
                if is_valid_subset(current_subset):
                    new_mask = current_mask
                    for index in current_subset:
                        new_mask |= (1 << index)
                    
                    incompatibility = get_incompatibility(current_subset)
                    remaining_incompatibility = dp(new_mask)
                    if remaining_incompatibility != float('inf'):
                        min_incompatibility = min(min_incompatibility, incompatibility + remaining_incompatibility)
                return

            for i in range(start_index, len(available_indices)):
                index = available_indices[i]
                generate_subsets(i + 1, current_subset + [index], current_mask)
        
        generate_subsets(0, [], mask)
        

        return min_incompatibility

    result = dp(0)
    return result if result != float('inf') else -1
	```