LeetCode Solutions Blog

Minimum Incompatibility

Number: 1799

Difficulty: Hard

Paid? No

Companies: Microsoft

Problem Description

Given an integer array and an integer k, you are required to partition the array into k subsets of equal size such that no subset contains duplicate elements. The incompatibility of a subset is defined as the difference between its maximum and minimum elements. The goal is to minimize the total sum of incompatibilities among all k subsets. If it is not possible to form such k subsets, return -1.

Key Insights

The total number of elements is n and must be divisible by k (subset size = n / k).
For each subset, if there are duplicate values the subset is invalid.
Precompute all possible valid subsets (using bit masks) that have exactly subset size elements and store their incompatibility (max - min).
Use dynamic programming with bitmasking where each bit represents whether an element from the original array has been used.
For each state (mask) in DP, try to add one of the valid groups (bit masks) that do not conflict with the current state.
The answer will be the minimum sum computed when all elements are used; otherwise return -1.

Space and Time Complexity

Time Complexity: O(2^n * V) where V is the number of valid groups (which is bounded by C(n, n/k)), making it feasible for n up to 16. Space Complexity: O(2^n) for the DP table and precomputed valid groups.

Solution

The solution involves two main phases:

Precomputation: Iterate over all bit masks of size n. For subsets of size equal to n/k, check for duplicates. If a subset is valid, compute its incompatibility (max - min) and store it.
Dynamic Programming: Use a DP array of size 2^n, where each index represents a subset of chosen indices. Start from the empty subset (mask = 0) and try to add valid group masks which do not intersect with the current state. Update the DP state to reflect the minimal incompatibility sum to reach that state. The answer is found when the DP state for mask = (1<<n) - 1 is reached.

Code Solutions

# Python solution with line-by-line comments

class Solution:
    def minimumIncompatibility(self, nums, k):
        from math import inf
        n = len(nums)
        group_size = n // k
        
        # Precompute all valid groups (subsets) that have exactly group_size elements.
        valid_groups = {}
        # There are 2^n possible masks. Iterate over each one.
        for mask in range(1 << n):
            # Only consider masks with exactly group_size selected elements.
            if bin(mask).count("1") == group_size:
                seen = set()
                group_nums = []
                valid = True
                # Collect numbers corresponding to indices set in mask.
                for i in range(n):
                    if mask & (1 << i):
                        if nums[i] in seen:
                            valid = False
                            break
                        seen.add(nums[i])
                        group_nums.append(nums[i])
                if valid:
                    # Calculate incompatibility as max - min.
                    valid_groups[mask] = max(group_nums) - min(group_nums)

        # DP using bitmask (dp[mask] = min incompatibility sum for state represented by mask)
        dp = [inf] * (1 << n)
        dp[0] = 0
        
        # Iterate over all possible states.
        for mask in range(1 << n):
            # If current state is not reachable, skip.
            if dp[mask] == inf:
                continue
            # available holds indices not chosen yet.
            available = ((1 << n) - 1) ^ mask
            # Iterate over all valid groups.
            for group_mask, incompat in valid_groups.items():
                # Check if the valid group can be added to the current state (no conflict)
                if (available & group_mask) == group_mask:
                    new_mask = mask | group_mask
                    dp[new_mask] = min(dp[new_mask], dp[mask] + incompat)
        
        # If the final state is still inf, then partitioning was not possible.
        return dp[(1 << n) - 1] if dp[(1 << n) - 1] != inf else -1

# Example usage:
# sol = Solution()
# print(sol.minimumIncompatibility([1,2,1,4], 2))