LeetCode 1681 - Minimum Incompatibility

LeetCode Problem 1681

Difficulty: 🔴 Hard
Topics: Array, Hash Table, Dynamic Programming, Bit Manipulation, Bitmask

Solution

Problem Understanding

The problem is asking us to partition an integer array nums into exactly 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 values. Our goal is to find the minimum possible sum of these incompatibilities across all subsets after an optimal partition, or return -1 if no valid partition exists.

The input nums represents an array of integers where each element is between 1 and nums.length. The integer k determines the number of subsets we need to form, and the constraint nums.length % k == 0 ensures that each subset will have equal size n = nums.length / k.

Key constraints and implications are:

• 1 <= nums.length <= 16 means the array is small enough to consider bitmasking or dynamic programming over subsets.
• nums[i] <= nums.length allows us to potentially use frequency counts or set-based checks efficiently.
• Subsets cannot have repeated elements; thus, if any element occurs more than k times, a valid distribution is impossible.

Important edge cases include:

• Arrays with duplicates exceeding k occurrences, which must return -1.
• Arrays where all elements are identical, which may also lead to an impossible distribution.
• Minimal size arrays, e.g., nums.length == k, where each subset will contain exactly one element, making the incompatibility sum zero.

Approaches

A brute-force approach would involve generating all possible partitions of nums into k subsets and calculating the sum of incompatibilities for each valid partition. This is guaranteed to be correct but highly inefficient because the number of partitions grows factorially with nums.length. With nums.length up to 16, this is computationally infeasible.

The key insight for a better solution is to leverage bitmask dynamic programming. Each state in the DP represents a subset of elements we have already used, encoded as a bitmask. We iterate over all combinations of subset_size = n/k elements, precompute the incompatibility of valid subsets, and update the DP state if the subset can be added to the current mask without repeating elements. This reduces the problem from factorial complexity to something manageable using 2^n states combined with efficient subset enumeration.

Algorithm Walkthrough

1. Compute the size of each subset: subset_size = len(nums) // k.
2. Check if any element occurs more than k times. If so, return -1 immediately, as no valid partition exists.
3. Precompute all valid subsets of size subset_size without duplicates. For each valid subset, calculate its incompatibility (max - min) and store it in a dictionary keyed by the bitmask representing that subset.
4. Initialize a DP array of size 2^n with inf, where dp[mask] represents the minimum sum of incompatibilities for the subset of elements indicated by mask. Set dp[0] = 0.
5. Iterate over each DP mask. For each mask, try adding a valid subset that does not overlap with already used elements (using bitwise AND to check). Update the DP value of the new mask with the sum of incompatibilities.
6. After processing all masks, the DP value at the mask with all elements included ((1 << n) - 1) gives the minimum sum. If it is still inf, return -1.

Why it works: The DP guarantees that each state only considers non-overlapping, valid subsets. By iterating over all possible subsets and updating DP incrementally, we explore all feasible partitions while avoiding duplicates within subsets. The minimum sum is therefore guaranteed.

Python Solution

from typing import List
from itertools import combinations

class Solution:
    def minimumIncompatibility(self, nums: List[int], k: int) -> int:
        from collections import Counter
        n = len(nums)
        subset_size = n // k
        
        # Early check for impossible cases
        if any(v > k for v in Counter(nums).values()):
            return -1
        
        # Map each valid subset (bitmask) to its incompatibility
        nums_index = list(range(n))
        valid_subsets = {}
        for combo in combinations(nums_index, subset_size):
            elements = [nums[i] for i in combo]
            if len(set(elements)) == subset_size:  # ensure no duplicates
                mask = sum(1 << i for i in combo)
                valid_subsets[mask] = max(elements) - min(elements)
        
        # Initialize DP
        INF = float('inf')
        dp = [INF] * (1 << n)
        dp[0] = 0
        
        for mask in range(1 << n):
            if dp[mask] == INF:
                continue
            for submask, incompat in valid_subsets.items():
                if mask & submask == 0:
                    dp[mask | submask] = min(dp[mask | submask], dp[mask] + incompat)
        
        return dp[(1 << n) - 1] if dp[(1 << n) - 1] != INF else -1

Explanation: We first check for impossible distributions using element frequency. Next, all valid subsets of the required size are precomputed using combinations, ensuring no duplicates. We then use bitmask DP to explore all combinations of these subsets without overlap, updating the minimum incompatibility sum incrementally. The final DP state represents using all elements.

Go Solution

package main

import (
	"math"
)

func minimumIncompatibility(nums []int, k int) int {
	n := len(nums)
	subsetSize := n / k

	// Early check for impossible cases
	counts := make([]int, n+1)
	for _, v := range nums {
		counts[v]++
		if counts[v] > k {
			return -1
		}
	}

	// Precompute all valid subsets of size subsetSize
	validSubsets := map[int]int{}
	for mask := 0; mask < (1 << n); mask++ {
		if bitCount(mask) == subsetSize {
			seen := map[int]bool{}
			minVal, maxVal := math.MaxInt32, math.MinInt32
			valid := true
			for i := 0; i < n; i++ {
				if mask&(1<<i) != 0 {
					if seen[nums[i]] {
						valid = false
						break
					}
					seen[nums[i]] = true
					if nums[i] < minVal {
						minVal = nums[i]
					}
					if nums[i] > maxVal {
						maxVal = nums[i]
					}
				}
			}
			if valid {
				validSubsets[mask] = maxVal - minVal
			}
		}
	}

	// DP
	INF := math.MaxInt32
	dp := make([]int, 1<<n)
	for i := range dp {
		dp[i] = INF
	}
	dp[0] = 0

	for mask := 0; mask < (1 << n); mask++ {
		if dp[mask] == INF {
			continue
		}
		for submask, incompat := range validSubsets {
			if mask&submask == 0 {
				newMask := mask | submask
				if dp[newMask] > dp[mask]+incompat {
					dp[newMask] = dp[mask] + incompat
				}
			}
		}
	}

	if dp[(1<<n)-1] == INF {
		return -1
	}
	return dp[(1<<n)-1]
}

func bitCount(x int) int {
	count := 0
	for x > 0 {
		count += x & 1
		x >>= 1
	}
	return count
}

Go-specific notes: Go requires manual bit counting for subset size validation. We use map[int]int and map[int]bool for frequency and duplicate checks, and slices for DP. math.MaxInt32 is used to represent infinity.

Worked Examples

Example 1: nums = [1,2,1,4], k = 2

• Subset size = 2

• Valid subsets: [1,2], [1,4], [2,4]

• DP iteration:

• Start with mask 0, dp[0] = 0

• Add [1,2] → mask = 0b0011, dp[0b0011] = 1

• Add [1,4] → mask = 0b0101, dp[0b0101] = 3

• Combine non-overlapping subsets → dp[0b1111] = 4

• Return 4

Example 2: nums = [6,3,8,1,3,1,2,2], k = 4

• Subset size = 2
• Valid subsets without duplicates: [6,8], [1,2], [1,3], [2,3]
• DP constructs