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Minimum Incompatibility - Solution & Explanation

HardArrayDynamic ProgrammingBit ManipulationBitmask18 min readAsked at: Microsoft
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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

Approach Overview

Problem Overview: You are given an array nums and an integer k. Split the array into k subsets of equal size. The incompatibility of a subset is max(subset) - min(subset). Each subset must contain unique values. The goal is to minimize the total incompatibility across all subsets.

Approach 1: Backtracking with Pruning (Exponential Time)

This approach builds the k groups recursively. Sort nums first so duplicates are easy to detect. During recursion, try placing each unused number into the current subset while ensuring no duplicate value exists in that subset. Track the current subset's min and max to compute incompatibility when the subset reaches size n/k. Pruning drastically reduces the search space: skip duplicate placements, stop when partial cost already exceeds the best answer, and avoid building identical subsets. Time complexity is roughly O(n!) in the worst case due to permutations, with O(n) auxiliary space for recursion and tracking used elements.

This method is straightforward to implement and works well when pruning eliminates large portions of the search tree. It demonstrates strong reasoning about constraints and subset construction.

Approach 2: Bitmask Dynamic Programming (Optimal for n ≀ 16)

The optimal solution uses bitmask representation and dynamic programming. Represent the used elements with a mask of length n. Each state dp[mask] stores the minimum incompatibility for selecting the elements represented by that mask. Precompute all valid subsets of size groupSize = n/k that contain unique numbers. For each valid subset, calculate its incompatibility max - min.

During DP transitions, iterate through masks whose bit count is divisible by groupSize. For the remaining unused elements, try adding a precomputed valid subset (another mask). Update dp[newMask] = min(dp[newMask], dp[mask] + subsetCost). Bit operations make membership checks and merges efficient. The overall complexity is about O(3^n) time with O(2^n) space, which works because n ≀ 16. This technique heavily relies on efficient subset enumeration and bit manipulation on the array indices.

Recommended for interviews: Interviewers typically expect the bitmask DP approach. A brute-force or backtracking explanation shows you understand subset generation, but the DP solution demonstrates mastery of state compression and optimization techniques commonly used in hard combinatorial problems.

Approach 1: Backtracking with Pruning

This 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.

Code

PythonJavaScript

Python

JavaScript

Complexity

Time Complexity: O(k^n/k) due to combinatorial subset partitioning.

Space Complexity: O(n) for storing states and recursion stack.

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Approach 2: Bitmask and Dynamic Programming

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:

  • Initialize a DP table with size 2^n to store minimum incompatibility given a state of subset usage.
  • Iterate through all possible masks, filtering invalid group sizes.
  • Use a bitmask to select available indices and compute transitions in a subset.

Code

C++Java

C++

Java

Complexity

Time Complexity: O(3^n) due to subset enumeration with bitmask optimization.

Space Complexity: O(2^n) to store DP table states.

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Approach 3: Preprocessing + State Compression + Dynamic Programming

Let's assume that the size of each subset after partitioning is m, so m=\frac{n}{k}, where n is the length of the array.

We can enumerate all subsets i, where i \in [0, 2^n), if the binary representation of subset i has m ones, and the elements in subset i are not repeated, then we can calculate the incompatibility of subset i, denoted as g[i], i.e., g[i]=max_{j \in i} {nums[j]} - min_{j \in i} {nums[j]}.

Next, we can use dynamic programming to solve.

We define f[i] as the minimum sum of incompatibilities when the current partitioned subset state is i. Initially, f[0]=0, which means no elements are partitioned into the subset, and the rest f[i]=+infty.

For state i, we find all undivided and non-repeated elements, represented by a state mask. If the number of elements in state mask is greater than or equal to m, then we enumerate all subsets j of mask, and satisfy j \subset mask, then f[i \cup j]=min {f[i \cup j], f[i]+g[j]}.

Finally, if f[2^n-1]=+infty, it means that it cannot be partitioned into k subsets, return -1, otherwise return f[2^n-1].

The time complexity is O(3^n), and the space complexity is O(2^n). Here, n is the length of the array.

Code

PythonJavaC++GoTypeScriptC#

Python

Java

C++

Go

TypeScript

C#

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Complexity Comparison

ApproachComplexity
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.

Preprocessing + State Compression + Dynamic Programmingβ€”

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Backtracking with PruningO(n!) worst caseO(n)When exploring subset construction logic or when pruning drastically reduces the search space
Bitmask Dynamic ProgrammingO(3^n)O(2^n)Optimal for n ≀ 16; preferred interview solution using state compression

Video Solution

θŠ±θŠ±ι…± LeetCode 1681. Minimum Incompatibility - εˆ·ι’˜ζ‰Ύε·₯作 EP374 β€’ Hua Hua β€’ 1,765 views views

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Frequently Asked Questions

Is Minimum Incompatibility easy or hard?
Minimum Incompatibility is classified as a Hard problem on LeetCode. It combines subset partitioning, duplicate handling, and state-compression dynamic programming, which requires strong understanding of bitmask techniques and combinatorial optimization.
Minimum Incompatibility Python/Java solution
Python solutions often implement backtracking with pruning because recursion and sets are convenient for duplicate checks. Java or C++ implementations commonly use bitmask dynamic programming for better performance, storing results in arrays indexed by mask states.
How to solve Minimum Incompatibility in O(n)?
Minimum Incompatibility cannot be solved in O(n) because the task requires evaluating combinations of elements across subsets. The optimal solution uses bitmask dynamic programming with O(3^n) complexity, which is feasible because the problem constraint limits n to at most 16.
What is the best approach for Minimum Incompatibility?
The most effective approach is Bitmask Dynamic Programming. Represent selected elements using a bitmask and store the minimum incompatibility for each mask state. Precompute valid subsets of size n/k without duplicate values and transition between masks. This reduces redundant work and solves the problem in about O(3^n) time with O(2^n) space.
Is Minimum Incompatibility asked at Google/Amazon/Meta?
Minimum Incompatibility represents the type of state-compression dynamic programming problem commonly used in interviews at companies like Google, Amazon, and Meta. While the exact question may vary, similar bitmask DP and subset partitioning problems appear frequently in senior-level interviews.
What data structure is used in Minimum Incompatibility?
The solution primarily uses bitmasks to represent subsets of the array. Combined with dynamic programming arrays or hash maps, bit operations allow efficient checks for used elements, subset merges, and enumeration of candidate groups.
What is the time complexity of Minimum Incompatibility?
The optimal bitmask DP solution runs in roughly O(3^n) time with O(2^n) memory, where n ≀ 16. Backtracking solutions can degrade to O(n!) in the worst case, although pruning and sorting can significantly reduce the search space in practice.

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