LeetCode #1681 — HARD

Minimum Incompatibility

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Dynamic Programming · Bit Manipulation

Example 1

[1,2,1,4]
2

Example 2

[6,3,8,1,3,1,2,2]
4

Example 3

[5,3,3,6,3,3]
3

Step 02

Core Insight

What unlocks the optimal approach

The constraints are small enough for a backtrack solution but not any backtrack solution
If we use a naive n^k don't you think it can be optimized

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1681: Minimum Incompatibility
class Solution {
    public int minimumIncompatibility(int[] nums, int k) {
        int n = nums.length;
        int m = n / k;
        int[] g = new int[1 << n];
        Arrays.fill(g, -1);
        for (int i = 1; i < 1 << n; ++i) {
            if (Integer.bitCount(i) != m) {
                continue;
            }
            Set<Integer> s = new HashSet<>();
            int mi = 20, mx = 0;
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    if (!s.add(nums[j])) {
                        break;
                    }
                    mi = Math.min(mi, nums[j]);
                    mx = Math.max(mx, nums[j]);
                }
            }
            if (s.size() == m) {
                g[i] = mx - mi;
            }
        }
        int[] f = new int[1 << n];
        final int inf = 1 << 30;
        Arrays.fill(f, inf);
        f[0] = 0;
        for (int i = 0; i < 1 << n; ++i) {
            if (f[i] == inf) {
                continue;
            }
            Set<Integer> s = new HashSet<>();
            int mask = 0;
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 0 && !s.contains(nums[j])) {
                    s.add(nums[j]);
                    mask |= 1 << j;
                }
            }
            if (s.size() < m) {
                continue;
            }
            for (int j = mask; j > 0; j = (j - 1) & mask) {
                if (g[j] != -1) {
                    f[i | j] = Math.min(f[i | j], f[i] + g[j]);
                }
            }
        }
        return f[(1 << n) - 1] == inf ? -1 : f[(1 << n) - 1];
    }
}

// Accepted solution for LeetCode #1681: Minimum Incompatibility
func minimumIncompatibility(nums []int, k int) int {
	n := len(nums)
	m := n / k
	const inf = 1 << 30
	f := make([]int, 1<<n)
	g := make([]int, 1<<n)
	for i := range g {
		f[i] = inf
		g[i] = -1
	}
	for i := 1; i < 1<<n; i++ {
		if bits.OnesCount(uint(i)) != m {
			continue
		}
		s := map[int]struct{}{}
		mi, mx := 20, 0
		for j, x := range nums {
			if i>>j&1 == 1 {
				if _, ok := s[x]; ok {
					break
				}
				s[x] = struct{}{}
				mi = min(mi, x)
				mx = max(mx, x)
			}
		}
		if len(s) == m {
			g[i] = mx - mi
		}
	}
	f[0] = 0
	for i := 0; i < 1<<n; i++ {
		if f[i] == inf {
			continue
		}
		s := map[int]struct{}{}
		mask := 0
		for j, x := range nums {
			if _, ok := s[x]; !ok && i>>j&1 == 0 {
				s[x] = struct{}{}
				mask |= 1 << j
			}
		}
		if len(s) < m {
			continue
		}
		for j := mask; j > 0; j = (j - 1) & mask {
			if g[j] != -1 {
				f[i|j] = min(f[i|j], f[i]+g[j])
			}
		}
	}
	if f[1<<n-1] == inf {
		return -1
	}
	return f[1<<n-1]
}

# Accepted solution for LeetCode #1681: Minimum Incompatibility
class Solution:
    def minimumIncompatibility(self, nums: List[int], k: int) -> int:
        n = len(nums)
        m = n // k
        g = [-1] * (1 << n)
        for i in range(1, 1 << n):
            if i.bit_count() != m:
                continue
            s = set()
            mi, mx = 20, 0
            for j, x in enumerate(nums):
                if i >> j & 1:
                    if x in s:
                        break
                    s.add(x)
                    mi = min(mi, x)
                    mx = max(mx, x)
            if len(s) == m:
                g[i] = mx - mi
        f = [inf] * (1 << n)
        f[0] = 0
        for i in range(1 << n):
            if f[i] == inf:
                continue
            s = set()
            mask = 0
            for j, x in enumerate(nums):
                if (i >> j & 1) == 0 and x not in s:
                    s.add(x)
                    mask |= 1 << j
            if len(s) < m:
                continue
            j = mask
            while j:
                if g[j] != -1:
                    f[i | j] = min(f[i | j], f[i] + g[j])
                j = (j - 1) & mask
        return f[-1] if f[-1] != inf else -1

// Accepted solution for LeetCode #1681: Minimum Incompatibility
struct Solution;

use std::collections::HashSet;

impl Solution {
    fn minimum_incompatibility(nums: Vec<i32>, k: i32) -> i32 {
        let k = k as usize;
        let n = nums.len();
        let mut sets: Vec<(u32, i32)> = vec![];
        let mut visited: HashSet<i32> = HashSet::new();
        Self::dfs(0, n / k, 0, &mut visited, &mut sets, &nums, n);
        let mut memo: Vec<Option<Option<i32>>> = vec![None; 1 << n];
        if let Some(sum) = Self::dp((1 << n) - 1, &mut memo, &sets, &nums, n, k) {
            sum
        } else {
            -1
        }
    }

    fn dp(
        cur: u32,
        memo: &mut Vec<Option<Option<i32>>>,
        sets: &[(u32, i32)],
        nums: &[i32],
        n: usize,
        k: usize,
    ) -> Option<i32> {
        if cur == 0 {
            Some(0)
        } else {
            if let Some(res) = memo[cur as usize] {
                res
            } else {
                let mut min_sum = std::i32::MAX;
                for (set, incompatibility) in sets {
                    if (set & cur).count_ones() as usize == n / k {
                        if let Some(sum) = Self::dp(cur & !set, memo, sets, nums, n, k) {
                            min_sum = min_sum.min(incompatibility + sum);
                        }
                    }
                }
                let res = if min_sum == std::i32::MAX {
                    None
                } else {
                    Some(min_sum)
                };
                memo[cur as usize] = Some(res);
                res
            }
        }
    }

    fn dfs(
        start: usize,
        size: usize,
        cur: u32,
        visited: &mut HashSet<i32>,
        all: &mut Vec<(u32, i32)>,
        nums: &[i32],
        n: usize,
    ) {
        if size == 0 {
            all.push((
                cur,
                visited.iter().max().unwrap() - visited.iter().min().unwrap(),
            ));
        } else {
            for i in start..n {
                if visited.insert(nums[i]) {
                    Self::dfs(i + 1, size - 1, cur | 1 << i, visited, all, nums, n);
                    visited.remove(&nums[i]);
                }
            }
        }
    }
}

#[test]
fn test() {
    let nums = vec![1, 2, 1, 4];
    let k = 2;
    let res = 4;
    assert_eq!(Solution::minimum_incompatibility(nums, k), res);
    let nums = vec![6, 3, 8, 1, 3, 1, 2, 2];
    let k = 4;
    let res = 6;
    assert_eq!(Solution::minimum_incompatibility(nums, k), res);
    let nums = vec![5, 3, 3, 6, 3, 3];
    let k = 3;
    let res = -1;
    assert_eq!(Solution::minimum_incompatibility(nums, k), res);
    let nums = vec![14, 4, 6, 6, 4, 14, 13, 12, 3, 1, 7, 14, 3, 10, 5];
    let k = 1;
    let res = -1;
    assert_eq!(Solution::minimum_incompatibility(nums, k), res);
}

// Accepted solution for LeetCode #1681: Minimum Incompatibility
function minimumIncompatibility(nums: number[], k: number): number {
    const n = nums.length;
    const m = Math.floor(n / k);
    const g: number[] = Array(1 << n).fill(-1);
    for (let i = 1; i < 1 << n; ++i) {
        if (bitCount(i) !== m) {
            continue;
        }
        const s: Set<number> = new Set();
        let [mi, mx] = [20, 0];
        for (let j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                if (s.has(nums[j])) {
                    break;
                }
                s.add(nums[j]);
                mi = Math.min(mi, nums[j]);
                mx = Math.max(mx, nums[j]);
            }
        }
        if (s.size === m) {
            g[i] = mx - mi;
        }
    }
    const inf = 1e9;
    const f: number[] = Array(1 << n).fill(inf);
    f[0] = 0;
    for (let i = 0; i < 1 << n; ++i) {
        if (f[i] === inf) {
            continue;
        }
        const s: Set<number> = new Set();
        let mask = 0;
        for (let j = 0; j < n; ++j) {
            if (((i >> j) & 1) === 0 && !s.has(nums[j])) {
                s.add(nums[j]);
                mask |= 1 << j;
            }
        }
        if (s.size < m) {
            continue;
        }
        for (let j = mask; j; j = (j - 1) & mask) {
            if (g[j] !== -1) {
                f[i | j] = Math.min(f[i | j], f[i] + g[j]);
            }
        }
    }
    return f[(1 << n) - 1] === inf ? -1 : f[(1 << n) - 1];
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(3^n)

Space

O(2^n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.