Leetcode #1681: Minimum Incompatibility
In this guide, we solve Leetcode #1681 Minimum Incompatibility in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Dynamic Programming, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.