LeetCode 1681: Minimum Incompatibility Solution

Master LeetCode problem 1681 (Minimum Incompatibility), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1681. Minimum Incompatibility

Hard

Array Dynamic Programming Bit Manipulation Bitmask

Problem Explanation

Explanation:

To solve this problem, we can use backtracking with some optimizations. We first sort the input array nums. Then, we iterate through all possible subsets of size n/k where n is the length of nums and k is the number of subsets. For each subset, we check if it is a valid subset (no duplicates) and calculate the incompatibility. We keep track of the minimum incompatibility found so far.

To optimize the process, we can pre-calculate the incompatibility of subsets of size n/k and store them in a map to avoid redundant calculations. We also use bit manipulation to efficiently generate all possible subsets.

The time complexity of this approach is O(2^n * n^2) where n is the length of nums, and the space complexity is O(2^n * n).

Solution Code

class Solution {
    public int minimumIncompatibility(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        int groupSize = n / k;
        
        Map<Integer, Integer> subsetIncompatibility = new HashMap<>();
        for (int mask = 0; mask < (1 << n); mask++) {
            if (Integer.bitCount(mask) == groupSize) {
                List<Integer> subset = new ArrayList<>();
                boolean validSubset = true;
                for (int i = 0; i < n; i++) {
                    if ((mask & (1 << i)) > 0) {
                        subset.add(nums[i]);
                        if (i > 0 && nums[i] == nums[i - 1] && (mask & (1 << (i - 1))) == 0) {
                            validSubset = false;
                            break;
                        }
                    }
                }
                if (validSubset) {
                    int incompatibility = subset.get(subset.size() - 1) - subset.get(0);
                    subsetIncompatibility.put(mask, incompatibility);
                }
            }
        }
        
        int[] dp = new int[1 << n];
        Arrays.fill(dp, -1);
        dp[0] = 0;
        
        for (int mask = 0; mask < (1 << n); mask++) {
            if (Integer.bitCount(mask) % groupSize == 0) {
                for (int subset : subsetIncompatibility.keySet()) {
                    if ((subset & mask) == subset && dp[mask ^ subset] != -1) {
                        if (dp[mask] == -1 || dp[mask] > dp[mask ^ subset] + subsetIncompatibility.get(subset)) {
                            dp[mask] = dp[mask ^ subset] + subsetIncompatibility.get(subset);
                        }
                    }
                }
            }
        }
        
        return dp[(1 << n) - 1];
    }
}

Try It Yourself

Loading code editor...

Frequently Asked Questions

How to solve LeetCode 1681 (Minimum Incompatibility)?

This page provides optimized solutions for LeetCode problem 1681 (Minimum Incompatibility) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 1681 (Minimum Incompatibility)?

The time complexity for LeetCode 1681 (Minimum Incompatibility) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 1681 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 1681 in Java, C++, or Python.

Back to LeetCode Solutions