#1681Hard~50 min

Minimum Incompatibility

Time O(max(n * 2^n, 3^n)) · Space O(2^n) · Official statement on LeetCode

Solutions

// Time:  O(max(n * 2^n, 3^n))
// Space: O(2^n)

class Solution {
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        const vector<int> candidates = findCandidates(nums, k);  // Time: O(n * 2^n)
        const int total = (1 << size(nums)) - 1;
        vector<int> dp(total + 1, numeric_limits<int>::max());
        dp[0] = 0;
        for (int mask = 0; mask <= total; ++mask) {  // Time: O(3^n)
            if (__builtin_popcount(mask) % (size(nums) / k) != 0) {
                continue;
            }
            // submask enumeration:
            // => sum(nCr(n, k) * 2^k for k in xrange(n+1)) = (1 + 2)^n = 3^n
            // => Time: O(3^n), see https://cp-algorithms.com/algebra/all-submasks.html
            for (int submask = mask; submask; submask = (submask - 1) & mask) {
                if (candidates[submask] == numeric_limits<int>::max() ||
                    dp[mask - submask] == numeric_limits<int>::max()) {
                    continue;
                }
                dp[mask] = min(dp[mask], dp[mask - submask] + candidates[submask]);
            }
        }
        return dp.back() != numeric_limits<int>::max() ? dp.back() : -1;
    }

private:                        
    vector<int> findCandidates(const vector<int>& nums, int k) {
        const int total = (1 << size(nums)) - 1;
        vector<int> result(total + 1, numeric_limits<int>::max());
        for (int mask = 0; mask <= total; ++mask) {
            if (__builtin_popcount(mask) != size(nums) / k) {
                continue;
            }
            int lookup = 0;
            int mx = 0, mn = numeric_limits<int>::max();
            bool duplicated = false;
            for (int i = 0; i < size(nums); ++i) {
                if ((mask & (1 << i)) == 0) {
                    continue;
                }
                if (lookup & (1 << nums[i])) {
                    duplicated = true;
                    break;
                }
                lookup |= (1 << nums[i]);
                mx = max(mx, nums[i]);
                mn = min(mn, nums[i]);
            }
            if (duplicated) {
                continue;
            }
            result[mask] = mx - mn;
        }
        return result;
    }
};

// Time:  O(sum(i*d * nCr(i*d, d) * nCr(n, i*d) for i in xrange(1, k+1))) < O(sum(n * 2^m * nCr(n, m) for m in xrange(n+1))) = O(n * 3^n)
// Space: O(n * k)
class Solution_TLE {
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        unordered_map<string, int> lookup;
        int result = backtracking(nums, size(nums) / k, &lookup);
        return result != numeric_limits<int>::max() ? result : -1;
    }

private:
    int backtracking(const vector<int>& nums, int d, unordered_map<string, int> *lookup) {
        if (empty(nums)) {
            return 0;
        }
        string k = key(nums);
        if (!lookup->count(k)) {
            int ret = numeric_limits<int>::max();
            combinations(size(nums), d,
                         [this, &nums, &d, &lookup, &ret](const vector<int>& idxs) {
                             unordered_set<int> new_nums_set;
                             for (const auto& idx : idxs) {
                                 new_nums_set.emplace(nums[idx]);
                             }
                             if (size(new_nums_set) < d) {
                                 return;
                             }
                             int mx = *max_element(cbegin(new_nums_set), cend(new_nums_set));
                             int mn = *min_element(cbegin(new_nums_set), cend(new_nums_set));
                             vector<int> left;
                             for (const auto& num : nums) {
                                 if (new_nums_set.count(num)) {
                                     new_nums_set.erase(num);
                                     continue;
                                 }
                                 left.emplace_back(num);
                             }
                             int result = backtracking(left, d, lookup);
                             if (result != numeric_limits<int>::max()) {
                                 ret = min(ret, mx - mn + result);
                             }
                         });
            (*lookup)[k] = ret;
        }
        return (*lookup)[k];
    }

    string key(const vector<int>& nums) {
        string result;
        for (const auto& num : nums) {
            result += to_string(num);
            result.push_back(',');
        }
        return result;
    }

    void combinations(int n, int k, const function<void (const vector<int>&)>& callback) {
        static const auto& next_pos =
            [](const auto& n, const auto& k, const auto& idxs) {
                int i = k - 1;
                for (; i >= 0; --i) {
                    if (idxs[i] != i + n - k) {
                        break;
                    }
                }
                return i;
            };
    
        vector<int> idxs(k);
        iota(begin(idxs), end(idxs), 0);
        callback(idxs);
        for (int i; (i = next_pos(n, k, idxs)) >= 0;) {
            ++idxs[i];
            for (int j = i + 1; j < k; ++j) {
                idxs[j] = idxs[j - 1] + 1;
            }
            callback(idxs);
        }
    }
};

// Time:  O(nlogn)
// Space: O(n)
// wrong with greedy solution
// nums = [15, 9, 7, 10, 15, 14, 12, 2, 10, 8, 10, 13, 4, 11, 2]
// k = 5
// greedy  => [[2, 4, 7], [2, 8, 9], [10, 11, 12], [10, 13, 15], [10, 14, 15]] => 24
// correct => [[2, 4, 7], [2, 8, 10], [9, 10, 11], [10, 12, 15], [13, 14, 15]] => 22
// optimized from Solution_Wrong_Greedy
class Solution_Wrong_Greedy_Map {
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        return min(greedy<less<int>>(nums, k), greedy<greater<int>>(nums, k));  // two possible minimas
    }

private:
    template<typename T>
    int greedy(const vector<int>& nums, int k) {
        map<int, int, T> count;
        for (const auto& num : nums) {
            ++count[num];
        }
        unordered_map<int, list<int>> freq_to_nodes;
        unordered_map<int, list<int>::iterator> key_to_nodeit;
        for (const auto& [x, cnt] : count) {
            freq_to_nodes[cnt].emplace_back(x);
            key_to_nodeit[x] = prev(end(freq_to_nodes[cnt]));
            if (cnt > k) {
                return -1;
            }
        }
        vector<vector<int>> stks(k);
        int curr = 0;
        while (!empty(count)) {  // the while loop runs O(k) times
            if (freq_to_nodes.count(size(stks) - curr)) {  // fill the deterministic elements into the remaining subsets
                for (const auto& x : freq_to_nodes[size(stks) - curr]) {  // total time = O(n)
                    for (int i = curr; i < size(stks); ++i) {
                        stks[i].emplace_back(x);
                    }
                    key_to_nodeit.erase(x);
                    count.erase(x);
                }
                freq_to_nodes.erase(size(stks) - curr);
            }
            // greedily fill the contiguous ordered elements into the first vacant subset until it is full,
            // otherwise, the result sum would get larger => in fact, this is wrong
            vector<int> to_delete;
            for (auto& [x, cnt] : count) {
                stks[curr].emplace_back(x);
                freq_to_nodes[cnt].erase(key_to_nodeit[x]);
                if (empty(freq_to_nodes[cnt])) {
                    freq_to_nodes.erase(cnt);
                }
                --cnt;  // total time = O(n)
                if (!cnt) {
                    key_to_nodeit.erase(x);
                    to_delete.emplace_back(x);
                } else {
                    freq_to_nodes[cnt].emplace_back(x);
                    key_to_nodeit[x] = prev(end(freq_to_nodes[cnt]));
                }
                if (size(stks[curr]) == size(nums) / k) { 
                    ++curr;
                    break;
                }
            }
            for (const auto& x : to_delete) {
                count.erase(x);  // total time = O(nlogn)
            }
        }
        return accumulate(cbegin(stks), cend(stks), 0,
                          [](const auto& a, const auto& b) {
                              return a + (*max_element(cbegin(b), cend(b)) - *min_element(cbegin(b), cend(b)));
                          });
    }
};

// Time:  O(nlogn + k * n)
// Space: O(n)
// wrong with greedy solution
// nums = [15, 9, 7, 10, 15, 14, 12, 2, 10, 8, 10, 13, 4, 11, 2]
// k = 5
// greedy  => [[2, 4, 7], [2, 8, 9], [10, 11, 12], [10, 13, 15], [10, 14, 15]] => 24
// correct => [[2, 4, 7], [2, 8, 10], [9, 10, 11], [10, 12, 15], [13, 14, 15]] => 22
class Solution_Wrong_Greedy {
public:
    int minimumIncompatibility(vector<int>& nums, int k) {
        return min(greedy<less<int>>(nums, k), greedy<greater<int>>(nums, k));  // two possible minimas
    }

private:
    template<typename T>
    int greedy(const vector<int>& nums, int k) {
        map<int, int, T> count;
        for (const auto& num : nums) {
            ++count[num];
        }
        for (const auto& [_, cnt] : count) {
            if (cnt > k) {
                return -1;
            }
        }
        vector<vector<int>> stks(k);
        int curr = 0;
        int remain = size(nums);
        while (remain) {  // the while loop runs O(k) times, and the inner loops runs O(n) times
            for (auto& [x, cnt] : count) {  // fill the deterministic elements into the remaining subsets
                if (cnt != size(stks) - curr) {
                    continue;
                }
                for (int i = curr; i < size(stks); ++i) {
                    stks[i].emplace_back(x);
                }
                remain -= cnt;
                cnt = 0;
            }
            // greedily fill the contiguous ordered elements into the first vacant subset until it is full,
            // otherwise, the result sum would get larger => in fact, this is wrong
            for (auto& [x, cnt] : count) {
                if (!cnt) {
                    continue;
                }
                stks[curr].emplace_back(x);
                --remain;
                --cnt;
                if (size(stks[curr]) == size(nums) / k) { 
                    ++curr;
                    break;
                }
            }
        }
        return accumulate(cbegin(stks), cend(stks), 0,
                          [](const auto& a, const auto& b) {
                              return a + (*max_element(cbegin(b), cend(b)) - *min_element(cbegin(b), cend(b)));
                          });
    }
};

Beginner Explanation

What is Minimum Incompatibility?

Minimum Incompatibility (LeetCode #1681) is a Hard problem that primarily trains dynamic programming.

How to think about it

Restate the goal in your own words before coding.
Work a tiny example by hand so the invariant becomes obvious.
Identify the pattern — this problem aligns with dfs backtracking.
Only then translate the idea into code.

Why this problem matters

Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: Combinations, Backtracking, Submask Enumeration.

AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.

Interview Walkthrough

Interview approach for Minimum Incompatibility

Opening (30–60 seconds)

Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
State a brute force so the interviewer knows you can solve it naively.
Propose the optimal direction tied to dfs backtracking.

Core solution narrative

Define the state you track (pointers, DP cell, set membership, stack top, etc.).
Explain the transition when you process the next element.
Call out time (O(max(n * 2^n, 3^n))) and space (O(2^n)) before coding.
Code cleanly; narrate variable names.

What interviewers listen for

Correctness on edge cases
Complexity honesty
Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)

Follow-up questions they may ask

Can you solve it with less memory?
What if the input stream is infinite / doesn't fit in RAM?
How would tests look for adversarial inputs?

Optimized Approach

Optimized solution notes

The reference solutions on AlgoForge target O(max(n * 2^n, 3^n)) time and O(2^n) space.

Pattern focus: dfs backtracking

Use the pattern as a checklist:

dfs backtracking — confirm the invariant holds after each step

Multiple methods appear in the source solutions — compare them and explain when each is preferable.

Implementation tips

Prefer readable names over micro-optimizations in interviews.
Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
After AC-level logic, re-scan for off-by-one and null checks.

Complexity Analysis

Complexity

Measure	Bound
Time	*O(max(n 2^n, 3^n))**
Space	O(2^n)

How to justify this in an interview

Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
Space: include hash maps, recursion stack, and output allocation when the problem asks for it.

If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.

Common Mistakes

Common mistakes on Minimum Incompatibility

Skipping edge cases — empty collections, single-element inputs, max constraints.
Wrong invariant for dfs backtracking — updating state too early or too late.
Mutating input unexpectedly when the problem forbids it.
Off-by-one in windows, ranges, or binary search bounds.
Ignoring overflow / precision for integer arithmetic problems.
Overengineering — jumping to an advanced structure when a simpler approach works.

Alternative Approaches

Alternatives

The source file includes more than one method. Compare:

Primary optimized path — best complexity for typical interviews.
Secondary approach — often brute force, sorting-based, or space-optimized variant.

Practice articulating when you would pick each (constraints, readability, follow-ups).

Edge Cases

Edge cases checklist

Minimum input size
Maximum input size / time limits
Duplicates and already-sorted input
Negative numbers / zeros (if applicable)
Disconnected structures (graphs/trees)
Single path vs branching recursion depth

Pattern Recognition

Spotting this pattern

Signal phrases that point to dfs backtracking:

Sorted input or ability to sort without changing the answer class
Need for contiguous subarray / substring → consider sliding window
Need for O(1) membership → hash set/map
Optimal substructure + overlapping subproblems → DP
Connectivity / components → graph DFS/BFS or Union-Find

Primary topics: dynamic programming.

Follow-up Interview Questions

Follow-ups

How does the solution change if the input is a stream?
Can you solve it in-place?
What if duplicates must be handled differently?
How would you parallelize the approach?
Design tests that would break a buggy implementation.

Practice Recommendations

What to practice next

Re-solve Minimum Incompatibility in a second language (cpp, python).
Drill 3–5 more problems tagged dynamic programming.
Teach the solution out loud in under 5 minutes.
Add this problem to your revision calendar in 3 days and 14 days.

Visualization

Conceptual diagram for Minimum Incompatibility: show input structure (dynamic programming), highlight the moving parts of the dfs backtracking approach, and annotate each step with the maintained invariant and complexity.

Study checklist

Read the official problem statement on LeetCode
Solve on paper / whiteboard first
Implement the dfs backtracking approach
Verify edge cases from the checklist
State time and space complexity aloud
Compare with the AlgoForge reference solution
Schedule a revision session

Revision notes

Minimum Incompatibility (#1681) — Hard. Pattern: dfs backtracking. Complexity: O(max(n * 2^n, 3^n)) time / O(2^n) space. Re-derive the invariant before coding.

FAQs

What is the time complexity of Minimum Incompatibility?+

The reference solutions aim for O(max(n * 2^n, 3^n)) time and O(2^n) space. Always re-derive complexity from the code you write in the interview.

What pattern does Minimum Incompatibility use?+

It primarily maps to dfs backtracking, within the broader topic of dynamic programming.

Is Minimum Incompatibility good for interviews?+

Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.

Where can I read the official statement?+

Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/minimum-incompatibility/

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