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Problem solution · C++

Minimum Incompatibility

Minimum Incompatibility: a C++ solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
72 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For Minimum Incompatibility, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 72 lines of C++ from the credited upstream file 1681.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 5 loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Incompatibility · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minimumIncompatibility(vector<int>& nums, int k) {    constexpr int kMaxCompatibility = (16 - 1) * (16 / 2);    const int n = nums.size();    const int subsetSize = n / k;    const int maxMask = 1 << n;    const vector<int> incompatibilities =        getIncompatibilities(nums, subsetSize);    // dp[i] := the minimum possible sum of incompatibilities of the subset    // of numbers represented by the bitmask i    vector<int> dp(maxMask, kMaxCompatibility);    dp[0] = 0;     for (unsigned mask = 1; mask < maxMask; ++mask) {      // The number of 1s in `mask` isn't a multiple of `subsetSize`.      if (popcount(mask) % subsetSize != 0)        continue;      // https://cp-algorithms.com/algebra/all-submasks.html      for (int submask = mask; submask > 0; submask = (submask - 1) & mask)        if (incompatibilities[submask] != -1)  // valid subset          dp[mask] =              min(dp[mask], dp[mask - submask] + incompatibilities[submask]);    }     return dp.back() == kMaxCompatibility ? -1 : dp.back();  }  private:  static constexpr int kMaxNum = 16;   // Returns an incompatibilities array where  // * incompatibilities[i] := the incompatibility of the subset of numbers  //   represented by the bitmask i  // * incompatibilities[i] := -1 if the number of 1s in the bitmask i is not  //   `subsetSize`  vector<int> getIncompatibilities(const vector<int>& nums, int subsetSize) {    const int maxMask = 1 << nums.size();    vector<int> incompatibilities(maxMask, -1);    for (unsigned mask = 0; mask < maxMask; ++mask)      if (popcount(mask) == subsetSize && isUnique(nums, mask, subsetSize))        incompatibilities[mask] = getIncompatibility(nums, mask);    return incompatibilities;  }   // Returns true if the numbers selected by `mask` are unique.  //  // e.g. If we call isUnique(0b1010, 2, [1, 2, 1, 4]), `used` variable  // will be 0b1, which only has one 1 (less than `subsetSize`). In this case,  // we should return false.  bool isUnique(const vector<int>& nums, int mask, int subsetSize) {    unsigned used = 0;    for (int i = 0; i < nums.size(); ++i)      if (mask >> i & 1)        used |= 1 << nums[i];    return popcount(used) == subsetSize;  }   // Returns the incompatibility of the selected numbers represented by the  // `mask`.  int getIncompatibility(const vector<int>& nums, int mask) {    int mn = kMaxNum;    int mx = 0;    for (int i = 0; i < nums.size(); ++i)      if (mask >> i & 1) {        mx = max(mx, nums[i]);        mn = min(mn, nums[i]);      }    return mx - mn;  }};

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