CPMagical CP Library

Problem solution · C++

Find the Number of Subsequences With Equal GCD

Find the Number of Subsequences With Equal GCD: a C++ solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Disjoint set union
Source
walkccc LeetCode Solutions
Length
33 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

Disjoint set union

For Find the Number of Subsequences With Equal GCD, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 33 lines of C++ from the credited upstream file 3336.cpp.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

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 codeFind the Number of Subsequences With Equal GCD · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int subsequencePairCount(vector<int>& nums) {    const int maxNum = ranges::max(nums);    vector<vector<vector<int>>> mem(        nums.size(),        vector<vector<int>>(maxNum + 1, vector<int>(maxNum + 1, -1)));    return subsequencePairCount(nums, 0, 0, 0, mem);  }  private:  static constexpr int kMod = 1'000'000'007;   // Returns the number of disjoint pairs `seq1` and `seq2` of nums[i..n - 1],  // where GCD(seq1) == x and GCD(seq2) == y.  int subsequencePairCount(const vector<int>& nums, int i, int x, int y,                           vector<vector<vector<int>>>& mem) {    if (i == nums.size())      return x > 0 && x == y;    if (mem[i][x][y] != -1)      return mem[i][x][y];    // 1. Skip nums[i].    const int skip = subsequencePairCount(nums, i + 1, x, y, mem);    // 2. Pick nums[i] in the first subsequence.    const int take1 =        subsequencePairCount(nums, i + 1, gcd(x, nums[i]), y, mem);    // 3. Pick nums[i] in the second subsequence.    const int take2 =        subsequencePairCount(nums, i + 1, x, gcd(y, nums[i]), mem);    return mem[i][x][y] = (static_cast<long>(skip) + take1 + take2) % kMod;  }};

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗