Algorithm (Counter)

1. Let $n$ denote the length of $\texttt{nums}$ and $m$ be the maximum duplicate count of nums. In order to the divided subsequences to be disjoint and each has length at least $K,$ we would require $n\geq mK.$
2. We show that this is in fact also a sufficient condition constructively as follows: we create $m$ buckets $B_{i}$ and assign num[i] to $B_{i\mod m}$. It is easy to see that $B_{i}$’s will form sequences that are disjoint.

Code

class Solution {
 public:
  bool canDivideIntoSubsequences(vector<int>& nums, int K) {
    const auto max_duplicate = [&] {
      const auto counter = [&](unordered_map<int, int> self = {}) {
        for (const auto x : nums)
          self[x]++;
        return self;
      }();

      const auto [val, count] = *max_element(begin(counter), end(counter), [&](const auto &x, const auto &y) {
        return x.second < y.second;
      });
      return count;
    }();

    const auto solution = size(nums) >= max_duplicate * K;

    return solution;
  }
};