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Problem solution · C++

Maximize Count of Distinct Primes After Split

Maximize Count of Distinct Primes After Split: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
Source
Kamyu LeetCode Solutions
Length
149 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

Segment tree or range structure

For Maximize Count of Distinct Primes After Split, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 149 lines of C++ from the credited upstream file maximize-count-of-distinct-primes-after-split.cpp.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 8 loop blocks detected.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

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

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximize Count of Distinct Primes After Split · C++C++
Use this to learn the idea, then write your own version.
// Time:  O(r + nlogn + qlogn), r = max(nums)// Space: O(r + n) // number theory, bst, segment treevector<int> linear_sieve_of_eratosthenes(int n) {  // Time: O(n), Space: O(n)    vector<int> spf(n + 1, -1);    vector<int> primes;    for (int i = 2; i <= n; ++i) {        if (spf[i] == -1) {            spf[i] = i;            primes.emplace_back(i);        }        for (const auto& p : primes) {            if (i * p > n || p > spf[i]) {                break;            }            spf[i * p] = p;        }    }    return spf;} const int MAX_N = 1e5;const auto& SPF = linear_sieve_of_eratosthenes(MAX_N);class Solution {public:    vector<int> maximumCount(vector<int>& nums, vector<vector<int>>& queries) {        unordered_map<int, set<int>> lookup;        SegmentTree st(size(nums) - 1);        const auto& add = [&](int i, int d) {            const auto& x = nums[i];            if (SPF[x] != x) {                return;            }            if (d == 1) {                lookup[x].emplace(i);            }            if (size(lookup[x]) == 1) {                st.update(0, size(nums) - 2, d);            } else if (i == *begin(lookup[x])) {                st.update(i, *next(begin(lookup[x])) - 1, d);            } else if (i == *rbegin(lookup[x])) {                st.update(*next(rbegin(lookup[x])), i - 1, d);            }            if (d == -1) {                lookup[x].erase(i);            }        };         for (int i = 0; i < size(nums); ++i) {            add(i, +1);        }        vector<int> result(size(queries));        for (int i = 0; i < size(queries); ++i) {            const int idx = queries[i][0], x = queries[i][1];            if (nums[idx] != x) {                add(idx, -1);                nums[idx] = x;                add(idx, +1);            }            result[i] = st.tree[1];  // st.query(0, size(nums) - 2);        }        return result;    } private:    class SegmentTree {    public:        explicit SegmentTree(int N)         :  base_(N > 1 ? 1 << (__lg(N - 1) + 1) : 1),            lazy_(base_),            tree(N > 1 ? 1 << (__lg(N - 1) + 2) : 2) {         }         void update(int L, int R, const int val) {            L += base_;            R += base_;            // push(L);  // enable if range assignment            // push(R);  // enable if range assignment            int L0 = L, R0 = R;            for (; L <= R; L >>= 1, R >>= 1) {                if ((L & 1) == 1) {                    apply(L++, val);                }                if ((R & 1) == 0) {                    apply(R--, val);                }            }            pull(L0);            pull(R0);        }         int query(int L, int R) {            if (L > R) {                return 0;            }            L += base_;            R += base_;            push(L);            push(R);            int left = 0, right = 0;            for (; L <= R; L >>= 1, R >>= 1) {                if ((L & 1) == 1) {                    left = max(left, tree[L++]);                }                if ((R & 1) == 0) {                    right = max(tree[R--], right);                }            }            return max(left, right);        }         vector<int> tree;     private:        void apply(int x, const int val) {            tree[x] += val;            if (x < base_) {                lazy_[x] += val;            }        }         void pull(int x) {            while (x > 1) {                x >>= 1;                tree[x] = max(tree[x << 1], tree[(x << 1) + 1]);                if (lazy_[x]) {                    tree[x] += lazy_[x];                }            }        }         void push(int x) {            for (int h = __lg(x) - 1; h > 0; --h) {                int y = x >> h;                if (lazy_[y]) {                    apply(y << 1, lazy_[y]);                    apply((y << 1) + 1, lazy_[y]);                    lazy_[y] = 0;                }            }        }         int base_;        vector<int> lazy_;    };};

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