所有操作都是O(logn)

class TaskManager {
public:
    TaskManager(vector<vector<int>>& tasks) {
      for (const auto& task : tasks)
        add(task[0], task[1], task[2]);
    }

    void add(int userId, int taskId, int priority) {
      m_[taskId] = pq_.emplace(make_pair(priority, taskId), userId).first;
    }
    
    void edit(int taskId, int newPriority) {
      const int userId = m_[taskId]->second;
      rmv(taskId);
      add(userId, taskId, newPriority);
    }
    
    void rmv(int taskId) {
      auto it = m_.find(taskId);
      pq_.erase(it->second);
      m_.erase(it);
    }
    
    int execTop() {
      if (pq_.empty()) return -1;
      auto it = pq_.rbegin();
      const int userId = it->second;
      const int taskId = (it->first.second);
      rmv(taskId);
      return userId;
    }
  private:
    map<std::pair<int,int>, int> pq_; // (priority, taskId) -> userId;
    unordered_map<int, map<std::pair<int,int>, int>::iterator> m_; // taskId -> pq iterator
};

The post 花花酱 LeetCode 3408. Design Task Manager appeared first on Huahua's Tech Road.

Return an integer array answer of size n, where answer[i] is the number of flowers that are in full bloom when the ith person arrives.

Example 1:

Input: flowers = [[1,6],[3,7],[9,12],[4,13]], persons = [2,3,7,11]
Output: [1,2,2,2]
Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.

Example 2:

Input: flowers = [[1,10],[3,3]], persons = [3,3,2]
Output: [2,2,1]
Explanation: The figure above shows the times when the flowers are in full bloom and when the people arrive.
For each person, we return the number of flowers in full bloom during their arrival.

Constraints:

• 1 <= flowers.length <= 5 * 104
• flowers[i].length == 2
• 1 <= starti <= endi <= 109
• 1 <= persons.length <= 5 * 104
• 1 <= persons[i] <= 109

Solution: Prefix Sum + Binary Search

Use a treemap to store the counts (ordered by time t), when a flower begins to bloom at start, we increase m[start], when it dies at end, we decrease m[end+1]. prefix_sum[t] indicates the # of blooming flowers at time t.

For each people, use binary search to find the latest # of flowers before his arrival.

Time complexity: O(nlogn + mlogn)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  vector<int> fullBloomFlowers(vector<vector<int>>& flowers, vector<int>& persons) {
    map<int, int> m{{0, 0}};
    for (const auto& f : flowers)
      ++m[f[0]], --m[f[1] + 1];    
    int sum = 0;
    for (auto& [t, c] : m)
      c = sum += c;    
    vector<int> ans;    
    for (int t : persons)      
      ans.push_back(prev(m.upper_bound(t))->second);
    return ans;
  }
};

The post 花花酱 LeetCode 2251. Number of Flowers in Full Bloom appeared first on Huahua's Tech Road.

You are also given a 0-indexed integer array queries. For each queries[j], you want to determine the maximum beauty of an item whose price is less than or equal to queries[j]. If no such item exists, then the answer to this query is 0.

Return an array answer of the same length as queries where answer[j] is the answer to the jth query.

Example 1:

Input: items = [[1,2],[3,2],[2,4],[5,6],[3,5]], queries = [1,2,3,4,5,6]
Output: [2,4,5,5,6,6]
Explanation:
- For queries[0]=1, [1,2] is the only item which has price <= 1. Hence, the answer for this query is 2.
- For queries[1]=2, the items which can be considered are [1,2] and [2,4]. 
  The maximum beauty among them is 4.
- For queries[2]=3 and queries[3]=4, the items which can be considered are [1,2], [3,2], [2,4], and [3,5].
  The maximum beauty among them is 5.
- For queries[4]=5 and queries[5]=6, all items can be considered.
  Hence, the answer for them is the maximum beauty of all items, i.e., 6.

Example 2:

Input: items = [[1,2],[1,2],[1,3],[1,4]], queries = [1]
Output: [4]
Explanation: 
The price of every item is equal to 1, so we choose the item with the maximum beauty 4. 
Note that multiple items can have the same price and/or beauty.  

Example 3:

Input: items = [[10,1000]], queries = [5]
Output: [0]
Explanation:
No item has a price less than or equal to 5, so no item can be chosen.
Hence, the answer to the query is 0.

Constraints:

• 1 <= items.length, queries.length <= 105
• items[i].length == 2
• 1 <= pricei, beautyi, queries[j] <= 109

Solution: Prefix Max + Binary Search

Sort items by price. For each price, use a treemap to store the max beauty of an item whose prices is <= p. Then use binary search to find the max beauty whose price is <= p.

Time complexity: Pre-processing O(nlogn) + query: O(qlogn)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
    sort(begin(items), end(items));
    map<int, int> m;
    int best = 0;
    for (const auto& item : items) {
      best = max(best, item[1]);
      m[item[0]] = best;
    }
    vector<int> ans;
    for (const int q: queries) {
      auto it = m.upper_bound(q);
      if (it == begin(m))
        ans.push_back(0);
      else
        ans.push_back(prev(it)->second);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 2070. Most Beautiful Item for Each Query appeared first on Huahua's Tech Road.

• 0 if it is a batch of buy orders, or
• 1 if it is a batch of sell orders.

Note that orders[i] represents a batch of amounti independent orders with the same price and order type. All orders represented by orders[i] will be placed before all orders represented by orders[i+1] for all valid i.

There is a backlog that consists of orders that have not been executed. The backlog is initially empty. When an order is placed, the following happens:

• If the order is a buy order, you look at the sell order with the smallest price in the backlog. If that sell order’s price is smaller than or equal to the current buy order’s price, they will match and be executed, and that sell order will be removed from the backlog. Else, the buy order is added to the backlog.
• Vice versa, if the order is a sell order, you look at the buy order with the largest price in the backlog. If that buy order’s price is larger than or equal to the current sell order’s price, they will match and be executed, and that buy order will be removed from the backlog. Else, the sell order is added to the backlog.

Return the total amount of orders in the backlog after placing all the orders from the input. Since this number can be large, return it modulo 109 + 7.

Example 1:

Input: orders = [[10,5,0],[15,2,1],[25,1,1],[30,4,0]]
Output: 6
Explanation: Here is what happens with the orders:
- 5 orders of type buy with price 10 are placed. There are no sell orders, so the 5 orders are added to the backlog.
- 2 orders of type sell with price 15 are placed. There are no buy orders with prices larger than or equal to 15, so the 2 orders are added to the backlog.
- 1 order of type sell with price 25 is placed. There are no buy orders with prices larger than or equal to 25 in the backlog, so this order is added to the backlog.
- 4 orders of type buy with price 30 are placed. The first 2 orders are matched with the 2 sell orders of the least price, which is 15 and these 2 sell orders are removed from the backlog. The 3rd order is matched with the sell order of the least price, which is 25 and this sell order is removed from the backlog. Then, there are no more sell orders in the backlog, so the 4th order is added to the backlog.
Finally, the backlog has 5 buy orders with price 10, and 1 buy order with price 30. So the total number of orders in the backlog is 6.

Example 2:

Input: orders = [[7,1000000000,1],[15,3,0],[5,999999995,0],[5,1,1]]
Output: 999999984
Explanation: Here is what happens with the orders:
- 109 orders of type sell with price 7 are placed. There are no buy orders, so the 109 orders are added to the backlog.
- 3 orders of type buy with price 15 are placed. They are matched with the 3 sell orders with the least price which is 7, and those 3 sell orders are removed from the backlog.
- 999999995 orders of type buy with price 5 are placed. The least price of a sell order is 7, so the 999999995 orders are added to the backlog.
- 1 order of type sell with price 5 is placed. It is matched with the buy order of the highest price, which is 5, and that buy order is removed from the backlog.
Finally, the backlog has (1000000000-3) sell orders with price 7, and (999999995-1) buy orders with price 5. So the total number of orders = 1999999991, which is equal to 999999984 % (109 + 7).

Constraints:

• 1 <= orders.length <= 105
• orders[i].length == 3
• 1 <= pricei, amounti <= 109
• orderTypei is either 0 or 1.

Solution: Treemap / PriorityQueue / Heap

buy backlog: max heap
sell backlog: min heap
Trade happens between the tops of two queues.

Time complexity: O(nlogn)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  int getNumberOfBacklogOrders(vector<vector<int>>& orders) {  
    constexpr int kMod = 1e9 + 7;
    map<int64_t, int64_t> buys;
    map<int64_t, int64_t> sells;
    for (const auto& order : orders) {      
      auto& m = order[2] == 0 ? buys : sells;
      m[order[0]] += order[1];
      while (buys.size() && sells.size()) {
        auto b = rbegin(buys);
        auto s = begin(sells);
        if (b->first < s->first) break;
        const int k = min(b->second, s->second);
        if (!(b->second -= k)) buys.erase((++b).base());
        if (!(s->second -= k)) sells.erase(s);
      }
    }
    int64_t ans = 0;
    for (const auto& [p, c] : buys) ans += c;
    for (const auto& [p, c] : sells) ans += c;    
    return ans % kMod;
  }
};

The post LeetCode 1801. Number of Orders in the Backlog appeared first on Huahua's Tech Road.

• For example, the beauty of "abaacc" is 3 - 1 = 2.

Given a string s, return the sum of beauty of all of its substrings.

Example 1:

Input: s = "aabcb"
Output: 5
Explanation: The substrings with non-zero beauty are ["aab","aabc","aabcb","abcb","bcb"], each with beauty equal to 1.

Example 2:

Input: s = "aabcbaa"
Output: 17

Constraints:

• 1 <= s.length <=500
• s consists of only lowercase English letters.

Solution: Treemap

Time complexity: O(n²log26)
Space complexity: O(26)

// Author: Huahua
class Solution {
public:
  int beautySum(string_view s) {
    const int n = s.size();
    int ans = 0;
    vector<int> f(26);
    map<int, int> m;
    for (int i = 0; i < n; ++i) {
      fill(begin(f), end(f), 0);
      m.clear();
      for (int j = i; j < n; ++j) {
        const int c = ++f[s[j] - 'a'];
        ++m[c];
        if (c > 1) {
          auto it = m.find(c - 1);
          if (--it->second == 0)
            m.erase(it);
        }
        ans += rbegin(m)->first - begin(m)->first;  
      }
    }
    return ans;
  }
};

The post 花花酱 LeetCode 1781. Sum of Beauty of All Substrings appeared first on Huahua's Tech Road.