Line sweep Technique / 2251. Number of Flowers in Full Bloom

Problem Statement

You are given a 0-indexed 2D integer array flowers, where flowers[i] = [starti, endi] means the ith flower will be in full bloom from starti to endi (inclusive). You are also given a 0-indexed integer array people of size n, where people[i] is the time that the ith person will arrive to see the flowers.

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]], poeple = [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]], poeple = [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 <= people.length <= 5 * 104
1 <= people[i] <= 109

Intuition

Line sweep -> sorting + heap


Approach:

Interval type of problem, use sorting and hence prioirity queue for sorting people 

At each juncture we maintain, how much flowers using the count, We check at start 
heap if current time (people) is less than, than flower is bloomed count++

Hence check in end heap to do count--

Links

https://leetcode.com/problems/number-of-flowers-in-full-bloom/description/

Video Links

https://www.youtube.com/watch?v=zY3Uty9IwvY&ab_channel=NeetCodeIO

Approach 1:

C++

typedef pair<int,int> pii;

class Solution {
public:
    vector<int> fullBloomFlowers(vector<vector<int>>& flowers, vector<int>& people) {
        vector<int> ans(people.size(),0);
        priority_queue<int, vector<int>, greater<int>> start,end;
        vector<pii> pl;

        for(int i=0; i<people.size(); i++)
            pl.push_back({people[i], i});
        
        sort(pl.begin(), pl.end());

        for(auto &it: flowers){
            start.push(it[0]);
            end.push(it[1]);
        }

        int count = 0;

        for(auto &it: pl){
            int time = it.first;
            int idx = it.second;
            while(!start.empty() and start.top() <= time){
                count ++;
                start.pop();
            }
            while(!end.empty() and end.top() < time){
                count--;
                end.pop();
            }

            ans[idx] = count;
        }

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++