Number of Flowers in Full Bloom

class Solution {
public:
    vector<int> countFlowersInBloom(vector<vector<int>>& bloomPeriods, vector<int>& persons) {
        vector<int> startTime;
        vector<int> endTime;
    
        for (vector<int>& bloom : bloomPeriods) {
            startTime.push_back(bloom[0]);
            endTime.push_back(bloom[1] + 1);
        }
    
        sort(startTime.begin(), startTime.end());
        sort(endTime.begin(), endTime.end());
        vector<int> result;
    
        for (int person : persons) {
            int startCount = upper_bound(startTime.begin(), startTime.end(), person) - startTime.begin();
            int endCount = upper_bound(endTime.begin(), endTime.end(), person) - endTime.begin();
            result.push_back(startCount - endCount);
        }
    
        return result;
    }
};

The solution to the problem "Number of Flowers in Full Bloom" involves calculating how many flowers are blooming at the times queried by each person. The solution uses C++ for its implementation. The main idea is to employ sorting and binary search techniques to efficiently count the number of blooms for each query time.

Here's a breakdown of the solution:

• Extract start and end times of blooming periods into separate vectors, adjusting the end time by adding one to account for the inclusive nature of the blooming period end.
• Sort both the start and end time vectors. This sorting is crucial as it enables the use of binary search techniques.
• For each person's query time, determine how many blooms have started and not yet ended using std::upper_bound. The upper_bound function is leveraged here to find the first element in the sorted list that is greater than the queried time, which provides a count of blooms up to that point.
• The difference between the counts from the start and end times vectors gives the number of flowers in bloom at the person’s specific time.
• Results are aggregated and returned.

This approach is efficient with respect to time complexity due to the use of sorting and binary search, both of which are significantly faster for large input sizes compared to naive comparisons. This makes the solution well-suited for scenarios where there are many queries and bloom periods to process.

class Solution {
    public int[] evaluateFlowersStatus(int[][] flowers, int[] visitors) {
        List<Integer> beginTimes = new ArrayList();
        List<Integer> endTimes = new ArrayList();
            
        for (int[] flower: flowers) {
            beginTimes.add(flower[0]);
            endTimes.add(flower[1] + 1);
        }
            
        Collections.sort(beginTimes);
        Collections.sort(endTimes);
            
        int[] result = new int[visitors.length];
            
        for (int idx = 0; idx < visitors.length; idx++) {
            int visitorTime = visitors[idx];
            int startCount = customBinarySearch(beginTimes, visitorTime);
            int endCount = customBinarySearch(endTimes, visitorTime);
            result[idx] = startCount - endCount;
        }
            
        return result;
    }
        
    public int customBinarySearch(List<Integer> sortedList, int targetTime) {
        int low = 0;
        int high = sortedList.size();
            
        while (low < high) {
            int mid = (low + high) / 2;
            if (targetTime < sortedList.get(mid)) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
            
        return low;
    }
}

The solution provided in Java tackles the problem of determining how many flowers are in full bloom at given times. The task is managed with the following steps:

1. Initialize two lists, beginTimes and endTimes, to store the start and end times of flower blooming periods respectively. The end time is adjusted to (flower[1] + 1) to handle the inclusion of end time in the blooming period.

2. Populate these lists with the respective times from the flowers array and then sort both lists to facilitate efficient searching.

3. Prepare an array result to store the count of flowers in full bloom at each visitor's time.

4. Implement a customBinarySearch method tailored to find the position where the visitor’s time would fit into the beginTimes and endTimes lists, helping determine how many flowers start blooming before or at the visitor's time and how many stop blooming just before the visitor's time. This method efficiently finds the correct count using binary search logic.

5. For each time in the visitors array, calculate the difference between the count of flowers that have started to bloom and those that have ceased blooming by the respective times. This difference gives the number of flowers in full bloom at each visitor's time.

6. Return the result array containing counts of flowers in bloom for all visitor times as determined by the successive computation and search steps.

The overall approach leverages efficient sorting and binary search techniques to make the solution capable of handling larger sets of data effectively.

class Solution:
    def flowersInBloom(self, bloom_periods: List[List[int]], queries: List[int]) -> List[int]:
        opening = []
        closing = []
            
        for open_time, close_time in bloom_periods:
            opening.append(open_time)
            closing.append(close_time + 1)
                
        opening.sort()
        closing.sort()
        results = []
    
        for query in queries:
            start_idx = bisect_right(opening, query)
            end_idx = bisect_right(closing, query)
            results.append(start_idx - end_idx)
            
        return results

In this solution to the "Number of Flowers in Full Bloom" problem, the code efficiently determines how many flowers are in bloom at each queried time. It uses a powerful approach leveraging sorting and binary search mechanisms in Python3. Here's a quick breakdown of how the provided solution works:

• Create lists to track when flowers start blooming (opening) and when they stop blooming (closing).
• Separate the bloom periods into open and close events, making a note to adjust the closing time to accommodate checks exclusive of the endpoint.
• Sort both opening and closing lists to enable efficient search operations.
• Process each query time to count how many flowers are in bloom by determining:
  • how many blooms have started (using bisect_right on opening)
  • how many blooms have ended (using bisect_right on closing)
• Calculate the difference between started blooms and ended blooms for each query to get the count of flowers in full bloom at that quarried moment.

This elegant solution ensures that you get accurate results with improved efficiency, crucial for handling a large set of data or queries without increasing computational load significantly.