Codestin Search App

Number of Flowers in Full Bloom - Problem

Array Hash Table Binary Search Sorting Prefix Sum Ordered Set Hard

Imagine you're organizing a beautiful flower garden tour! 🌸 You have a garden where different flowers bloom at different times throughout the year, and visitors arrive at various times to see the spectacle.

You're given a flowers array where flowers[i] = [start_i, end_i] represents the i-th flower's blooming period from day start_i to day end_i (both days inclusive). You're also given a people array where people[i] is the day when the i-th person arrives to visit your garden.

Your mission: For each visitor, determine exactly how many flowers they'll see in full bloom when they arrive! Return an array where answer[i] is the number of flowers blooming when person i visits.

Example: If flowers bloom during periods [[1,6],[3,7],[9,12],[4,13]] and people arrive on days [2,3,7,11], then visitors will see [1,2,2,2] flowers respectively.

Input & Output

example_1.py — Basic Garden Tour

$ Input: flowers = [[1,6],[3,7],[9,12],[4,13]] people = [2,3,7,11]

› Output: [1,2,2,2]

💡 Note: Person at day 2: Only flower [1,6] is blooming (1 flower). Person at day 3: Flowers [1,6] and [3,7] are blooming (2 flowers). Person at day 7: Flowers [3,7] and [4,13] are blooming (2 flowers). Person at day 11: Flowers [9,12] and [4,13] are blooming (2 flowers).

example_2.py — Single Flower

$ Input: flowers = [[1,10]] people = [3,3,2]

› Output: [1,1,1]

💡 Note: There's only one flower blooming from day 1 to 10. All three people arrive during this period, so each person sees exactly 1 flower in bloom.

example_3.py — No Overlapping Blooms

$ Input: flowers = [[1,3],[4,6],[7,9]] people = [2,5,8,10]

› Output: [1,1,1,0]

💡 Note: Each person arrives during different non-overlapping bloom periods. The last person arrives after all flowers have finished blooming, so they see 0 flowers.

Visualization

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Understanding the Visualization

Timeline Setup

Create a timeline showing all flower blooming periods

Event Extraction

Extract start and end events from flower intervals

Binary Search Magic

Use binary search to instantly count active blooms for any query time

Efficient Queries

Answer all visitor queries in O(log n) time each

Key Takeaway

🎯 Key Insight: By converting flower intervals into sorted start/end events and using binary search, we can answer bloom count queries in O(log n) time instead of checking every flower individually.

Time & Space Complexity

Time Complexity

⏱️

O(n*m)

For each of n people, we check all m flowers, resulting in n*m operations

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for counters and variables

✓ Linear Space

Constraints

1 ≤ flowers.length ≤ 5 × 10⁴
flowers[i].length == 2
1 ≤ start_i ≤ end_i ≤ 10⁹
1 ≤ people.length ≤ 5 × 10⁴
1 ≤ people[i] ≤ 10⁹
Time limit: 2 seconds

Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28 Apple 22

The optimal solution uses an event-based approach with binary search. Instead of checking each flower individually, we extract all bloom start and end times, sort them, then use binary search to efficiently count how many flowers are blooming at any given time. This reduces time complexity from O(n×m) to O(n log m), making it suitable for large inputs.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check Every Flower)	O(n*m)	O(1)	For each person, iterate through all flowers to count how many are blooming
Event-Based + Binary Search (Optimal)	O(n log n)	O(m)	Convert flowers to start/end events, then use binary search to count active blooms

Brute Force (Check Every Flower) — Algorithm Steps

For each person in the people array
Initialize a counter to 0
For each flower in the flowers array
Check if person's time is between flower's start and end (inclusive)
If yes, increment the counter
Add the counter to the result array

Visualization

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Step-by-Step Walkthrough

Person arrives

A person arrives at time t

Check each flower

Go through every flower and check if it's blooming at time t

Count blooms

Count flowers where start ≤ t ≤ end

Repeat

Repeat this process for every person

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int* fullBloomFlowers(int** flowers, int flowersSize, int* flowersColSize, int* people, int peopleSize, int* returnSize) {
    int* result = (int*)malloc(peopleSize * sizeof(int));
    *returnSize = peopleSize;
    
    for (int i = 0; i < peopleSize; i++) {
        int count = 0;
        for (int j = 0; j < flowersSize; j++) {
            if (flowers[j][0] <= people[i] && people[i] <= flowers[j][1]) {
                count++;
            }
        }
        result[i] = count;
    }
    
    return result;
}

int main() {
    int flowers_data[][2] = {{1,6},{3,7},{9,12},{4,13}};
    int* flowers[4];
    for (int i = 0; i < 4; i++) flowers[i] = flowers_data[i];
    int flowersColSize[] = {2,2,2,2};
    int people[] = {2,3,7,11};
    int returnSize;
    
    int* result = fullBloomFlowers(flowers, 4, flowersColSize, people, 4, &returnSize);
    
    for (int i = 0; i < returnSize; i++) {
        printf("%d ", result[i]);
    }
    printf("\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n*m)

For each of n people, we check all m flowers, resulting in n*m operations

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for counters and variables

✓ Linear Space

Constraints

1 ≤ flowers.length ≤ 5 × 10⁴
flowers[i].length == 2
1 ≤ start_i ≤ end_i ≤ 10⁹
1 ≤ people.length ≤ 5 × 10⁴
1 ≤ people[i] ≤ 10⁹
Time limit: 2 seconds

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Check Every Flower) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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