What is the time complexity of Number of Flowers in Full Bloom?

The optimal solution has a time complexity of O(nlogn + mlogn).

What data structures are used to solve Number of Flowers in Full Bloom?

This problem uses Array, Hash Table, Binary Search, Sorting, Prefix Sum, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Flowers in Full Bloom?

Insight 1: The brute-force approach of iterating through flowers for each person to check if the arrival time falls within the bloom period would be too slow (O(n*m) where n is the number of flowers and m is the number of people). Insight 2: Binary search can efficiently find the number of flowers that have started blooming before or at a given time and the number of flowers that have ended blooming before a given time.

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Number of Flowers in Full Bloom - LeetCode 2251 Solution

Q: Which companies ask LeetCode 2251. Number of Flowers in Full Bloom in interviews?

This problem is frequently asked at Capital One, Databricks, Netflix, PhonePe, Roblox and 1 more companies.

Number of Flowers in Full Bloom - Complete Solution Guide

Number of Flowers in Full Bloom is LeetCode problem 2251, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed 2D integer array flowers , where flowers[i] = [start i , end i ] means the i th flower will be in full bloom from start i to end i ( inclusive ). You are also given a 0-indexed integer array people of size n , where people[i] is the time that the i th 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 i th person arrives. Example 1: Input: flowers = [[1,6],[3,7],[9,12

Detailed Explanation

The problem asks us to determine, for each person arriving at a specific time, how many flowers are in full bloom. We are given a list of flowers where each flower has a start and end time indicating when it's in full bloom (inclusive). We are also given a list of people and their arrival times. The goal is to return a list where each element is the number of flowers in bloom at the corresponding person's arrival time.

Solution Approach

The solution involves first extracting the start and end times of the flowers into separate arrays. These arrays are then sorted to allow for binary searches. For each person, we perform two binary searches: one on the sorted start times to find the number of flowers that have started blooming by the person's arrival time, and another on the sorted end times to find the number of flowers that have ended blooming by the person's arrival time. The difference between these two counts gives us the number of flowers in bloom at that person's arrival time. This process is repeated for each person, and the results are stored in an answer array.

Step-by-Step Algorithm

Step 1: Extract the start times and end times from the 'flowers' array into two separate arrays, 'starts' and 'ends'.
Step 2: Sort both the 'starts' and 'ends' arrays in ascending order.
Step 3: Initialize an empty 'ans' array to store the results for each person.
Step 4: Iterate through the 'people' array. For each person's arrival time 'p':
Step 5: Perform a binary search on the 'starts' array to find the number of flowers that have started blooming before or at time 'p'. This is equivalent to finding the index of the first element greater than 'p' (bisect_right or upper_bound).
Step 6: Perform a binary search on the 'ends' array to find the number of flowers that have ended blooming before time 'p'. This is equivalent to finding the index of the first element not less than 'p' (bisect_left or lower_bound).
Step 7: Calculate the difference between the results of the two binary searches (started - ended), which represents the number of flowers in bloom at time 'p'.
Step 8: Append this difference to the 'ans' array.
Step 9: After iterating through all the people, return the 'ans' array.

Key Insights

Insight 1: The brute-force approach of iterating through flowers for each person to check if the arrival time falls within the bloom period would be too slow (O(n*m) where n is the number of flowers and m is the number of people).
Insight 2: Binary search can efficiently find the number of flowers that have started blooming before or at a given time and the number of flowers that have ended blooming before a given time.
Insight 3: Sorting the start and end times of the flowers separately enables us to use binary search to count the number of flowers blooming at any given time.
Insight 4: The number of flowers in bloom at time 't' is the number of flowers that have started blooming by time 't' minus the number of flowers that have already finished blooming by time 't'.

Complexity Analysis

Time Complexity: O(nlogn + mlogn)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Binary Search, Sorting, Prefix Sum, Ordered Set.

Companies

Asked at: Capital One, Databricks, Netflix, PhonePe, Roblox, Samsara.

Number of Flowers in Full Bloom | LeetCode 2251

Hard

Array Hash Table Binary Search Sorting Prefix Sum Ordered Set

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int[] fullBloomFlowers(int[][] flowers, int[] people) {
        int[] starts = new int[flowers.length];
        int[] ends = new int[flowers.length];
        
        for (int i = 0; i < flowers.length; i++) {
            starts[i] = flowers[i][0];
            ends[i] = flowers[i][1];
        }
        
        Arrays.sort(starts);
        Arrays.sort(ends);
        
        int[] ans = new int[people.length];
        for (int i = 0; i < people.length; i++) {
            int p = people[i];
            int started = binarySearchRight(starts, p);
            int ended = binarySearchLeft(ends, p);
            ans[i] = started - ended;
        }
        
        return ans;
    }
    
    private int binarySearchRight(int[] arr, int target) {
        int low = 0;
        int high = arr.length;
        
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] <= target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        
        return low;
    }
    
    private int binarySearchLeft(int[] arr, int target) {
        int low = 0;
        int high = arr.length;
        
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] < target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        
        return low;
    }
}

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Problem Info

Description

Problem 2251: Number of Flowers in Full Bloom - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Sorting Prefix Sum Ordered Set

Companies

Capital OneDatabricksNetflixPhonePeRobloxSamsara

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:2251

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Extract the start times and end times from the 'flowers' array into two separate arrays, 'starts' and 'ends'.
2
Step 2: Sort both the 'starts' and 'ends' arrays in ascending order.
3
Step 3: Initialize an empty 'ans' array to store the results for each person.
4
Step 4: Iterate through the 'people' array. For each person's arrival time 'p':
5
Step 5: Perform a binary search on the 'starts' array to find the number of flowers that have started blooming before or at time 'p'. This is equivalent to finding the index of the first element greater than 'p' (bisect_right or upper_bound).
6
Step 6: Perform a binary search on the 'ends' array to find the number of flowers that have ended blooming before time 'p'. This is equivalent to finding the index of the first element not less than 'p' (bisect_left or lower_bound).
7
Step 7: Calculate the difference between the results of the two binary searches (started - ended), which represents the number of flowers in bloom at time 'p'.
8
Step 8: Append this difference to the 'ans' array.
9
Step 9: After iterating through all the people, return the 'ans' array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(nlogn + mlogn)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sorting step and the binary searches. Sorting takes O(nlogn). For each person, we perform two binary searches on arrays of size n, so the total time for searches is O(mlogn). The space complexity is dominated by the storage required for the sorted 'starts' and 'ends' arrays which is O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Prefix Sum with Difference Array

Create a difference array representing changes in the number of blooming flowers at specific times. Calculate the prefix sum of this array to determine the number of blooming flowers at any given time.

Time: O(n + m + klogk) where k is number of distinct start/end times.Space: O(k)

Trade-offs: This approach can be more efficient if the range of start and end times is small, and there are many overlapping bloom periods. However, if the range is large and there are few overlapping periods, the binary search approach is preferable. It requires creating an intermediate array, which impacts space complexity compared to simply sorting and binary searching.

Segment Tree

Use a segment tree to represent the intervals of flower blooms. For each person's arrival time, query the segment tree to find the number of overlapping intervals.

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

Trade-offs: Segment trees offer flexibility for more complex range queries. However, for this specific problem, the overhead of building and querying a segment tree is not justified, as the binary search approach is simpler and more efficient.

✨

Key Insights

1Insight 1: The brute-force approach of iterating through flowers for each person to check if the arrival time falls within the bloom period would be too slow (O(n*m) where n is the number of flowers and m is the number of people).
2Insight 2: Binary search can efficiently find the number of flowers that have started blooming before or at a given time and the number of flowers that have ended blooming before a given time.
3Insight 3: Sorting the start and end times of the flowers separately enables us to use binary search to count the number of flowers blooming at any given time.
4Insight 4: The number of flowers in bloom at time 't' is the number of flowers that have started blooming by time 't' minus the number of flowers that have already finished blooming by time 't'.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: A person arrives before any flowers have started blooming. The binary search on 'starts' will return 0, and the binary search on 'ends' will also return 0. The difference will be 0, which is the correct answer.
Edge case 2: A person arrives after all flowers have finished blooming. The binary search on 'starts' will return n (number of flowers), and the binary search on 'ends' will also return n. The difference will be 0, which is the correct answer.
Edge case 3: A person arrives exactly at the start or end time of a flower. The bisect_right and bisect_left functions correctly handle these cases to ensure that the count of flowers in bloom is accurate.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the start and end times before performing the binary searches, which would lead to incorrect results.

✗Common mistake 2: Using the wrong type of binary search (e.g., bisect_left instead of bisect_right or vice-versa), which would result in incorrect counts of blooming flowers.

✗Common mistake 3: Integer overflow when calculating the difference between started and ended counts, especially if the number of flowers is very large. This is less of a concern with the given constraints.

✗Common mistake 4: Forgetting to handle the cases where a person arrives before any flowers start blooming or after all flowers have finished blooming.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Scheduling tasks with dependencies, where each task has a start and end time. Determine how many tasks are active at a given time.

🏦Real-world scenario 2: Analyzing website traffic to determine the number of active users at different points in time. Each user session has a start and end time, and you want to know how many users are logged in at a particular moment.

🎮Real-world scenario 3: Resource allocation, where each resource is available during a specific time interval. Determine the number of available resources at a given time.

📱Real-world scenario 4: Event management: Finding number of events happening at a given time, provided you have the event start and end times

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Flowers in Full Bloom"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?