LeetCode #2251 — HARD

Number of Flowers in Full Bloom

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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],[4,13]], people = [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]], people = [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 <= start_i <= end_i <= 10⁹
1 <= people.length <= 5 * 10⁴
1 <= people[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Binary Search · Segment Tree

Example 1

[[1,6],[3,7],[9,12],[4,13]]
[2,3,7,11]

Example 2

[[1,10],[3,3]]
[3,3,2]

Core Insight

What unlocks the optimal approach

Notice that for any given time t, the number of flowers blooming at time t is equal to the number of flowers that have started blooming minus the number of flowers that have already stopped blooming.
We can obtain these values efficiently using binary search.
We can store the starting times in sorted order, which then allows us to binary search to find how many flowers have started blooming for a given time t.
We do the same for the ending times to find how many flowers have stopped blooming at time t.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2251: Number of Flowers in Full Bloom
class Solution {
    public int[] fullBloomFlowers(int[][] flowers, int[] people) {
        int n = flowers.length;
        int[] start = new int[n];
        int[] end = new int[n];
        for (int i = 0; i < n; ++i) {
            start[i] = flowers[i][0];
            end[i] = flowers[i][1];
        }
        Arrays.sort(start);
        Arrays.sort(end);
        int m = people.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = search(start, people[i] + 1) - search(end, people[i]);
        }
        return ans;
    }

    private int search(int[] nums, int x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #2251: Number of Flowers in Full Bloom
func fullBloomFlowers(flowers [][]int, people []int) (ans []int) {
	n := len(flowers)
	start := make([]int, n)
	end := make([]int, n)
	for i, f := range flowers {
		start[i] = f[0]
		end[i] = f[1]
	}
	sort.Ints(start)
	sort.Ints(end)
	for _, p := range people {
		r := sort.SearchInts(start, p+1)
		l := sort.SearchInts(end, p)
		ans = append(ans, r-l)
	}
	return
}

# Accepted solution for LeetCode #2251: Number of Flowers in Full Bloom
class Solution:
    def fullBloomFlowers(
        self, flowers: List[List[int]], people: List[int]
    ) -> List[int]:
        start, end = sorted(a for a, _ in flowers), sorted(b for _, b in flowers)
        return [bisect_right(start, p) - bisect_left(end, p) for p in people]

// Accepted solution for LeetCode #2251: Number of Flowers in Full Bloom
use std::collections::BTreeMap;

impl Solution {
    #[allow(dead_code)]
    pub fn full_bloom_flowers(flowers: Vec<Vec<i32>>, people: Vec<i32>) -> Vec<i32> {
        let n = people.len();

        // First sort the people vector based on the first item
        let mut people: Vec<(usize, i32)> = people.into_iter().enumerate().map(|x| x).collect();

        people.sort_by(|lhs, rhs| lhs.1.cmp(&rhs.1));

        // Initialize the difference vector
        let mut diff = BTreeMap::new();
        let mut ret = vec![0; n];

        for f in flowers {
            let (left, right) = (f[0], f[1]);
            diff.entry(left)
                .and_modify(|x| {
                    *x += 1;
                })
                .or_insert(1);

            diff.entry(right + 1)
                .and_modify(|x| {
                    *x -= 1;
                })
                .or_insert(-1);
        }

        let mut sum = 0;
        let mut i = 0;
        for (k, v) in diff {
            while i < n && people[i].1 < k {
                ret[people[i].0] += sum;
                i += 1;
            }
            sum += v;
        }

        ret
    }
}

// Accepted solution for LeetCode #2251: Number of Flowers in Full Bloom
function fullBloomFlowers(flowers: number[][], people: number[]): number[] {
    const n = flowers.length;
    const start = new Array(n).fill(0);
    const end = new Array(n).fill(0);
    for (let i = 0; i < n; ++i) {
        start[i] = flowers[i][0];
        end[i] = flowers[i][1];
    }
    start.sort((a, b) => a - b);
    end.sort((a, b) => a - b);
    const ans: number[] = [];
    for (const p of people) {
        const r = search(start, p + 1);
        const l = search(end, p);
        ans.push(r - l);
    }
    return ans;
}

function search(nums: number[], x: number): number {
    let l = 0;
    let r = nums.length;
    while (l < r) {
        const mid = (l + r) >> 1;
        if (nums[mid] >= x) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((m + n)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.