LeetCode #1482 — MEDIUM

Minimum Number of Days to Make m Bouquets

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array bloomDay, an integer m and an integer k.

You want to make m bouquets. To make a bouquet, you need to use k adjacent flowers from the garden.

The garden consists of n flowers, the i^th flower will bloom in the bloomDay[i] and then can be used in exactly one bouquet.

Return the minimum number of days you need to wait to be able to make m bouquets from the garden. If it is impossible to make m bouquets return -1.

Example 1:

Input: bloomDay = [1,10,3,10,2], m = 3, k = 1
Output: 3
Explanation: Let us see what happened in the first three days. x means flower bloomed and _ means flower did not bloom in the garden.
We need 3 bouquets each should contain 1 flower.
After day 1: [x, _, _, _, _]   // we can only make one bouquet.
After day 2: [x, _, _, _, x]   // we can only make two bouquets.
After day 3: [x, _, x, _, x]   // we can make 3 bouquets. The answer is 3.

Example 2:

Input: bloomDay = [1,10,3,10,2], m = 3, k = 2
Output: -1
Explanation: We need 3 bouquets each has 2 flowers, that means we need 6 flowers. We only have 5 flowers so it is impossible to get the needed bouquets and we return -1.

Example 3:

Input: bloomDay = [7,7,7,7,12,7,7], m = 2, k = 3
Output: 12
Explanation: We need 2 bouquets each should have 3 flowers.
Here is the garden after the 7 and 12 days:
After day 7: [x, x, x, x, _, x, x]
We can make one bouquet of the first three flowers that bloomed. We cannot make another bouquet from the last three flowers that bloomed because they are not adjacent.
After day 12: [x, x, x, x, x, x, x]
It is obvious that we can make two bouquets in different ways.

Constraints:

bloomDay.length == n
1 <= n <= 10⁵
1 <= bloomDay[i] <= 10⁹
1 <= m <= 10⁶
1 <= k <= n

Step 01

Brute Force Baseline

Problem summary: You are given an integer array bloomDay, an integer m and an integer k. You want to make m bouquets. To make a bouquet, you need to use k adjacent flowers from the garden. The garden consists of n flowers, the ith flower will bloom in the bloomDay[i] and then can be used in exactly one bouquet. Return the minimum number of days you need to wait to be able to make m bouquets from the garden. If it is impossible to make m bouquets return -1.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

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

Example 2

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

Example 3

[7,7,7,7,12,7,7]
2
3

Core Insight

What unlocks the optimal approach

If we can make m or more bouquets at day x, then we can still make m or more bouquets at any day y > x.
We can check easily if we can make enough bouquets at day x if we can get group adjacent flowers at day x.

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

Upper-end input sizes

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 #1482: Minimum Number of Days to Make m Bouquets
class Solution {
    private int[] bloomDay;
    private int m, k;

    public int minDays(int[] bloomDay, int m, int k) {
        this.bloomDay = bloomDay;
        this.m = m;
        this.k = k;
        final int mx = (int) 1e9;
        int l = 1, r = mx + 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l > mx ? -1 : l;
    }

    private boolean check(int days) {
        int cnt = 0, cur = 0;
        for (int x : bloomDay) {
            cur = x <= days ? cur + 1 : 0;
            if (cur == k) {
                ++cnt;
                cur = 0;
            }
        }
        return cnt >= m;
    }
}

// Accepted solution for LeetCode #1482: Minimum Number of Days to Make m Bouquets
func minDays(bloomDay []int, m int, k int) int {
	mx := slices.Max(bloomDay)
	if l := sort.Search(mx+2, func(days int) bool {
		cnt, cur := 0, 0
		for _, x := range bloomDay {
			if x <= days {
				cur++
				if cur == k {
					cnt++
					cur = 0
				}
			} else {
				cur = 0
			}
		}
		return cnt >= m
	}); l <= mx {
		return l
	}
	return -1

}

# Accepted solution for LeetCode #1482: Minimum Number of Days to Make m Bouquets
class Solution:
    def minDays(self, bloomDay: List[int], m: int, k: int) -> int:
        def check(days: int) -> int:
            cnt = cur = 0
            for x in bloomDay:
                cur = cur + 1 if x <= days else 0
                if cur == k:
                    cnt += 1
                    cur = 0
            return cnt >= m

        mx = max(bloomDay)
        l = bisect_left(range(mx + 2), True, key=check)
        return -1 if l > mx else l

// Accepted solution for LeetCode #1482: Minimum Number of Days to Make m Bouquets
struct Solution;

impl Solution {
    fn min_days(bloom_day: Vec<i32>, m: i32, k: i32) -> i32 {
        let m = m;
        let k = k as usize;
        let n = bloom_day.len();
        if n < m as usize * k {
            return -1;
        }
        let mut left = 1;
        let mut right = std::i32::MAX;
        while left < right {
            let mid = left + (right - left) / 2;
            let mut group = 0;
            let mut count = 0;
            for i in 0..n {
                if bloom_day[i] > mid {
                    count = 0;
                } else {
                    count += 1;
                    if count >= k {
                        count = 0;
                        group += 1;
                    }
                }
            }
            if group < m {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        left
    }
}

#[test]
fn test() {
    let bloom_day = vec![1, 10, 3, 10, 2];
    let m = 3;
    let k = 1;
    let res = 3;
    assert_eq!(Solution::min_days(bloom_day, m, k), res);
    let bloom_day = vec![1, 10, 3, 10, 2];
    let m = 3;
    let k = 2;
    let res = -1;
    assert_eq!(Solution::min_days(bloom_day, m, k), res);
    let bloom_day = vec![7, 7, 7, 7, 12, 7, 7];
    let m = 2;
    let k = 3;
    let res = 12;
    assert_eq!(Solution::min_days(bloom_day, m, k), res);
    let bloom_day = vec![1000000000, 1000000000];
    let m = 1;
    let k = 1;
    let res = 1000000000;
    assert_eq!(Solution::min_days(bloom_day, m, k), res);
    let bloom_day = vec![1, 10, 2, 9, 3, 8, 4, 7, 5, 6];
    let m = 4;
    let k = 2;
    let res = 9;
    assert_eq!(Solution::min_days(bloom_day, m, k), res);
}

// Accepted solution for LeetCode #1482: Minimum Number of Days to Make m Bouquets
function minDays(bloomDay: number[], m: number, k: number): number {
    const mx = Math.max(...bloomDay);
    let [l, r] = [1, mx + 1];
    const check = (days: number): boolean => {
        let [cnt, cur] = [0, 0];
        for (const x of bloomDay) {
            cur = x <= days ? cur + 1 : 0;
            if (cur === k) {
                cnt++;
                cur = 0;
            }
        }
        return cnt >= m;
    };
    while (l < r) {
        const mid = (l + r) >> 1;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l > mx ? -1 : 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(log n)

Space

O(1)

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.

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.