What is the time complexity of Minimum Number of Days to Make m Bouquets?

The optimal solution has a time complexity of O(nlog(max(bloomDay))).

What data structures are used to solve Minimum Number of Days to Make m Bouquets?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Number of Days to Make m Bouquets?

Insight 1: The problem can be solved using binary search. We are searching for the minimum day to be able to make m bouquets. Binary search can be applied on the range of possible days (minimum and maximum bloomDay values). Insight 2: The feasibility check (can_make_bouquets function) is crucial. For a given number of days, we need to efficiently determine if we can make 'm' bouquets with 'k' adjacent flowers.

Which companies ask LeetCode 1482. Minimum Number of Days to Make m Bouquets in interviews?

This problem is frequently asked at Navi.

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Minimum Number of Days to Make m Bouquets - LeetCode 1482 Solution

Minimum Number of Days to Make m Bouquets - Complete Solution Guide

Minimum Number of Days to Make m Bouquets is LeetCode problem 1482, a Medium 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 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,

Detailed Explanation

The problem asks us to find the minimum number of days needed to create `m` bouquets of flowers, where each bouquet requires `k` adjacent flowers. We are given an array `bloomDay` representing the days on which each flower blooms. We need to determine the earliest day by which we can create the desired number of bouquets. If it is impossible to make the required number of bouquets, we return -1. The input consists of the `bloomDay` array, the number of bouquets `m`, and the number of adjacent flowers required per bouquet `k`. The output is the minimum number of days required or -1 if not possible. Constraints are provided on the sizes of the input and the ranges of values within the input.

Solution Approach

The solution employs a binary search algorithm. First, it checks if it's even possible to make 'm' bouquets of 'k' flowers each, by verifying if `m * k` is greater than the length of the `bloomDay` array. If it is, it returns -1. If it's possible, the solution performs binary search on the possible range of bloom days (from the minimum bloom day to the maximum bloom day). For each `mid` value during the binary search (representing a candidate day), it calls the `can_make_bouquets` function to check if it is possible to make `m` bouquets by that day. If it is possible, it means the answer may lie on the left side (smaller bloom days), so we record the `mid` value as a potential answer and narrow the search space. If it is not possible, it means we need to wait longer (larger bloom days), so we narrow the search space to the right side.

Step-by-Step Algorithm

Step 1: Check if `m * k > n`. If true, return -1 as it's impossible to make the bouquets.
Step 2: Find the minimum and maximum bloom days from the `bloomDay` array. These will be the lower and upper bounds for our binary search.
Step 3: Initialize `ans` to -1. This variable will hold our potential answer.
Step 4: Perform binary search between `low` (min bloom day) and `high` (max bloom day).
Step 5: In each iteration, calculate `mid = low + (high - low) // 2` (to prevent overflow).
Step 6: Call the `can_make_bouquets(mid)` function to check if we can make `m` bouquets by day `mid`.
Step 7: If `can_make_bouquets(mid)` returns true, update `ans = mid` and set `high = mid - 1` (search for a smaller day that also works).
Step 8: If `can_make_bouquets(mid)` returns false, set `low = mid + 1` (need to search for a larger day).
Step 9: After the binary search is complete (when `low > high`), return the value of `ans`.
Step 10: The `can_make_bouquets(day)` function iterates through the `bloomDay` array, counting adjacent flowers that have bloomed by the given `day`. It increments the `bouquets` count whenever `k` adjacent flowers are found, and returns true if `bouquets >= m`, otherwise it returns false.

Key Insights

Insight 1: The problem can be solved using binary search. We are searching for the minimum day to be able to make m bouquets. Binary search can be applied on the range of possible days (minimum and maximum bloomDay values).
Insight 2: The feasibility check (can_make_bouquets function) is crucial. For a given number of days, we need to efficiently determine if we can make 'm' bouquets with 'k' adjacent flowers.
Insight 3: Handling the edge case where the total number of flowers required (m * k) exceeds the total number of flowers available (bloomDay.length) is critical. If this condition is met, it's impossible to make the bouquets, and we should return -1.

Complexity Analysis

Time Complexity: O(nlog(max(bloomDay)))

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Navi.

Minimum Number of Days to Make m Bouquets | LeetCode 1482

Medium

Array Binary Search

Time: O(nlog(max(bloomDay)))Space: O(1)

Solution Code

Available in:

class Solution {
    public int minDays(int[] bloomDay, int m, int k) {
        int n = bloomDay.length;
        if (m * k > n) {
            return -1;
        }

        int low = Integer.MAX_VALUE;
        int high = Integer.MIN_VALUE;
        for (int b_day : bloomDay) {
            low = Math.min(low, b_day);
            high = Math.max(high, b_day);
        }


        int ans = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (can_make_bouquets(bloomDay, mid, m, k)) {
                ans = mid;
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }

        return ans;
    }

    private boolean can_make_bouquets(int[] bloomDay, int day, int m, int k) {
        int bouquets = 0;
        int flowers = 0;
        for (int b_day : bloomDay) {
            if (b_day <= day) {
                flowers += 1;
                if (flowers == k) {
                    bouquets += 1;
                    if (bouquets >= m) {
                        return true;
                    }
                    flowers = 0;
                }
            } else {
                flowers = 0;
            }
        }
        return false;
    }
}

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Problem Info

Description

Problem 1482: Minimum Number of Days to Make m Bouquets - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

Navi

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1482

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

10 steps

1
Step 1: Check if `m * k > n`. If true, return -1 as it's impossible to make the bouquets.
2
Step 2: Find the minimum and maximum bloom days from the `bloomDay` array. These will be the lower and upper bounds for our binary search.
3
Step 3: Initialize `ans` to -1. This variable will hold our potential answer.
4
Step 4: Perform binary search between `low` (min bloom day) and `high` (max bloom day).
5
Step 5: In each iteration, calculate `mid = low + (high - low) // 2` (to prevent overflow).
6
Step 6: Call the `can_make_bouquets(mid)` function to check if we can make `m` bouquets by day `mid`.
7
Step 7: If `can_make_bouquets(mid)` returns true, update `ans = mid` and set `high = mid - 1` (search for a smaller day that also works).
8
Step 8: If `can_make_bouquets(mid)` returns false, set `low = mid + 1` (need to search for a larger day).
9
Step 9: After the binary search is complete (when `low > high`), return the value of `ans`.
10
Step 10: The `can_make_bouquets(day)` function iterates through the `bloomDay` array, counting adjacent flowers that have bloomed by the given `day`. It increments the `bouquets` count whenever `k` adjacent flowers are found, and returns true if `bouquets >= m`, otherwise it returns false.

💡

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(nlog(max(bloomDay)))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search contributes the log(max(bloomDay)) factor. The can_make_bouquets function iterates through the bloomDay array in O(n) time during each binary search step. Since the space usage remains constant throughout the algorithm's execution, space complexity is O(1).

📖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

Linear Search

Iterate through each possible day from the minimum bloom day to the maximum bloom day and check if it's possible to make the bouquets using the `can_make_bouquets` function. Return the first day for which the function returns true.

Time: O(n * (max(bloomDay) - min(bloomDay)))Space: O(1)

Trade-offs: Binary search is significantly faster because it avoids checking every single day. Linear search becomes inefficient when the difference between the maximum and minimum bloom days is very large. Prefer binary search in almost all scenarios.

✨

Key Insights

1Insight 1: The problem can be solved using binary search. We are searching for the minimum day to be able to make m bouquets. Binary search can be applied on the range of possible days (minimum and maximum bloomDay values).
2Insight 2: The feasibility check (can_make_bouquets function) is crucial. For a given number of days, we need to efficiently determine if we can make 'm' bouquets with 'k' adjacent flowers.
3Insight 3: Handling the edge case where the total number of flowers required (m * k) exceeds the total number of flowers available (bloomDay.length) is critical. If this condition is met, it's impossible to make the bouquets, and we should return -1.

📚

Key Terms Explained

For Students

Array

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

Binary Search

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

Edge Cases

Edge case 1: `m * k > n`. The problem statement specifies that if it's impossible to make the bouquets, return -1. The code explicitly checks for this at the beginning and returns -1 immediately, preventing further unnecessary calculations.
Edge case 2: `bloomDay` is already sorted and allows making the bouquets on the first possible day. The binary search correctly identifies the minimum possible day due to the `high = mid - 1` update when a valid day is found. This ensures the leftmost possible day is chosen.
Edge case 3: All flowers bloom on the same day, and m * k <= n. The binary search will quickly converge on this single day, and the can_make_bouquets will return true, so the correct day is returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where `m * k > n`. Forgetting to check this condition will result in incorrect output or potential runtime errors.

✗Common mistake 2: Incorrect implementation of the binary search. If the conditions for updating `low` and `high` are reversed, the binary search may not converge to the correct answer, or it might lead to an infinite loop. Ensure that the search space is correctly narrowed based on the result of the `can_make_bouquets` function.

✗Common mistake 3: Integer overflow when calculating `mid`. Using `mid = (low + high) / 2` can lead to integer overflow when `low + high` exceeds the maximum integer value. Using `mid = low + (high - low) / 2` prevents overflow.

✅

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: Resource allocation with deadlines. Imagine scheduling tasks where each task requires a contiguous block of resources and each resource becomes available on a specific day. The problem is to find the minimum day we can complete a certain number of tasks.

🏦Real-world scenario 2: Manufacturing process optimization. Consider a production line where 'k' consecutive steps are required to produce a product and each step can be performed after a certain day. The problem is to find the minimum number of days to produce a certain quantity of products.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: July 13, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Number of Days to Make m Bouquets"?

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

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