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DSA Visualizations /

Overview of TechniquesOverview of Techniques14
❯

  1. 1Overview of DSA Problem Solving Techniques
  2. 2Brute Force Technique in DSA
  3. 3Greedy Algorithm Technique in DSA
  4. 4Divide and Conquer Technique in DSA
  5. 5Dynamic Programming Technique in DSA
  6. 6Backtracking Technique in DSA
  7. 7Recursion Technique in DSA
  8. 8Sliding Window Technique in DSA
  9. 9Two Pointers Technique
  10. 10Binary Search Technique
  11. 11Tree / Graph Traversal Technique in DSA
  12. 12Bit Manipulation Technique in DSA
  13. 13Hashing Technique
  14. 14Heaps Technique in DSA

SortingSorting10
❯

  1. 1Bubble Sort
  2. 2Selection Sort
  3. 3Insertion Sort
  4. 4Merge Sort
  5. 5Quick Sort
  6. 6Heap Sort
  7. 7Radix Sort
  8. 8Counting Sort
  9. 9Bucket Sort
  10. 10Shell Sort

ArraysArrays32
❯

  1. 1Find Maximum and Minimum in Array using Loop
  2. 2Find Second Largest in Array
  3. 3Find Second Smallest in Array
  4. 4Reverse Array using Two Pointers
  5. 5Check if Array is Sorted
  6. 6Remove Duplicates from Sorted Array
  7. 7Left Rotate an Array by One Place
  8. 8Left Rotate an Array by K Places
  9. 9Move Zeroes in Array to End
  10. 10Linear Search in Array
  11. 11Union of Two Arrays
  12. 12Find Missing Number in Array
  13. 13Max Consecutive Ones in Array
  14. 14Find Kth Smallest Element
  15. 15Longest Subarray with Given Sum (Positives)
  16. 16Longest Subarray with Given Sum (Positives and Negatives)
  17. 17Find Majority Element in Array (more than n/2 times)
  18. 18Find Majority Element in Array (more than n/3 times)
  19. 19Maximum Subarray Sum using Kadane's Algorithm
  20. 20Print Subarray with Maximum Sum
  21. 21Stock Buy and Sell
  22. 22Rearrange Array Alternating Positive and Negative Elements
  23. 23Next Permutation of Array
  24. 24Leaders in an Array
  25. 25Longest Consecutive Sequence in Array
  26. 26Count Subarrays with Given Sum
  27. 27Sort an Array of 0s, 1s, and 2s
  28. 28Two Sum Problem
  29. 29Three Sum Problem
  30. 304 Sum Problem
  31. 31Find Length of Largest Subarray with 0 Sum
  32. 32Find Maximum Product Subarray

Arrays - Binary SearchArrays - Binary Search28
❯

  1. 1Binary Search in Array<br><span class="highlight">Using Iteration</span>
  2. 2Find Lower Bound in Sorted Array
  3. 3Find Upper Bound in Sorted Array
  4. 4Search Insert Position in Sorted Array (Lower Bound Approach)
  5. 5Floor and Ceil in Sorted Array
  6. 6First Occurrence in a Sorted Array
  7. 7Last Occurrence in a Sorted Array
  8. 8Count Occurrences in Sorted Array
  9. 9Search Element in a Rotated Sorted Array
  10. 10Search in Rotated Sorted Array with Duplicates
  11. 11Minimum in Rotated Sorted Array
  12. 12Find Rotation Count in Sorted Array
  13. 13Search Single Element in Sorted Array
  14. 14Find Peak Element in Array
  15. 15Square Root using Binary Search
  16. 16Nth Root of a Number using Binary Search
  17. 17Koko Eating Bananas
  18. 18Minimum Days to Make M Bouquets
  19. 19Find the Smallest Divisor Given a Threshold
  20. 20Capacity to Ship Packages within D Days
  21. 21Kth Missing Positive Number
  22. 22Aggressive Cows Problem
  23. 23Allocate Minimum Number of Pages
  24. 24Split Array - Minimize Largest Sum
  25. 25Painter's Partition Problem
  26. 26Minimize Maximum Distance Between Gas Stations
  27. 27Median of Two Sorted Arrays of Different Sizes
  28. 28K-th Element of Two Sorted Arrays

StringsStrings2
❯

  1. 1Remove Outermost Parentheses in String
  2. 2Edit Distance Problem

MatrixMatrix3
❯

  1. 1Set Matrix Zeroes
  2. 2Rotate Matrix by 90 Degrees Clockwise
  3. 3Print Matrix in Spiral Manner

Matrix - Binary SearchMatrix - Binary Search5
❯

  1. 1Row with Maximum Number of 1's
  2. 2Search in a Sorted 2D Matrix
  3. 3Search in Row and Column-wise Sorted Matrix
  4. 4Find Target in 2D Matrix
  5. 5Matrix Median

Binary TreesBinary Trees36
❯

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

Binary Search TreesBinary Search Trees14
❯

  1. 1Find a Value in a Binary Search Tree
  2. 2Delete a Node in a Binary Search Tree
  3. 3Find the Minimum Value in a Binary Search Tree
  4. 4Find the Maximum Value in a Binary Search Tree
  5. 5Find the Inorder Successor in a Binary Search Tree
  6. 6Find the Inorder Predecessor in a Binary Search Tree
  7. 7Check if a Binary Tree is a Binary Search Tree
  8. 8Find the Lowest Common Ancestor of Two Nodes in a Binary Search Tree
  9. 9Convert a Binary Tree into a Binary Search Tree
  10. 10Balance a Binary Search Tree
  11. 11Merge Two Binary Search Trees
  12. 12Find the kth Largest Element in a Binary Search Tree
  13. 13Find the kth Smallest Element in a Binary Search Tree
  14. 14Flatten a Binary Search Tree into a Sorted List

Minimum Days to Make M Bouquets
Binary Search Optimal Solution

Topic Contents

ProblemExamplesSolutionCodeTime ComplexitySpace Complexity

Problem Statement

You are given an array bloomDay[] where bloomDay[i] represents the day on which the i-th flower will bloom. Your task is to determine the minimum number of days required to make m bouquets, where each bouquet requires k adjacent bloomed flowers.

If it's not possible to make m bouquets, return -1.

Examples❯

Bloom Day ArraymkOutputDescription
[1, 10, 3, 10, 2]313On day 3, flowers at indices 0, 2, and 4 are bloomed. 3 bouquets possible.
[1, 10, 3, 10, 2]32-1Not enough adjacent flowers to form 3 bouquets of 2 flowers.
[7, 7, 7, 7, 12, 7, 7]2312Day 12 is the earliest when two bouquets of 3 adjacent flowers can be made.
[1000000000, 1000000000]111000000000Only one flower needed, take the earliest available.
[1, 10, 2, 9, 3, 8]229Day 9 gives two pairs of adjacent bloomed flowers: (2,3) and (4,5)
[1]111One flower, one bouquet needed — return 1.
[1]21-1Only one flower, but two bouquets required — not possible.
[]11-1Empty array — can't make any bouquets.

Solution❯

This problem is about figuring out the minimum number of days required to make m bouquets, where each bouquet is formed from k adjacent flowers that have bloomed.

Understanding the Setup

We are given a list of bloom days for each flower. A flower can only be used to form a bouquet on or after the day it blooms. For every bouquet, we need exactly k adjacent flowers. If they're not next to each other, we cannot use them in the same bouquet.

How to Think About It

Imagine checking a particular day — let's say day 6. You can only use flowers that bloom on or before day 6. Now go through the array, and every time you see k or more adjacent bloomed flowers, you make one bouquet. Count how many bouquets you can make this way. If you can make at least m bouquets, then day 6 is a valid answer.

Our goal is to find the minimum such day. So, instead of checking each day one by one (which would be slow), we use binary search between the smallest and largest bloom days. At each step, we check if it’s possible to make enough bouquets by that day. If yes, we try an earlier day (move left); if not, we try a later day (move right).

Special Cases to Consider

  • Not Enough Flowers: If m × k is greater than the total number of flowers, it's impossible to make that many bouquets. We return -1.
  • All Flowers Bloom Late: If flowers bloom very late (e.g., all values are huge), the answer may be a large number, but binary search helps us get there quickly.
  • Empty Array: If the input array is empty, no bouquets can be made. We return -1.
  • Edge Size Matching: If the total number of flowers is exactly m × k, then we must use every flower and they must be positioned in a way that allows adjacency.

This approach gives us a clean and efficient solution with time complexity O(n × log(max bloom day)), where n is the number of flowers.

Algorithm Steps

  1. Set low = minimum bloom day, high = maximum bloom day.
  2. Perform binary search while low ≤ high.
  3. Set mid = (low + high) / 2.
  4. Check if it’s possible to make m bouquets in mid days using a helper function:
    • Traverse bloom days. Count consecutive bloomed roses (i.e., arr[i] ≤ mid).
    • Every time you collect k adjacent, increment bouquet count and reset count.
    • If bouquet count ≥ m, return true.
  5. If possible, store mid as answer and move high = mid - 1.
  6. Else move low = mid + 1.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class Bouquets {
  public static boolean canMakeBouquets(int[] bloomDay, int m, int k, int day) {
    int bouquets = 0, flowers = 0;
    for (int b : bloomDay) {
      if (b <= day) {
        flowers++;
        if (flowers == k) {
          bouquets++;
          flowers = 0;
        }
      } else {
        flowers = 0;  // Reset if bloom not ready
      }
    }
    return bouquets >= m;
  }

  public static int minDays(int[] bloomDay, int m, int k) {
    if ((long)m * k > bloomDay.length) return -1;
    int low = Integer.MAX_VALUE, high = Integer.MIN_VALUE;
    for (int day : bloomDay) {
      low = Math.min(low, day);
      high = Math.max(high, day);
    }
    int ans = -1;
    while (low <= high) {
      int mid = low + (high - low) / 2;
      if (canMakeBouquets(bloomDay, m, k, mid)) {
        ans = mid;
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    return ans;
  }

  public static void main(String[] args) {
    int[] bloom = {1, 10, 3, 10, 2};
    int m = 3, k = 1;
    System.out.println("Minimum Days: " + minDays(bloom, m, k));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)In the best case, the first binary search guess gives a valid answer after a single bouquet check.
Average CaseO(n log(max_day - min_day))Binary search runs in log(max_day - min_day) steps and each step checks the full array to count bouquets.
Average CaseO(n log(max_day - min_day))Binary search iterates across the entire range of possible days, checking all n roses each time.

Space Complexity

O(1)

Explanation: Only a few variables are used; no extra space is needed apart from input.

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