Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  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
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Minimum Days to Make M Bouquets Binary Search Optimal Solution

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.

  • You can only use a flower if it has bloomed on or before the day you're checking.
  • You must use exactly k adjacent flowers to make each bouquet.

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

Examples

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

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Solution

Understanding the Problem

We are given an array of integers where each element represents the day a flower will bloom. Our task is to make m bouquets, each consisting of exactly k adjacent bloomed flowers. A flower can only be used on or after the day it blooms.

The main question is: What is the minimum number of days needed to make m such bouquets?

Step-by-Step Solution with Example

Step 1: Think About a Single Day

Let’s say we choose a day, like day 7. On this day, only flowers that bloom on or before day 7 can be used. We scan the array and try to form bouquets using only these bloomed flowers. We can only form a bouquet if we find k adjacent bloomed flowers.

Step 2: Try to Simulate It

Let’s take an example:

flowers = [1, 10, 3, 10, 2], m = 1, k = 2

This means we need to make 1 bouquet using 2 adjacent bloomed flowers. On day 3, the bloom status is:

  • Day 1 → bloomed (1 ≤ 3)
  • Day 10 → not bloomed (10 > 3)
  • Day 3 → bloomed (3 ≤ 3)
  • Day 10 → not bloomed
  • Day 2 → bloomed (2 ≤ 3)

The bloomed flowers are at index 0, 2, and 4. None of them are adjacent, so we cannot form any bouquet on day 3. Try day 10 — now all flowers are bloomed, and we can form a bouquet with [3, 10] or [10, 2]. So day 10 works. We try to find the minimum such day.

Step 3: Use Binary Search to Optimize

Since trying all days linearly is slow, we use binary search on days from the minimum to the maximum bloom value.

At each mid-point day, we check if it's possible to make m bouquets. If it is, we try a smaller day. If it’s not, we try a bigger day. This helps us zero in on the minimum valid day efficiently.

Step 4: Bouquet Formation Check Logic

While scanning, we count how many adjacent bloomed flowers we have. If a flower is bloomed on or before the current day, we increase the count. If not, we reset the count to 0. Every time the count reaches k, we form a bouquet and reset the count.

Edge Cases

  • Not Enough Flowers: If m × k is greater than the total number of flowers, it's impossible to form the required bouquets. We should return -1 immediately.
  • All Flowers Bloom Late: Even if all flowers bloom on high days (like day 1000), binary search will still find the answer quickly.
  • Empty Array: If the array is empty, we cannot form any bouquet. Return -1.
  • Exactly Enough Flowers: If the array has exactly m × k flowers, we must use every flower — and they must be arranged in proper adjacent groups. This is a tight case to test our adjacency logic.

Finally

We solved this problem using a binary search approach, which significantly improves performance over a brute force solution. The key insights were:

  • Understanding how to simulate bouquet formation on a specific day
  • Applying binary search on the range of days to minimize the answer
  • Carefully handling edge cases, especially when adjacency is required

The time complexity is O(n × log(max bloom day)), where n is the number of flowers. This makes the solution both fast and scalable.

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.
Worst 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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