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

Painter's Partition Problem
Optimal Binary Search Solution



Problem Statement

Given an array where each element represents the length of a board, and an integer k representing the number of painters available, your task is to allocate the boards among the painters such that:

Assume that each unit length takes 1 unit of time to paint. Return the minimum possible time to paint all boards under these constraints.

If the input array is empty or k is zero, return 0.

Examples

BoardsK (Painters)Minimum TimeDescription
[10, 20, 30, 40]260Painter 1 → [10, 20, 30], Painter 2 → [40]. Max = 60
[5, 5, 5, 5]210Painter 1 → [5, 5], Painter 2 → [5, 5]. Max = 10
[10, 10, 10]310Each painter gets one board. Max = 10
[10, 20, 30, 40]1100Only one painter paints all boards. Total = 100
[10, 20, 30, 40]440Each painter paints one board. Max = 40
[]20Empty board list. Nothing to paint
[5, 10]00Zero painters. Return 0
[10, 20, 30]530More painters than boards. Assign each board individually. Max = largest board = 30

Solution

The Painter’s Partition Problem is all about assigning tasks (boards) to workers (painters) in a way that balances the load as efficiently as possible. Each painter must paint a continuous sequence of boards, and our goal is to minimize the time the most-burdened painter spends.

Understanding the Problem

Let’s say we have a list of board lengths like [10, 20, 30, 40] and two painters. One approach is to divide the boards evenly by count—but that won’t always give the best result. The key is to find a way to divide the array so that the painter who ends up with the most work has as little as possible.

This is where Binary Search helps. Instead of checking all ways to split the array (which is too slow), we search for the minimum possible time using a helper function that checks whether a given maximum time is achievable with the available painters.

Different Scenarios Explained

  • Normal Case: When the number of painters is less than the number of boards, we need to group boards cleverly so that all painters are used and the work is balanced. Binary search helps us test mid-values as the upper limit of work a painter can do, narrowing down to the optimal time.
  • One Painter: The only painter must paint all the boards. So the answer is just the sum of all board lengths.
  • Same Number of Painters and Boards: Each painter gets exactly one board. The result is the maximum board length.
  • More Painters Than Boards: Since we can’t split a board among painters, some painters won’t be used. The result is still the maximum single board length.
  • Empty Input: If the board list is empty, there’s nothing to paint—so we return 0.
  • Zero Painters: If no painter is available, the task can’t begin. We return 0 as a safe default.

By combining Binary Search with this understanding of constraints, we can efficiently find the best answer without testing every possible way to split the work.

Efficiency

This binary search method runs in O(N log S), where N is the number of boards and S is the range between the largest board and total sum. This makes it feasible for large arrays.

Visualization

Algorithm Steps

  1. Set low = max(board), high = sum(board).
  2. While low ≤ high:
    1. Calculate mid = (low + high) / 2.
    2. Check if it's possible to paint all boards within this time limit using ≤ K painters.
    3. If possible, store the result and move high = mid - 1.
    4. If not, move low = mid + 1.
  3. Return the result found as the minimum time.

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Php
def is_possible(boards, k, max_time):
    painters = 1
    curr_time = 0
    for length in boards:
        if length > max_time:
            return False  # Single board exceeds max time
        if curr_time + length > max_time:
            painters += 1
            curr_time = length  # Assign new painter
        else:
            curr_time += length
    return painters <= k

def painters_partition(boards, k):
    low = max(boards)
    high = sum(boards)
    result = high

    while low <= high:
        mid = (low + high) // 2
        if is_possible(boards, k, mid):
            result = mid
            high = mid - 1
        else:
            low = mid + 1

    return result

boards = [10, 20, 30, 40]
k = 2
print("Minimum time to paint all boards:", painters_partition(boards, k))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(N log S)Where N is the number of boards and S is the sum of board lengths. Binary search runs in log(S) and each feasibility check in O(N).
Average CaseO(N log S)Feasibility check takes linear time, and binary search runs in logarithmic range space.
Average CaseO(N log S)Worst-case is same due to deterministic binary search iterations and linear scan in check.

Space Complexity

O(1)

Explanation: Only constant extra space is used for counters and limits, no additional data structures.



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