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

Smallest Divisor Greater Than Threshold
Optimal Binary Search Approach



Problem Statement

You are given a list of positive integers nums and a positive integer threshold. Your task is to find the smallest positive integer divisor such that the sum of all elements in the array when each is divided by this divisor (rounded up) is less than or equal to threshold.

If nums is empty or threshold is not a valid positive number, return -1.

Examples

Input ArrayThresholdSmallest DivisorDescription
[1, 2, 5, 9]65Dividing with 5: ceil(1/5)+ceil(2/5)+ceil(5/5)+ceil(9/5) = 1+1+1+2 = 5 ≤ 6
[2, 3, 5, 7, 11]113Dividing with 3 gives total sum 9, which is ≤ threshold
[19]54We need to divide 19 so that ceil(19/divisor) ≤ 5
[1, 2, 3]31Any number divided by 1 will result in original value, sum = 6, so must search for larger divisor
[100, 200, 300]3300Only a very large divisor can reduce the sum below threshold
[1]11Only one number, and we need sum ≤ 1, so divisor = 1
[]5-1Empty array, nothing to divide
[1, 2, 3]0-1Threshold is invalid (non-positive)

Solution

To solve this problem, we need to understand what it means to divide all numbers by a divisor and take the ceiling of each result. The goal is to find the smallest such divisor that keeps the sum of all these rounded-up divisions within the specified threshold.

Understanding the Problem

For example, if the array is [1, 2, 5, 9] and we divide each number by 5, we get:

  • ceil(1/5) = 1
  • ceil(2/5) = 1
  • ceil(5/5) = 1
  • ceil(9/5) = 2

So the sum is 1 + 1 + 1 + 2 = 5, which is within the threshold of 6. The challenge is to find the smallest such divisor that still keeps this sum within limits.

Different Situations to Consider

  • If the divisor is small (e.g., 1): Then each number is divided by 1, so the sum of the array stays the same (no reduction).
  • If the divisor is large (e.g., more than the max value in array): Then all divisions will round up to 1, which may help reduce the total sum.

The key is to find a balance — a divisor large enough to reduce the sum, but as small as possible.

What Happens in Edge Cases?

  • If the array is empty, we can't divide anything. So we return -1.
  • If the threshold is zero or negative, there's no way to achieve a valid result, so again return -1.

How Do We Search Efficiently?

Instead of checking each divisor one by one, we can use binary search from 1 to max(nums). For each candidate divisor, we calculate the sum of ceilings, and decide if we need a larger or smaller divisor. This drastically reduces the number of checks and makes it efficient.

At the end of this process, we’ll have the smallest divisor that satisfies the condition. If no such divisor is found (which happens only if input is invalid), we return -1.

This method ensures optimal performance and is suitable even when the input array is very large.

Visualization

Algorithm Steps

  1. Initialize low = 1 and high = max(nums).
  2. While low ≤ high:
  3. → Calculate mid as the potential divisor.
  4. → Compute sum = sum(ceil(num / mid) for num in nums).
  5. → If sum ≤ threshold: store mid as potential answer, and search for smaller divisor by setting high = mid - 1.
  6. → Else: search higher by setting low = mid + 1.
  7. Return the smallest valid divisor found.

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
import math

def smallestDivisor(nums, threshold):
    def compute_sum(divisor):
        return sum(math.ceil(num / divisor) for num in nums)

    low, high = 1, max(nums)
    answer = high

    while low <= high:
        mid = (low + high) // 2
        if compute_sum(mid) <= threshold:
            answer = mid        # Valid divisor, try smaller one
            high = mid - 1
        else:
            low = mid + 1       # Too large sum, increase divisor

    return answer

nums = [1, 2, 5, 9]
threshold = 6
print("Smallest Divisor:", smallestDivisor(nums, threshold))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)In each binary search step, computing the sum involves iterating through the entire array.
Average CaseO(n log m)Binary search over range [1, max(nums)], and each step takes O(n) time to compute the sum.
Average CaseO(n log m)Same as average case — n elements, log(max(nums)) search steps.

Space Complexity

O(1)

Explanation: Only a few variables are used; no extra space proportional to input size.



Welcome to ProgramGuru

Sign up to start your journey with us

Support ProgramGuru.org

Mention your name, and programguru.org in the message. Your name shall be displayed in the sponsers list.

PayPal

UPI

PhonePe QR

MALLIKARJUNA M