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

Minimize Maximum Distance Between Gas Stations
Binary Search Optimization



Problem Statement

You are given a sorted array of integers representing the positions of existing gas stations along a road. You are also given an integer k, which represents the number of additional gas stations you are allowed to add anywhere between the existing ones.

Your task is to minimize the maximum distance between any two adjacent gas stations after placing the k new stations optimally.

Return the smallest possible maximum distance between gas stations after the optimal placement, rounded to 6 decimal places if required.

Examples

Gas StationskMinimized Max DistanceDescription
[1, 2, 3, 4, 5]11.000000All gaps are equal (1), adding 1 station doesn’t improve max gap
[1, 2, 3, 4, 5]20.666667We can split two of the 1-length gaps to reduce max gap
[10, 20, 30]110.000000We split either gap [10,20] or [20,30], max gap becomes 10
[10, 30]26.666667Gap is 20, splitting into 3 parts gives max 6.66
[0, 100]910.000000We can add 9 stations to divide 100 into 10 segments
[0, 100]0100.000000No stations added, max distance is the initial gap
[]20.000000No stations exist, no meaningful distance to minimize
[5]30.000000Only one station means no gaps, so max distance is 0

Solution

This problem challenges us to minimize the longest distance between any two gas stations after adding k new ones. Let’s explore this in a more intuitive way for beginners.

Understanding the Problem

Imagine you’re driving along a highway and gas stations are placed at certain positions. Some parts of the highway may have long gaps between stations. If you’re allowed to place new gas stations, you would want to put them in the widest gaps to make your journey smoother by reducing the largest distance you’d need to travel without fuel.

Goal

After adding k new gas stations, the goal is to make the largest gap as small as possible. This is not necessarily about equalizing all distances — it's about shrinking the biggest one as much as possible.

Key Observations

  • The array is sorted, so we can easily compute all existing gaps (differences between consecutive stations).
  • Each gap can be split into smaller segments by placing stations in between.
  • To minimize the largest segment after splitting, we should distribute the k stations into the largest gaps.

What Does a Valid Answer Look Like?

If a number d is a valid answer, that means we can place k or fewer stations in such a way that no distance between any two gas stations exceeds d. So we're searching for the smallest possible such d.

Different Scenarios

  • No stations at all: If the array is empty, there's no gap to minimize. We return 0.000000.
  • Only one station: Again, no gaps exist, so the max distance is 0.
  • k = 0: We cannot add any station, so the answer is simply the longest existing gap.
  • Enough stations: If we have many stations, we can split the longest gaps multiple times. For example, a 100-length gap can be split into 10 parts with 9 stations — making the max distance 10.

Approach to Solve

We apply binary search on the answer (the possible distance). We start with the range of [0, max gap] and repeatedly check whether it is possible to achieve a certain distance by placing ≤ k stations. If possible, we try smaller values; if not, we go higher. We stop when we’ve narrowed the answer to a small precision like 1e-6.

This technique is often called binary search with greedy checking — binary search helps us find the minimum possible value, and greedy logic verifies if our guess is feasible.

The result is a floating point number with six decimal precision representing the minimum possible maximum distance between any two stations.

Visualization

Algorithm Steps

  1. Initialize low = 0 and high = max(arr[i+1] - arr[i]).
  2. While high - low > 1e-6:
  3. → Compute mid = (low + high) / 2.
  4. → Check if it's possible to place ≤ k stations so that max distance ≤ mid.
  5. → If yes, update high = mid.
  6. → Else, update low = mid.
  7. Return high (smallest valid max distance).

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
def is_possible(arr, k, dist):
    count = 0
    for i in range(len(arr) - 1):
        gap = arr[i+1] - arr[i]
        count += int(gap // dist)  # Number of stations needed in this gap
    return count <= k

def minimize_max_distance(arr, k):
    low = 0.0
    high = max(arr[i+1] - arr[i] for i in range(len(arr)-1))
    eps = 1e-6

    while high - low > eps:
        mid = (low + high) / 2
        if is_possible(arr, k, mid):
            high = mid  # Try to minimize max distance further
        else:
            low = mid

    return round(high, 6)

# Example
arr = [1, 2, 8]
k = 1
print("Minimum Max Distance:", minimize_max_distance(arr, k))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n × log(max_gap/eps))We perform binary search over precision range and for each guess, count stations in O(n).
Average CaseO(n × log(max_gap/eps))Each step of binary search tests feasibility by iterating through the array.
Average CaseO(n × log(max_gap/eps))In the worst case, we perform many precision steps, each involving scanning the entire array.

Space Complexity

O(1)

Explanation: No extra space is used apart from a few variables for binary search and counters.



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