BS-20. Minimising Maximum Distance between Gas Stations

1. Problem Statement & Core Objective

The problem requires us to place $K$ new gas stations among $n$ existing gas stations (whose coordinates are given in a sorted array) such that the maximum distance between any two adjacent gas stations is minimized.

Key Rules and Constraints

• Sorted Input: The initial coordinates of the $n$ existing gas stations are provided in strictly ascending order (though not necessarily at consecutive integer steps; e.g., [1, 13, 17, 23]).
• Fixed Extremes: You cannot place new gas stations outside the outermost existing stations (to the left of the first station or to the right of the last station) because doing so would not reduce the maximum distance gap.
• Decimal Coordinates: New gas stations can be placed at floating-point (decimal) coordinates.
• Precision Acceptance: Because answers are calculated in floating-point formats (double or long double), online judges typically accept outputs if they are within $10^{-6}$ (or $10^{-7}$) of the official setter's solution.

2. Real-Life Analogy & Mental Model

Real-Life Analogy: The Highway Toll Booths / Rest Stops

Imagine a long stretch of highway where existing rest stops are located at specific mile markers. You are given a budget to build $K$ brand-new rest stops between them.

• If you clump all new rest stops in one small region, the rest of the highway suffers from massive gaps between stops.
• To provide the best driver experience, you must distribute the new rest stops evenly across the sections that currently have the widest gaps, ensuring the longest single gap on the entire highway is made as short as humanly possible.

3. Understanding Precision and Time Limit Exceeded (TLE) in Floating-Point Binary Search

When returning double or long double values, outputs can contain extremely long decimal expansions (e.g., 0.65421678124...).

Why TLE Happens: If your binary search or iterative algorithm tries to resolve differences down to the absolute limit of machine float representation, it will loop millions of times, resulting in a Time Limit Exceeded (TLE).

The Fix: Judges explicitly specify that an answer within $10^{-6}$ of the actual answer is accepted. This means the loop can terminate as soon as high - low < 10^{-6}, avoiding endless precision calculations.

4. Approach 1: Naive / Brute Force Solution

The Intuition

For each of the $K$ new gas stations we need to place:

1. Scan all existing sections between adjacent gas stations.
2. Determine the section currently having the maximum distance.
3. Place a new gas station inside that specific section to split it.
4. Repeat this $K$ times. Finally, re-scan all sections to find the maximum distance.

Algorithm Walkthrough

• Maintain an array howManyPlaced[n-1] to track how many new stations have been placed in each section.
• Run a loop $K$ times (for each new gas station):
  • Find the section i where $\frac{\text{Distance}_i}{\text{howManyPlaced}[i] + 1}$ is maximized.
  • Increment howManyPlaced[maxIndex]++.
• After placing all $K$ stations, calculate the final maximum section length across all intervals and return it.

Complexity Analysis

• Time Complexity: $O(K \times n)$ because we iterate $K$ times, and in each iteration, we scan $n$ intervals. For large constraints, this results in quadratic time and causes a TLE.
• Space Complexity: $O(n)$ to store the howManyPlaced array.

5. Approach 2: Optimized Solution Using Priority Queue (Heap)

The Core Idea

Instead of scanning all $n$ intervals in $O(n)$ time during every single one of the $K$ placements, we can store all intervals in a Max-Heap (Priority Queue) ordered by their current effective section length.

• Max-Heap Property: The top of the priority queue always gives us the interval with the largest current distance in $O(1)$ time, while insertions take $O(\log n)$ time.

Step-by-Step Execution

1. Push all initial intervals into a max-heap as pairs: (section_length, section_index).
2. Keep an array howManyPlaced initialized to $0$ for all sections.
3. Repeat $K$ times:
   • Extract the top element from the priority queue (the largest section).
   • Increment the placement count for that section index: howManyPlaced[index]++.
   • Calculate the new divided section length:

• Push the new length and index back into the priority queue.

4. After $K$ extractions and insertions, the top of the priority queue will hold our answer.

Code Implementation (C++ Concept)

#include <vector>
#include <queue>
#include <utility>

using namespace std;

long double minimiseMaxDistance(vector<int>& arr, int K) {
    int n = arr.size();
    // Max heap storing pairs of {current_section_length, section_index}
    priority_queue<pair<long double, int>> pq;
    vector<int> howManyPlaced(n - 1, 0);

    for (int i = 0; i < n - 1; i++) {
        pq.push({(long double)(arr[i+1] - arr[i]), i});
    }

    for (int gasStations = 1; gasStations <= K; gasStations++) {
        auto top = pq.top();
        pq.pop();
        
        int secIndex = top.second;
        howManyPlaced[secIndex]++;
        
        long double initialDist = arr[secIndex + 1] - arr[secIndex];
        long double newSectionLength = initialDist / (howManyPlaced[secIndex] + 1);
        
        pq.push({newSectionLength, secIndex});
    }

    return pq.top().first;
}

Complexity Analysis

• Time Complexity:
  • Initializing the Heap: $O(n \log n)$
  • Performing $K$ heap operations: $O(K \log n)$
  • Total Time: $O(n \log n + K \log n)$
• Space Complexity: $O(n)$ for the priority queue and tracking array.

6. Approach 3: Optimal Solution Using Binary Search on Answers

Why Binary Search Works

Whenever a problem asks us to "minimize the maximum" or "maximize the minimum", Binary Search on the Answer space is a prime candidate.

• If a specific maximum distance mid is possible to achieve with $\le K$ gas stations, then any distance greater than mid is also possible.
• If mid is not possible (requires $> K$ stations), then any distance smaller than mid is also impossible.
• This monotonic property allows us to apply Binary Search over the range of possible distances $[0, \text{Max\_Initial\_Distance}]$.

Defining the Search Space

• low: 0 (or a very small positive decimal value).
• high: The maximum distance between any two adjacent existing gas stations in the input array.

The Check Function (numberOfGasStationsRequired)

For a guessed distance mid, how many gas stations do we need to place so that no adjacent distance exceeds mid?

• For an interval of length diff = arr[i+1] - arr[i]:
  • Number of stations needed = $\lfloor \frac{\text{diff}}{\text{mid}} \rfloor$
  • Edge Case / Exact Division: If diff is perfectly divisible by mid (e.g., diff / mid is an exact integer), we actually need (diff / mid) - 1 stations because placing that exact number would leave a remainder of 0 extra stations.

int numberOfGasStationsRequired(long double mid, const vector<int>& arr) {
    int count = 0;
    for (int i = 1; i < arr.size(); i++) {
        long double diff = arr[i] - arr[i - 1];
        int stationsInBetween = diff / mid;
        // If exact division, subtract 1 to account for overlapping boundaries
        if ((diff / mid) == stationsInBetween) {
            stationsInBetween--;
        }
        count += stationsInBetween;
    }
    return count;
}

Binary Search Termination for Floating-Point Numbers

Unlike integer binary searches (while (low <= high) with mid + 1 / mid - 1), floating-point binary search runs until our search window is smaller than our precision threshold ($10^{-6}$).

long double low = 0;
long double high = 0;
for (int i = 0; i < n - 1; i++) {
    high = max(high, (long double)(arr[i+1] - arr[i]));
}

long double ans = high;

while (high - low > 1e-6) {
    long double mid = low + (high - low) / 2.0;
    if (numberOfGasStationsRequired(mid, arr) > K) {
        low = mid;     // Too many stations needed; increase distance threshold
    } else {
        ans = mid;     // Feasible; try to find an even smaller maximum distance
        high = mid;
    }
}
return ans;

7. Comparative Synthesis of All 3 Approaches

8. Summary and Key Takeaways

• Problem Classification: Minimizing maximum distances points directly to Binary Search on Answers.
• Fixed Extremes Rule: New gas stations should always be placed between existing stations rather than outside the outermost coordinates.
• Precision Handling: When dealing with long double binary search, condition the loop on high - low > 10^{-6} instead of integer bounds (low <= high).
• Mid Adjustments: In floating-point binary search, do not use mid + 1 or mid - 1; update pointers using low = mid and high = mid.
• Exact Division Trap: In the check function, check if diff / mid divides evenly to avoid over-counting required gas stations.
• Complexity Leap: Moving from a nested linear scan $O(K \cdot n)$ to a Max-Heap $O(K \log n)$ solves TLE, while Binary Search reduces auxiliary space to $O(1)$.