Minimise Max Distance to Gas Stations¶

Question¶

Given a sorted integer array arr representing positions of \(N\) existing gas stations on the X-axis, and an integer \(K\), place \(K\) new gas stations anywhere on the X-axis. Find the minimum possible value of the maximum distance between adjacent gas stations after all \(K\) new stations are added. Your answer must be within \(10^{-6}\) of the true value.

Solution¶

Pattern¶

Floating-Point Binary Search on Answer We cannot efficiently compute the exact optimal placements iteratively if \(K\) is massive. However, we know the optimal maximum distance lies somewhere between \(0.0\) and the maximum current gap. We binary search this continuous floating-point range. For a "guessed" maximum distance, we greedily calculate how many new stations are required to break up all existing gaps to be \(\le\) our guess.

How to Identify¶

The problem asks to minimize a maximum (minimize the maximum gap).
The solution requires high precision (e.g., \(10^{-6}\)), dictating a continuous floating-point search space rather than discrete integers.
The constraint on \(K\) can be extremely large, rendering \(O(K \log N)\) Priority Queue (Max-Heap) approaches unviable.

Description¶

Step-by-step explanation:

Establish Bounds: - Lower Bound (left): \(0.0\). The distance can theoretically be infinitesimal.
Upper Bound (right): The maximum distance between any two adjacent stations in the given array.
Floating-Point Loop: Loop while the difference between right and left is strictly greater than the requested precision (1e-6).
Calculate Midpoint: mid = left + (right - left) / 2.0. This represents our guessed optimal maximum gap.
Greedy Validation (canPlaceStations):
Iterate through every adjacent pair of stations.
Calculate the distance between them: dist = arr[i] - arr[i-1].
The number of new stations required to ensure no resulting segment is larger than mid is determined by: (int)((dist - 1e-9) / mid). The - 1e-9 is crucial for exact divisions (e.g., gap of 2 divided by max limit of 1 should require 1 internal station, not 2).
Sum these required stations across all gaps.
If total required stations \(> K\), our guess mid was too strict. Return false.
Adjust Bounds (CRITICAL):
Because the search space is continuous, do not use mid + 1 or mid - epsilon.
If valid: right = mid.
If invalid: left = mid.
Return: When the gap closes to \(< 10^{-6}\), right holds the precise minimized maximum distance.

The Intuition¶

Imagine you have a highway with several very long stretches without gas stations. You are given a budget of \(K\) new stations to place. Your goal is to ensure drivers never drive more than \(X\) miles between stations. You guess \(X = 50.0\) miles. You inspect a 120-mile stretch. To break it into pieces of 50 or less, you must place 2 stations (creating 3 pieces of 40 miles). You repeat this math for every gap on the highway. If you used \(10\) stations but your budget was only \(5\), your guess of \(50.0\) was too small. You need to allow a larger gap. If you only used \(3\) stations, your guess worked! But maybe you can squeeze it even tighter, so you try \(40.0\) miles. You halve the guess repeatedly until converging on the exact decimal value that perfectly utilizes your budget.

Complexity¶

Label	Worst	Average
Time Complexity	\(O(N \log \frac{M}{\epsilon})\)	\(O(N \log \frac{M}{\epsilon})\)
Space Complexity	\(O(1)\)	\(O(1)\)

Time Complexity¶

Let \(M\) be the maximum initial distance between any two stations, and \(\epsilon\) be the precision tolerance \(10^{-6}\). The search space halves in a continuous domain. It takes exactly \(\approx \log_2(\frac{M}{10^{-6}})\) steps to converge. Inside the loop, we execute a linear \(O(N)\) scan. Total time is \(O(N \log \frac{M}{\epsilon})\). Given constraints, this is roughly 50 iterations, making it highly efficient.

Space Complexity¶

We only allocate primitive variables (left, right, mid, requiredStations). Auxiliary space is strictly \(O(1)\).

Code¶

class Solution {
    public double minimiseMaxDistance(int[] arr, int k) {
        if (arr == null || arr.length < 2) return 0.0;

        double left = 0;
        double right = 0;

        // Establish the upper bound (maximum initial gap)
        for (int i = 1; i < arr.length; i++) {
            right = Math.max(right, arr[i] - arr[i - 1]);
        }

        // Search space is continuous; define precision threshold (10^-6 precision)
        double diff = 1e-6;

        while (right - left > diff) {
            double mid = left + (right - left) / 2.0;

            if (canPlaceStations(arr, k, mid)) {
                // mid is valid, attempt to find a smaller maximum distance
                right = mid;
            } else {
                // mid is too small, requires more stations than k
                left = mid;
            }
        }

        return right;
    }

    private boolean canPlaceStations(int[] arr, int targetStations, double maxDistance) {
        int requiredStations = 0;
        double epsilon = 1e-9;

        for (int i = 1; i < arr.length; i++) {
            double gap = arr[i] - arr[i - 1];

            // The -epsilon handles perfectly divisible gaps.
            // E.g., gap 2.0 / maxDist 1.0 requires 1 station, not 2.
            // (2.0 - 1e-9) / 1.0 = 1.999999999 -> truncates to 1.
            requiredStations += (int) ((gap - epsilon) / maxDistance);

            // Fast fail
            if (requiredStations > targetStations) return false;
        }

        return true;
    }
}

Caveats¶

Floating-Point Boundaries: A fatal mistake in floating-point binary searches is treating them like integer array indices (e.g., left = mid + 0.0001). If the optimal answer is 3.141592 and mid is 3.1415, adding 0.0001 forces left to 3.1416, permanently skipping over the true answer. Always use strict left = mid and right = mid.
IEEE 754 Precision Errors: Using Math.ceil(dist / maxDist) - 1 seems mathematically perfect. However, due to binary representation artifacts, \(2.0 / 1.0\) might evaluate to 2.0000000000000004. Math.ceil bumps this to \(3\), causing the required stations to overestimate and erroneously fail the validation. Using (int) truncation safely sidesteps this.
Modulo Operator: Never use % on double values to check divisibility; it is unreliable.
Scale of K: If \(k\) is small, a Max-Heap approach \(O(k \log N)\) is faster, but Binary Search is the general-purpose solution for large \(k\)
Exact Division Trap: On paper, an exact integer division (e.g., Gap is 2.0, maxDistance is 1.0) mathematically returns 2. However, to break a gap of 2 into pieces of 1, you only need to place 1 internal station. Subtracting an infinitesimally small epsilon from the gap (gap - 1e-9) before integer truncation elegantly fixes this.

Concepts to Think About¶

The Priority Queue (Max-Heap) Alternative: You can solve this by pushing all initial gaps into a Max-Heap. You pop the largest gap, increment its internal station count, recalculate its new split gap size, and push it back, repeating \(K\) times. While conceptually intuitive, the time complexity is \(O(K \log N)\). If \(K = 10^9\), this approach will Time Limit Exceed (TLE). Binary search is universally superior here because its runtime is decoupled from \(K\)'s magnitude.
Numerical Stability: Dealing with floating-point underflow/overflow.
Fixed Iterations vs. Epsilon Loop: Trade-offs in termination conditions.
Greedy vs. BS: When to use a Priority Queue vs. Binary Search on Answer.
NP-Hardness: How "contiguous" requirements (like in Book Allocation) simplify problems that would otherwise be NP-hard (like Partition).

Logical Follow-up¶

Question: If we use (int)(gap / maxDistance), won't it fail if the gap is exactly divisible? (e.g., Gap is 2, maxDistance is 1. (int)(2/1) = 2, but we only need 1 station!) Solution: On paper, exact integer division fails this truncation check. However, because we are executing a continuous binary search with a tolerance of \(10^{-6}\), the algorithm naturally tests values infinitesimally larger than the integer (e.g., mid = 1.0000001). When it evaluates (int)(2.0 / 1.0000001), it safely truncates to 1, which correctly satisfies the budget constraint. The continuous nature of the binary search automatically "saves" the division from its exact-integer flaw, rendering the algorithm perfectly accurate within the accepted tolerance.

Question: If you were given a very small \(k\) (e.g., \(k=10\)) but \(n=1,000,000\), would this approach still be the best? Solution: No. In that case, an \(O(k \log n)\) approach using a Max-Heap would be faster. You would store all current gaps in the heap and repeatedly split the largest one.

Question: Can we parallelize this? Solution: Yes. The predicate function getRequiredStations is a simple summation over an array, which can be easily distributed across multiple cores or even using MapReduce for massive datasets.