774. Minimize Max Distance to Gas Station


Problem Description

In this problem, we are dealing with an optimization problem involving gas stations positioned linearly along an x-axis. The positions of these existing gas stations are provided as an integer array stations. We are tasked with adding k new gas stations to this line, and these new stations can be placed at any point along the x-axis, which doesn't need to be an integer value.

The goal is to minimize the "penalty," which is defined as the maximum distance between any two adjacent gas stations after all k new stations have been added. The final output should be the smallest possible value of this maximum distance, and an answer that is within 10^-6 of the actual answer is considered valid.

Intuition

To solve this problem, we can use a binary search approach to find the smallest possible maximum distance that satisfies the conditions. The main intuition here is that if we have a certain maximum distance D, we can determine whether it's possible to add k or fewer gas stations such that no two stations are more than D miles apart.

The binary search is performed over the range of possible distances which could be the penalty. We start by setting the left end (minimum possible penalty) to 0 and the right end (maximum possible penalty) to 1e8, a sufficiently large number that is guaranteed to be larger than any possible maximum distance.

The check function is used within the binary search to determine if a given maximum distance x is feasible. If it is, that means we can place k new gas stations in such a way that the maximum distance between any two adjacent stations is not greater than x. To do this, for any two adjacent existing stations, we calculate how many new stations would need to be added between them to ensure that the distance between adjacent stations is less than or equal to x. If the sum of required stations for all gaps does not exceed k, the function returns True.

Using this check function, we continue to narrow down the range in the binary search until the difference between the left and right ends is less than 10^-6, which means we have found the minimum possible penalty to the desired precision. We then return this value as the smallest possible value of penalty().

Solution Approach

The solution employs a binary search algorithm to efficiently find the smallest possible "penalty", which is the maximum distance between adjacent gas stations after adding k new stations. The binary search operates over a range of possible values for the penalty, rather than searching through a list of items.

Here's an explanation of how the algorithm works, referring to the solution code:

Binary Search Range

  • Initial Range: We define an initial range for the binary search with the left end set to 0 and the right end set to 1e8. This wide range ensures that the actual minimum penalty is included.

check Function

  • Purpose: It determines whether it's possible to add stations such that the maximum distance between any two stations doesn't exceed x.
  • Implementation: It iterates through each pair of adjacent stations (using pairwise which groups the array elements in pairs, provided by the Python standard library), and computes how many additional stations need to be added in between to ensure that the distance between stations is at most x. The computation (b - a) / x gives the number of stations required between stations located at a and b. Summing these up for all pairs and comparing it to k tells us if x is a feasible penalty.

Iterative Binary Search

  • Loop: The while loop runs until the left and right ends of the search range are within 10^-6 of each other, meaning we have reached the required precision.
  • Midpoint Calculation: In each iteration, it calculates the midpoint mid of the current search range by averaging left and right.
  • Feasibility Check: It calls the check function with this midpoint value.
    • If check(mid) returns True, it means that we can achieve the penalty mid by adding k or fewer stations, so we update the right end to mid.
    • If check(mid) returns False, we need a larger penalty to fit k stations, so we update the left end to mid.

Convergence and Result

  • Precision: The search loop continues until the precision condition is met. This ensures our answer is within 10^-6 of the actual minimum penalty.
  • Return Value: The loop exits when the left and right ends are sufficiently close, and left is returned as it represents the smallest tested penalty that can be achieved with k gas stations.

This approach efficiently zooms in on the correct answer using logarithmic time complexity, which is typical of binary search algorithms. This is much more optimal than a linear search or brute force approach, which could be time-prohibitive given the potentially large search space.

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Example Walkthrough

Consider a scenario where there are three gas stations along a road at positions stations = [1, 5, 10] and we want to add k = 2 new gas stations. We will illustrate how the solution approach using binary search helps us find the minimum possible penalty.

Initial Setup

The current maximum distance between adjacent stations is 5, which is between the first and second stations (positions 1 and 5). We aim to minimize this maximum distance by adding 2 new stations.

Binary Search Range Setup

We'll set up our binary search with an initial range for the penalty, from left = 0 to right = 1e8.

First Iteration

  • We calculate the midpoint mid of left and right, which is mid = (0 + 1e8) / 2 = 5e7. Obviously, this is an overestimate.
  • We run the check function with mid. For each pair of stations, we calculate how many new stations would need to be placed to ensure no adjacent stations are more than 5e7 miles apart. Clearly, zero stations are needed since the largest gap is 5 which is much less than 5e7.
  • Since check(mid) would return true (no stations needed is less than or equal to k = 2), we update right to mid.

Subsequent Iterations

  • Next mid is mid = (0 + 5e7) / 2 = 2.5e7, and the checks and updates continue similarly. The midpoint is still much larger than the maximum gap, and the right will continue to decrease dramatically.
  • As we keep iterating, the mid value will come down significantly, closer to the actual gap size.

Narrowing Down

Eventually, after several iterations, mid will be close to the actual maximum distance we need. Suppose mid comes down to 2.5. At this distance, we check how many new stations are needed:

  • Between stations at 1 and 5 (gap = 4), we need at most 1 new station to ensure the gap is no larger than 2.5.
  • Between stations at 5 and 10 (gap = 5), we need 2 new stations.

Since we only need 2 stations and we are allowed to add 2, this mid value is feasible. If mid were smaller, we might need more than 2 stations, and the check would return false, prompting us to adjust the left end of the range.

Final Iteration and Result

As we continue the binary search, updating left and right, the interval of left and right narrows down to within 10^-6 of each other. Suppose through the binary search we reach left = 2.499999 and right = 2.500001, we then stop as this is within our acceptable precision.

At this point, if check(2.499999) returns true, that means we can place 2 new stations such that no segment is longer than 2.499999 miles, and thus, this is our final answer. Since our target is a precision within 10^-6, the minimum penalty, or the smallest maximum distance between adjacent gas stations after adding the new ones, is approximately 2.5.

This illustrates the effectiveness of the binary search technique for our optimization problem, where instead of exhaustively trying every possible location for new stations, the binary search algorithm quickly zeroes in on the smallest possible penalty, the maximum distance between adjacent stations after adding the new ones.

Python Solution

1from typing import List
2from itertools import pairwise
3
4class Solution:
5    def minmaxGasDist(self, stations: List[int], k: int) -> float:
6        # Helper function to calculate if it's possible to have maximum
7        # distance no more than 'x' between two adjacent gas stations after
8        # adding 'k' new gas stations
9        def is_possible(x):
10            # Count the number of stations needed to ensure that no segment
11            # between the existing stations is longer than 'x'
12            additional_stations_needed = sum(int((b - a) / x) for a, b in pairwise(stations))
13            # Return True if the calculated number of stations needed is
14            # less than or equal to 'k', False otherwise
15            return additional_stations_needed <= k
16
17        # Initialize the binary search bounds
18        left, right = 0, 1e8  # Start with a wide range since distances could be large
19
20        # Perform binary search with precision up to 1e-6
21        while right - left > 1e-6:
22            mid = (left + right) / 2
23            # Check if the current middle value can satisfy the condition
24            if is_possible(mid):
25                right = mid  # If it is possible, the answer is less than or equal to 'mid'
26            else:
27                left = mid  # If not possible, the answer is greater than 'mid'
28
29        # Once the loop ends, 'left' will be our answer to the smallest possible maximum distance
30        # that fulfills the requirements, with the required precision
31        return left
32

Java Solution

1class Solution {
2    // Finds the minimum possible distance between gas stations after adding k extra stations.
3    public double minmaxGasDist(int[] stations, int K) {
4        // Define the range for the possible solution (left is minimum distance, right is maximum distance)
5        double left = 0, right = 1e8;
6      
7        // Continue searching while the precision is greater than 1e-6
8        while (right - left > 1e-6) {
9            // Take the middle of the current range as a guess
10            double mid = (left + right) / 2.0;
11          
12            // If it's possible to place gas stations such that the maximum distance is less than or equal to 'mid',
13            // then we update right to the current 'mid' to see if we can find an even smaller maximum distance
14            if (isPossible(mid, stations, K)) {
15                right = mid;
16            } else {
17                // If it's not possible, we update left to be 'mid' to look for solutions greater than 'mid'
18                left = mid;
19            }
20        }
21      
22        // Since the left and right converge to the point where right - left <= 1e-6, left is the most accurate answer
23        return left;
24    }
25
26    // Helper method to check if it's possible to have a maximum gas station distance less than x after adding K gas stations.
27    private boolean isPossible(double x, int[] stations, int K) {
28        int count = 0; // the number of gas stations we need to add to satisfy the condition
29      
30        // Go through all pairs of adjacent stations
31        for (int i = 0; i < stations.length - 1; ++i) {
32            // Find the number of additional stations needed for this segment so that the distance between stations is <= x
33            count += (int) ((stations[i + 1] - stations[i]) / x);
34        }
35      
36        // Check if the count of additional stations required is less than or equal to K (the count we can add)
37        return count <= K;
38    }
39}
40

C++ Solution

1#include <vector>
2using namespace std;
3
4class Solution {
5public:
6    double minmaxGasDist(vector<int>& stations, int k) {
7        double minDist = 0, maxDist = 1e8; // Initializing the range for possible answers
8      
9        // Lambda function to check if a given maximum distance 'maxDistBetweenStations' can be achieved
10        // by adding at most 'k' gas stations
11        auto isPossible = [&](double maxDistBetweenStations) {
12            int requiredStations = 0;
13          
14            // Count the number of additional stations required for each interval between stations
15            for (int i = 0; i < stations.size() - 1; ++i) {
16                requiredStations += (int)((stations[i + 1] - stations[i]) / maxDistBetweenStations);
17            }
18          
19            // Return true if the number of additional stations to maintain the maximum distance
20            // is less than or equal to 'k'
21            return requiredStations <= k;
22        };
23      
24        // Binary search to find the smallest possible maximum distance
25        while (maxDist - minDist > 1e-6) { // 1e-6 is used as the precision for the answer
26            double midDist = (minDist + maxDist) / 2.0;
27          
28            // If it is possible to achieve this maximum distance, we try to find a smaller one
29            if (isPossible(midDist)) {
30                maxDist = midDist;
31            } else {
32                // If it is not possible, we need to increase the maximum distance
33                minDist = midDist;
34            }
35        }
36      
37        // After the loop, 'minDist' will be our answer to the problem, rounded to the nearest 1e-6
38        return minDist;
39    }
40};
41

Typescript Solution

1function minmaxGasDist(stations: number[], k: number): number {
2    let minDist: number = 0, maxDist: number = 1e8; // Initialize the range for possible answers
3  
4    // Function to check if a given maximum distance 'maxDistBetweenStations' can be achieved
5    // by adding at most 'k' gas stations
6    const isPossible = (maxDistBetweenStations: number): boolean => {
7        let requiredStations: number = 0;
8      
9        // Count the number of additional stations required for each interval between stations
10        for (let i = 0; i < stations.length - 1; ++i) {
11            requiredStations += Math.ceil((stations[i + 1] - stations[i]) / maxDistBetweenStations) - 1;
12        }
13      
14        // Return true if the number of additional stations to maintain the max distance is less than or equal to 'k'
15        return requiredStations <= k;
16    };
17  
18    // Binary search to find the smallest possible maximum distance
19    while (maxDist - minDist > 1e-6) { // 1e-6 is the precision for the answer
20        let midDist: number = (minDist + maxDist) / 2.0;
21      
22        // If it is possible to achieve this max distance, we try to find a smaller one
23        if (isPossible(midDist)) {
24            maxDist = midDist;
25        } else {
26            // If it is not possible, we increase the maximum distance
27            minDist = midDist;
28        }
29    }
30  
31    // After the loop, 'minDist' will be our answer to the problem, rounded to the nearest 1e-6
32    return minDist;
33}
34

Time and Space Complexity

The time complexity of the provided code is primarily determined by the while loop that performs a binary search over a range of possible answers from left to right to find the minimum possible maximum gas station distance with a precision of 1e-6. During each iteration of the binary search, a check function is called which takes linear time relative to the number of stations (O(N)), as it iteratively calculates the number of additional gas stations needed between each pair of consecutive stations. The binary search will run for O(log((right-left)/precision)) iterations, right-left being the initial search range and precision being 1e-6. Therefore, the total time complexity of the algorithm is O(N * log((right-left)/precision)). In this case, the initial search range right-left is fixed at 1e8, so we can consider this a constant and the log term simplifies to O(log(1/precision)), resulting in a final time complexity of O(N * log(1/precision)).

The space complexity of the code is O(1) which means constant space complexity, as only a fixed number of variables are used and there are no data structures that grow with the input size. The space used by the check function and the binary search variables left, right, and mid does not scale with the number of stations.

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