Problem Description

You have gas stations positioned along a straight line (x-axis), represented by an integer array stations where each element indicates the position of a gas station.

Your task is to add exactly k new gas stations. These new stations can be placed at any position on the x-axis, including non-integer positions (like 3.5 or 7.25).

After adding all k new stations, you need to minimize the "penalty", which is defined as the maximum distance between any two adjacent gas stations (including both original and newly added stations).

For example, if you have stations at positions [1, 5, 10] and the distances between adjacent stations are 4 (between 1 and 5) and 5 (between 5 and 10), the penalty would be 5 (the maximum distance). By adding new stations strategically, you can reduce this maximum distance.

The goal is to find the optimal placement of k new stations such that the penalty (maximum distance between adjacent stations) is as small as possible. Return this minimum possible penalty value.

The answer will be accepted if it's within 10^-6 of the actual answer, allowing for floating-point precision tolerance.

Intuition

The key insight is recognizing that this is an optimization problem where we want to minimize the maximum value - a classic scenario for binary search on the answer.

Think about it this way: if someone tells you "the maximum distance between adjacent stations must be at most x", you can calculate exactly how many new stations you'd need to achieve this. For each existing gap between stations, if the gap is larger than x, you'll need to add stations to break it down into smaller segments of size at most x.

For a gap of size d, the number of new stations needed is floor(d / x). Why? Because we're essentially dividing the gap into segments of size x, and the number of divisions tells us how many stations to insert.

This creates a monotonic relationship: as x decreases (we want smaller maximum distance), the number of required stations increases. Conversely, as x increases (we allow larger maximum distance), fewer stations are needed.

Since we have exactly k stations to add, we're looking for the smallest value of x such that we need at most k stations to ensure no gap exceeds x. This is perfect for binary search:

• If a certain maximum distance x requires more than k stations, then x is too small
• If it requires k or fewer stations, then x is feasible and we can try an even smaller value

The search space is bounded between 0 (theoretical minimum) and the largest existing gap (or some large value like 10^8). We keep narrowing down this range until we converge on the optimal maximum distance, using a precision threshold of 10^-6 to determine when to stop.

Learn more about Binary Search patterns.

Solution Approach

The solution implements binary search on the answer space to find the minimum possible maximum distance between adjacent gas stations.

Implementation Details:

1. Binary Search Setup:

   • Initialize left = 0 (minimum possible distance) and right = 1e8 (a sufficiently large upper bound for maximum distance)
   • Continue searching while right - left > 1e-6 to achieve the required precision

2. Check Function:

   • The check(x) function determines if we can achieve a maximum distance of x with at most k new stations
   • For each pair of adjacent stations (a, b) in the original array, calculate how many new stations are needed in that gap
   • The formula int((b - a) / x) computes the number of stations needed:
     • (b - a) is the gap size
     • Dividing by x tells us how many segments of size x we need
     • Using int() gives us the floor value, which represents the number of cuts (new stations) needed
   • Sum all required stations across all gaps and check if it's <= k

3. Binary Search Logic:

   • Calculate mid = (left + right) / 2
   • If check(mid) returns True (we can achieve distance mid with k or fewer stations):
     • Update right = mid to try for an even smaller maximum distance
   • Otherwise (we need more than k stations):
     • Update left = mid because mid is too small to be achievable

4. Using pairwise:

   • The pairwise(stations) function generates consecutive pairs from the stations array
   • For example, [1, 5, 10] produces pairs (1, 5) and (5, 10)
   • This elegantly iterates through all adjacent gaps in the original stations

5. Return Value:

   • After convergence, return left which represents the minimum achievable maximum distance
   • Both left and right will have converged to approximately the same value within the precision threshold

The time complexity is O(n * log(W/ε)) where n is the number of stations, W is the search range, and ε is the precision (1e-6). The space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through a concrete example with stations = [1, 7, 15] and k = 2 new stations to add.

Initial Setup:

• Original stations: [1, 7, 15]
• Gaps between stations: 7-1=6 and 15-7=8
• Current penalty (max gap): 8
• We need to add exactly 2 new stations to minimize this penalty

Binary Search Process:

Round 1:

• left = 0, right = 100 (using 100 as upper bound for simplicity)
• mid = 50
• Check if we can achieve max distance of 50 with 2 stations:
  • Gap [1, 7]: size 6, needs int(6/50) = 0 stations
  • Gap [7, 15]: size 8, needs int(8/50) = 0 stations
  • Total needed: 0 stations ≤ 2 ✓
• Since feasible, try smaller: right = 50

Round 2:

• left = 0, right = 50
• mid = 25
• Check if we can achieve max distance of 25:
  • Gap [1, 7]: needs int(6/25) = 0 stations
  • Gap [7, 15]: needs int(8/25) = 0 stations
  • Total: 0 stations ≤ 2 ✓
• Try smaller: right = 25

Round 3:

• left = 0, right = 25
• mid = 12.5
• Check if we can achieve max distance of 12.5:
  • Gap [1, 7]: needs int(6/12.5) = 0 stations
  • Gap [7, 15]: needs int(8/12.5) = 0 stations
  • Total: 0 stations ≤ 2 ✓
• Try smaller: right = 12.5

Round 4:

• left = 0, right = 12.5
• mid = 6.25
• Check if we can achieve max distance of 6.25:
  • Gap [1, 7]: needs int(6/6.25) = 0 stations
  • Gap [7, 15]: needs int(8/6.25) = 1 station
  • Total: 1 station ≤ 2 ✓
• Try smaller: right = 6.25

Round 5:

• left = 0, right = 6.25
• mid = 3.125
• Check if we can achieve max distance of 3.125:
  • Gap [1, 7]: needs int(6/3.125) = 1 station
  • Gap [7, 15]: needs int(8/3.125) = 2 stations
  • Total: 3 stations > 2 ✗
• Too ambitious, increase: left = 3.125

Round 6:

• left = 3.125, right = 6.25
• mid = 4.6875
• Check if we can achieve max distance of 4.6875:
  • Gap [1, 7]: needs int(6/4.6875) = 1 station
  • Gap [7, 15]: needs int(8/4.6875) = 1 station
  • Total: 2 stations = 2 ✓
• Try smaller: right = 4.6875

Continuing... The binary search continues, narrowing the range between left and right until they converge within 1e-6.

Final Result: The algorithm converges to approximately 4.0. With this maximum distance:

• We place 1 station in gap [1, 7] at position 4.5 → gaps become [1, 4.5] (size 3.5) and [4.5, 7] (size 2.5)
• We place 1 station in gap [7, 15] at position 11 → gaps become [7, 11] (size 4) and [11, 15] (size 4)
• Final stations: [1, 4.5, 7, 11, 15]
• All gaps: 3.5, 2.5, 4, 4
• Maximum gap (penalty): 4.0

This demonstrates how binary search efficiently finds the optimal placement strategy by testing different maximum distance thresholds and checking feasibility.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * log(M/ε)) where n is the number of stations, M is the maximum distance between consecutive stations (bounded by 1e8), and ε is the precision (1e-6).

• The binary search runs for log((right - left) / ε) = log(1e8 / 1e-6) = log(1e14) iterations, which is approximately 47 iterations.
• In each iteration of the binary search, the check function is called.
• The check function iterates through all consecutive pairs of stations using pairwise(stations), which takes O(n-1) = O(n) time.
• For each pair, it performs a constant time operation: int((b - a) / x).
• Therefore, each check call takes O(n) time.
• Total time complexity: O(n * log(M/ε))

Space Complexity: O(1) or O(n) depending on the implementation of pairwise.

• The binary search uses only a constant amount of extra space for variables left, right, and mid.
• The check function uses a generator expression with sum(), which doesn't create an intermediate list.
• However, pairwise(stations) may create an iterator that could use O(1) space if implemented efficiently, or O(n) space if it creates intermediate pairs.
• In Python 3.10+, itertools.pairwise uses O(1) extra space.
• Overall space complexity: O(1) auxiliary space (excluding input).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Station Counting Formula

Pitfall: Using ceil((b - a) / x) or (b - a) // x + 1 to count the number of new stations needed in a gap.

When you have a gap of size d and want segments of maximum size x, many developers incorrectly think they need ceil(d/x) new stations. However, this counts the number of segments, not the number of dividing points (stations).

Example:

• Gap between stations: 10 units (stations at position 0 and 10)
• Maximum allowed distance: 5 units
• Incorrect calculation: ceil(10/5) = 2 stations
• Correct calculation: int(10/5) = 2 stations

Both give 2 here, but consider:

• Gap: 10 units, max distance: 3 units
• Incorrect: ceil(10/3) = 4 stations (creates 5 segments - too many!)
• Correct: int(10/3) = 3 stations (creates 4 segments of sizes ≤3)

Solution: Use int(gap / max_dist) which correctly calculates the number of cuts (new stations) needed to ensure no segment exceeds max_dist.

2. Binary Search Boundary Issues

Pitfall: Setting the right boundary too small, like using the maximum gap in the original array as the upper bound.

# Incorrect:
right = max(stations[i+1] - stations[i] for i in range(len(stations)-1))

Problem: When k=0 (no new stations to add), this would fail because the answer IS the maximum gap, and binary search needs room to converge properly with floating-point precision.

Solution: Use a sufficiently large upper bound like 1e8 or explicitly set it to the maximum gap when k>0:

# Correct approach:
if k == 0:
    return float(max(stations[i+1] - stations[i] for i in range(len(stations)-1)))
else:
    right = max(stations[i+1] - stations[i] for i in range(len(stations)-1))
    # or simply use a large constant like 1e8

3. Floating-Point Precision Errors

Pitfall: Using exact equality or insufficient precision in the termination condition.

# Incorrect:
while left < right:  # May loop forever with floating-point numbers
    mid = (left + right) / 2
    ...

Solution: Always use a precision threshold:

# Correct:
while right - left > 1e-6:
    mid = (left + right) / 2
    ...

4. Off-by-One Error in Station Counting

Pitfall: Confusing the relationship between gaps, segments, and dividing points.

Remember:

• Number of gaps = n - 1 (for n original stations)
• Number of new stations in a gap = number of cuts needed
• Number of segments after k cuts = k + 1

Solution: Visualize with small examples and verify the formula with edge cases like when the gap is exactly divisible by the maximum distance.