Solution Approach

The solution implements binary search on the answer space using the boundary-finding template pattern, adapted for floating-point precision.

Defining the Feasible Function

For this problem, feasible(x) returns true if we can achieve a maximum distance of x using at most k new stations.

This creates a boolean pattern across possible distances:

distance: 0.1   0.5   1.0   2.0   3.0   4.0   ...
feasible: false false false true  true  true  ...

We want to find the first true value - the minimum distance that is feasible.

Binary Search Template (Floating-Point Adaptation)

Since this involves floating-point numbers, we adapt the template using precision-based termination:

1. Initialize left = 0, right = 1e8, and first_true_value = right
2. While right - left > 1e-6:
   • Calculate mid = (left + right) / 2
   • If feasible(mid) is true:
     • Record first_true_value = mid as a potential answer
     • Search left: right = mid (look for smaller feasible distance)
   • Else:
     • Search right: left = mid
3. Return first_true_value (or left, which converges to the same value)

Feasible Check Implementation

For each gap between adjacent stations of size d, we need floor(d / x) new stations to ensure no segment exceeds x:

• If the gap is 10 and max distance is 3, we need floor(10/3) = 3 stations
• This creates 4 segments of sizes ≤ 3

Sum across all gaps and check if total ≤ k.

Why This Works

The feasibility function is monotonic: larger x values require fewer stations. Once we find a feasible x, all larger values are also feasible. Binary search efficiently finds the boundary between infeasible and feasible regions.

Time Complexity: O(n × log(W/ε)) where n is stations count, W is the search range, ε is precision. Space Complexity: O(1)

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.

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. Using Wrong Binary Search Pattern for Floating-Point

For floating-point binary search, you cannot use the exact integer template (while left <= right with right = mid - 1). You must adapt for continuous values.

Incorrect (integer pattern):

while left <= right:  # Will not terminate for floats
    mid = (left + right) / 2
    if feasible(mid):
        right = mid - 1  # Wrong for floating-point
    else:
        left = mid + 1

Correct (floating-point adaptation):

while right - left > 1e-6:  # Precision-based termination
    mid = (left + right) / 2
    if feasible(mid):
        right = mid  # No -1 for floats
    else:
        left = mid  # No +1 for floats
return left

2. Incorrect Station Counting Formula

Pitfall: Using ceil((b - a) / x) to count new stations needed in a gap. This counts segments, not dividing points.

Example:

• 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 needed.

3. Floating-Point Precision Errors

Pitfall: Using exact equality or insufficient precision.

# Wrong: May loop forever
while left < right:
    ...

Solution: Always use a precision threshold:

while right - left > 1e-6:
    ...

4. Binary Search Boundary Issues

Pitfall: Setting right boundary too small when k=0.

Solution: Use a sufficiently large upper bound like 1e8, or handle k=0 as a special case returning the maximum gap directly.