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BS-20. Minimise Maximum Distance between Gas Stations | 3 Approaches | Heap | Binary Search

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Summary of the Video

Title: BS-20. Minimise Maximum Distance between Gas Stations | 3 Approaches | Heap | Binary Search

Problem Statement

  • Given an array of n sorted integers representing coordinates of existing gas stations.
  • Task: Place K new gas stations to minimize the maximum distance between any two consecutive gas stations.
  • New gas stations can be placed at decimal coordinates.
  • Return the minimized maximum distance, possibly as a floating-point number (double/long double).
  • Answers are accepted within a tolerance of 10^-6 decimal places.

Key Concepts & Insights

  • The problem focuses on minimizing the maximum gap between gas stations after adding K new stations.
  • New stations must be placed between existing stations, not outside the extremes.
  • The maximum gap can be reduced by evenly distributing new stations in the largest gaps.
  • Decimal placement is allowed, so distances can be fractional.
  • Answers should be precise up to 6 decimal places to avoid time limit exceeded (TLE) errors.

Approach 1: Brute Force (Naive)

  • Keep track of how many new stations are placed between each pair of existing stations.
  • Iteratively place one gas station in the section with the current maximum gap.
  • Recalculate the maximum gap after each placement.
  • Time complexity: O(K * N), which can lead to TLE for large inputs.

Steps:

  1. Initialize an array howMany to track new stations per section.
  2. For each new station (1 to K):
    • Find the section with the maximum current gap.
    • Place the station there (increment count).
  3. After placing all stations, compute the maximum gap by dividing each section length by howMany[section] + 1.
  4. Return the maximum gap.

Approach 2: Optimized with Priority Queue (Heap)

  • Use a max-heap (priority queue) to efficiently track and update the largest gap.
  • Store pairs (current_section_length, section_index) in the heap.
  • For each new station:
    • Extract the largest gap from the heap.
    • Place a station in that section, splitting it into smaller segments.
    • Calculate new segment length = original length / (number of stations placed + 1).
    • Push the updated segment length back into the heap.
  • Time complexity: O((N + K) log N)
  • Space complexity: O(N)

This approach avoids scanning all sections repeatedly, improving performance.

Approach 3: Binary Search on Answer (Most Optimal)

  • Solve the problem using binary search on the answer (the minimized maximum distance).

Range for answer:

  • Low = 0 (minimum possible distance)
  • High = maximum distance between any two existing gas stations

Binary search steps:

  1. Calculate mid = (low + high) / 2 (floating-point precision).
  2. Check if it is possible to place K stations so that no gap exceeds mid:
    • For each gap, calculate stations needed = floor(gap / mid) (subtract 1 if exact division).
  3. If stations needed > K → mid is too small, increase low.
  4. Else → mid is feasible, reduce high.
  5. Terminate when (high - low) < 10^-6 (precision requirement).
  6. Return high or the best feasible answer.
  • Time complexity: O(N log(range * 10^6)) due to binary search precision and linear checks.
  • Space complexity: O(N)

This approach is preferred for large inputs and interview settings.

Additional Notes

  • Precision handling is crucial; answers must be rounded or truncated to 6 decimal places.
  • Floating-point binary search differs from integer binary search:
    • Use condition while (high - low > 1e-6) instead of low <= high.
    • Update low = mid or high = mid (no ±1 increments).
  • Avoid placing stations outside the existing gas stations since it doesn’t reduce the maximum gap.
  • The problem is complex and considered a challenging binary search problem involving floating-point calculations.

Summary of Methodologies

Brute Force

  • Initialize placement counts to zero.
  • For each new station, place it in the section with the largest current gap.
  • Update gaps after each placement.
  • Return max gap after all placements.

Priority Queue Optimization

  • Push all initial gaps into a max-heap.
  • For each new station:
    • Pop the largest gap.
    • Place a station, split the gap.
    • Push the updated smaller gap back.
  • Return the largest gap after all placements.

Binary Search on Answer

  • Set low = 0, high = max gap.
  • While high - low > 1e-6:
    • mid = (low + high) / 2
    • Calculate stations needed to keep gaps ≤ mid.
    • If needed stations > K, low = mid (increase distance).
    • Else high = mid (try smaller max distance).
  • Return high as minimized max gap.

Speakers / Sources

  • The video features a single speaker (likely the channel owner or instructor), referred to as “Striver” by the narrator.
  • The speaker explains concepts, algorithms, and code implementation in a step-by-step manner.
  • No other speakers or external sources are mentioned.

This summary captures the problem, the three solution approaches, key algorithmic insights, and implementation details as explained in the video.

Original video