Solution Approach

The solution implements a binary search combined with a greedy validation function.

Step 1: Sorting First, we sort the start array in ascending order. This allows us to process intervals from left to right and make greedy decisions about where to place each integer.

Step 2: Binary Search Setup We set up binary search boundaries:

• Left boundary: left = 0 (minimum possible difference)
• Right boundary: right = start[-1] + d - start[0] (maximum possible difference between the rightmost point of the last interval and the leftmost point of the first interval)

Step 3: Binary Search Implementation We use the standard boundary-finding template to find the last true index - the largest minimum difference that satisfies our constraints:

left, right = 0, max_range
last_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):  # can we achieve minimum distance of mid?
        last_true_index = mid
        left = mid + 1  # mid works, try larger values
    else:
        right = mid - 1  # mid doesn't work, try smaller values

return last_true_index

Note: For "maximize minimum" problems, we're looking for the last position where the condition is true (the array looks like true, true, true, false, false...). When feasible, we record the answer and search right for a larger value.

Step 4: Validation Function feasible(min_distance) The core logic lies in the feasible function that determines if a minimum difference min_distance is achievable:

def feasible(min_distance: int) -> bool:
    last = -inf  # Track the position of the last placed integer
    for st in start:
        if last + min_distance > st + d:
            return False  # Cannot place an integer in this interval
        last = max(st, last + min_distance)  # Place at earliest valid position
    return True

The validation works as follows:

• Initialize last = -inf to represent that no integer has been placed yet
• For each interval [st, st + d]:
  • Check if we can place an integer at least min_distance from the previous one
  • If last + min_distance > st + d, it means even placing at the rightmost position st + d wouldn't maintain the minimum distance, so return False
  • Otherwise, place the integer at max(st, last + min_distance):
    • If last + min_distance ≤ st, place at the interval's start st
    • If last + min_distance > st, place at last + min_distance (the earliest position maintaining the minimum distance)
• If all intervals can accommodate an integer with the required spacing, return True

Time Complexity: O(n log(max_range)) where n is the number of intervals and max_range is the difference between the maximum and minimum possible positions.

Space Complexity: O(1) if we don't count the sorting space, otherwise O(n) for sorting.

Example Walkthrough

Let's walk through a concrete example with start = [2, 6, 13, 13] and d = 5.

Step 1: Understanding the intervals

• Interval 0: [2, 7] (from 2 to 2+5)
• Interval 1: [6, 11] (from 6 to 6+5)
• Interval 2: [13, 18] (from 13 to 13+5)
• Interval 3: [13, 18] (from 13 to 13+5)

Step 2: Sort the intervals After sorting by start positions: [2, 6, 13, 13] (already sorted)

Step 3: Set up binary search

• Left boundary: l = 0
• Right boundary: r = 13 + 5 - 2 = 16

Step 4: Binary search iterations

Iteration 1:

• mid = (0 + 16 + 1) // 2 = 8
• Check if minimum difference of 8 is achievable:
  • Place in interval [2, 7]: last = -∞, so place at max(2, -∞ + 8) = 2. Update last = 2
  • Place in interval [6, 11]: need at least 2 + 8 = 10, place at max(6, 10) = 10. Update last = 10
  • Place in interval [13, 18]: need at least 10 + 8 = 18, place at max(13, 18) = 18. Update last = 18
  • Place in interval [13, 18]: need at least 18 + 8 = 26, but interval ends at 18. Cannot place!
• Result: check(8) = False, so r = 7

Iteration 2:

• mid = (0 + 7 + 1) // 2 = 4
• Check if minimum difference of 4 is achievable:
  • Place in interval [2, 7]: place at max(2, -∞ + 4) = 2. Update last = 2
  • Place in interval [6, 11]: place at max(6, 2 + 4) = 6. Update last = 6
  • Place in interval [13, 18]: place at max(13, 6 + 4) = 13. Update last = 13
  • Place in interval [13, 18]: place at max(13, 13 + 4) = 17. Update last = 17
• Result: check(4) = True, so l = 4

Iteration 3:

• mid = (4 + 7 + 1) // 2 = 6
• Check if minimum difference of 6 is achievable:
  • Place in interval [2, 7]: place at 2. Update last = 2
  • Place in interval [6, 11]: place at max(6, 2 + 6) = 8. Update last = 8
  • Place in interval [13, 18]: place at max(13, 8 + 6) = 14. Update last = 14
  • Place in interval [13, 18]: need at least 14 + 6 = 20, but interval ends at 18. Cannot place!
• Result: check(6) = False, so r = 5

Iteration 4:

• mid = (4 + 5 + 1) // 2 = 5
• Check if minimum difference of 5 is achievable:
  • Place in interval [2, 7]: place at 2. Update last = 2
  • Place in interval [6, 11]: place at max(6, 2 + 5) = 7. Update last = 7
  • Place in interval [13, 18]: place at max(13, 7 + 5) = 13. Update last = 13
  • Place in interval [13, 18]: place at max(13, 13 + 5) = 18. Update last = 18
• Result: check(5) = True, so l = 5

Iteration 5:

• Now l = 5 and r = 5, so the loop terminates.

Final Answer: 5

The optimal selection is: [2, 7, 13, 18]

• Difference between 2 and 7: |2 - 7| = 5
• Difference between 7 and 13: |7 - 13| = 6
• Difference between 13 and 18: |13 - 18| = 5
• Difference between 2 and 13: |2 - 13| = 11
• Difference between 2 and 18: |2 - 18| = 16
• Difference between 7 and 18: |7 - 18| = 11

The minimum difference among all pairs is 5, which matches our answer.

Time and Space Complexity

Time Complexity: O(n × log M), where n is the length of the start array and M represents the maximum possible score range, which is start[-1] + d - start[0] after sorting.

• Sorting the start array takes O(n log n) time.
• The binary search performs O(log M) iterations, where M = start[-1] + d - start[0] is the search range.
• In each binary search iteration, the check function is called, which iterates through all n elements in the sorted start array, taking O(n) time.
• Therefore, the binary search contributes O(n × log M) to the time complexity.
• Since O(n × log M) typically dominates O(n log n) for large values of M, the overall time complexity is O(n × log M).

Space Complexity: O(1) if we don't count the space used for sorting (in-place sorting), or O(log n) if we consider the recursive stack space used by the sorting algorithm.

• The algorithm uses only a constant amount of extra space for variables like l, r, mid, and last in the check function.
• The sorting operation may use O(log n) space for the recursive call stack in typical implementations like Timsort (Python's default).
• Following the reference answer's convention, the space complexity is considered O(1).

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

Pitfall: Using a non-standard binary search template that doesn't properly track the answer. Many developers use while left < right with left = mid and right = mid - 1, which requires ceiling division to avoid infinite loops.

Incorrect (prone to infinite loop):

while left < right:
    mid = (left + right) // 2  # WRONG! Should use ceiling division
    if feasible(mid):
        left = mid
    else:
        right = mid - 1

Correct (using template with lastTrueIndex):

left, right = 0, max_range
last_true_index = 0  # Track the answer

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        last_true_index = mid
        left = mid + 1  # Try for larger value
    else:
        right = mid - 1

return last_true_index

2. Not Sorting the Intervals Before Processing

Pitfall: Forgetting to sort the start array before applying the greedy algorithm in the validation function.

Without sorting, the greedy approach fails because we can't guarantee that placing integers from left to right will give us an optimal solution. Consider:

• start = [10, 1], d = 2
• Without sorting, we'd try to place at interval [10, 12] first, then [1, 3]
• This makes validation unnecessarily complex and potentially incorrect

Solution: Always sort the intervals first: sorted_start = sorted(start)

3. Confusing First-True vs Last-True Binary Search

Pitfall: For "maximize minimum" problems, the feasibility pattern is true, true, true, false, false... We need to find the last true position (maximum feasible value), not the first true.

Key difference:

• First true (minimize): When feasible(mid), record answer and search left (right = mid - 1)
• Last true (maximize): When feasible(mid), record answer and search right (left = mid + 1)

This problem requires the "last true" pattern since we want to maximize the minimum distance.

4. Incorrect Greedy Placement in Validation

Pitfall: Placing integers at arbitrary positions within intervals instead of following the greedy principle of placing at the earliest valid position.

For example, placing at last_position + min_distance without checking if it exceeds the interval start:

# Wrong approach
last_position = last_position + min_distance

This fails when the calculated position is before the interval start.

Solution: Use last_position = max(start_pos, last_position + min_distance) to ensure we place at the earliest valid position within the interval.

5. Integer Overflow in Large Inputs

Pitfall: Not considering potential overflow when calculating positions or differences, especially with large values of start[i] and d.

Solution: In Python, this is generally not an issue due to arbitrary precision integers, but in other languages, use appropriate data types (e.g., long long in C++) or check for overflow conditions.