Compute the Eddington number from ride distances

Given a list of positive integers distances, where distances[i] is the distance traveled on day i, compute the cyclist’s Eddington number E:

E is the largest integer such that there are at least E days with distance ≥ E.

Return E.

Examples:

• distances = [1, 3, 3, 5, 6] → E = 3 (three days have distance ≥ 3)
• distances = [10, 8, 5, 4, 3] → E = 4 (four days have distance ≥ 4)
• distances = [1, 1, 1] → E = 1

Constraints: 1 ≤ n ≤ 2e5, 1 ≤ distances[i] ≤ 1e9. Aim for an efficient solution.

Compute the Eddington Number from Ride Distances

Given a list of positive integers distances, where distances[i] is the distance traveled on day i, compute the cyclist's Eddington number E.

E is the largest integer such that there are at least E days with distance ≥ E.

Return E.

Examples:

• distances = [1, 3, 3, 5, 6] → E = 3 (three or more days have distance ≥ 3)
• distances = [10, 8, 5, 4, 3] → E = 4 (four days have distance ≥ 4)
• distances = [1, 1, 1] → E = 1

Aim for an efficient solution: with up to 2×10^5 days and distances as large as 10^9, an O(n) counting approach is expected rather than sorting or per-candidate scanning.

Constraints

• 1 ≤ n ≤ 2×10^5 (n = number of days)
• 1 ≤ distances[i] ≤ 10^9
• The answer E satisfies 0 ≤ E ≤ n

Hints

1. The Eddington number E can never exceed the number of days n, so any distance larger than n can be treated as exactly n for counting purposes.
2. Build a count array bucket[k] = number of days with distance exactly k (capping at n). Then iterate e from n down to 1, accumulating the number of days with distance ≥ e, and return the first e where that running total is at least e.
3. This is the classic 'cap and count' technique, identical in spirit to computing the h-index. It runs in O(n) time and avoids sorting or scanning all distances for each candidate E.