Overview

This problem asks you to find the student ID number of the student with the highest score among those whose scores fall within the passing range \([L, R]\) (if there is a tie, choose the smallest student ID number). If there are no passing students, output -1.

Analysis

• We want to find the “maximum score among passing students” and “the smallest student ID number with that score.”
• Each student’s score is in the range \(0 \sim 100\), which is small, but \(N\) can be up to \(2\times 10^5\), so an \(O(N)\) approach that examines each student once is sufficiently fast.
• For the “find the person with the maximum score” problem, the standard technique is to scan through the data while updating the current best (maximum).
• For tie-breaking (when multiple people share the maximum score, choose the smallest student ID number): if we don’t update on ties, the first person found (= the one with the smallest student ID number) remains.
  • Example: If the scores within the passing range are [80(#1), 90(#2), 90(#3)],
    the maximum becomes 90 at student #2, and since student #3 has the same score, if we don’t update, the answer stays as #2 (which is correct as the smallest ID number).
• To handle the case where there are no passing students at all, it is simple to use an initial value of -1 to represent “not yet found.”

Algorithm

1. Initialize best_score = -1, best_idx = -1 (state where no passing student has been found yet).
2. Iterate through students in order \(i = 1 \ldots N\).
3. Only consider a student as a candidate when their score score satisfies the passing range \(L \le score \le R\).
4. If that score is greater than best_score, update best_score = score, best_idx = i.
   • When score == best_score, we do not update, so the smaller student ID number is preserved.
5. Finally, output best_idx (it remains -1 if no passing student was found).

Complexity

• Time complexity: \(O(N)\) (we only check each student once)
• Space complexity: \(O(1)\) (only a constant number of variables for the best value and index)

Implementation Notes

• Since student ID numbers are \(1\)-indexed, using something like enumerate(P, 1) to count from \(1\) helps prevent mistakes.

• To prioritize the smallest ID number in case of ties, the update condition should be score > best_score only (do not use >=).

• To handle the case where there are no passing students, initialize best_idx to -1 so it can be output directly.

  Source Code

import sys

def main():
    data = list(map(int, sys.stdin.buffer.read().split()))
    if not data:
        return
    N, L, R = data[0], data[1], data[2]
    P = data[3:3+N]

    best_score = -1
    best_idx = -1

    for i, score in enumerate(P, 1):
        if L <= score <= R:
            if score > best_score:
                best_score = score
                best_idx = i

    print(best_idx)

if __name__ == "__main__":
    main()

This editorial was generated by gpt-5.2-high.