Overview

This is a problem of finding the member number with the highest skill value among the members that Takahashi can beat (members whose skill value is at most \(L\)).

Analysis

Key Observations for Solving the Problem

In this problem, we need to find a member satisfying the following two conditions: 1. Winnable condition: \(D_i \leq L\) (skill value is at most Takahashi’s) 2. Optimal opponent: Among those satisfying condition 1, the member with the maximum \(D_i\) 3. Tiebreak: If \(D_i\) values are the same, choose the member with the smallest member number

Thinking Through a Concrete Example

For example, with \(N = 5\), \(L = 100\), and skill values \(D = [80, 120, 100, 100, 50]\): - Member 1: \(D_1 = 80 \leq 100\) → Can win ✓ - Member 2: \(D_2 = 120 > 100\) → Cannot win ✗ - Member 3: \(D_3 = 100 \leq 100\) → Can win ✓ - Member 4: \(D_4 = 100 \leq 100\) → Can win ✓ - Member 5: \(D_5 = 50 \leq 100\) → Can win ✓

The winnable members are {1, 3, 4, 5}, and among them, the ones with the maximum skill value are Member 3 and Member 4 (both 100). Since we choose the one with the smaller number, the answer is Member 3.

Approach Consideration

• Naive approach: Iterate through all members in order and find the member with the maximum value satisfying the condition
• This approach is already efficient enough (\(O(N)\)), so no special techniques are needed

Algorithm

1. Initialize variables best_index (the answer member number) and best_value (that member’s skill value)

   • best_index = -1 (the answer when no winnable member exists)
   • best_value = -1 (no candidate found yet)

2. Examine members in order starting from number 1 (index 0 in the array)

3. For each member \(i\):

   • If \(D_i \leq L\) and \(D_i > \text{best\_value}\), record this member as the new candidate
   • By using the strict inequality (>), when skill values are equal, the member found first (with the smaller number) is prioritized

4. Output the final best_index

Complexity

• Time complexity: \(O(N)\)
  • Since we check each member exactly once
• Space complexity: \(O(N)\)
  • For the array storing the input

Implementation Notes

1. Conversion to 1-indexed: In the problem, member numbers start from 1, but array indices start from 0. Therefore, when outputting the answer, we need to use i + 1.

2. Tiebreak handling: When multiple members have the same skill value, we choose the one with the smaller number. This is naturally achieved by using the strict inequality (>). The first member found is recorded, and subsequent members with the same value do not trigger an update.

3. When no winnable member exists: By initializing best_index to -1, if no member satisfies the condition, -1 is automatically output.

   Source Code

N, L = map(int, input().split())
D = list(map(int, input().split()))

best_index = -1
best_value = -1

for i in range(N):
    if D[i] <= L:
        if D[i] > best_value:
            best_value = D[i]
            best_index = i + 1  # 1-indexed

print(best_index)

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