Overview

This is a problem where you perform multiple updates on a list of student scores and count how many students have scores below a threshold at the end.

Analysis

Problem Breakdown

In this problem, we perform the following steps: 1. Receive the initial scores of \(N\) students 2. Apply \(M\) updates (rewriting specific students’ scores) in order 3. Count the number of students with scores below \(K\) at the end

Key Observations

• Note that the same student may be updated multiple times
• For example, if student 1’s score is updated “50→70→30” twice, the final score is 30
• In other words, by applying each update in order, the score is naturally overwritten to the latest value

Why This Problem is Easy

• No complex algorithms are needed; simply simulating (processing as instructed) is sufficient
• Since each update only rewrites the corresponding position in the array, it can be processed in \(O(1)\)

Algorithm

1. Initialization: Store the students’ scores in array Q

2. Update Processing: Process \(M\) updates in order

   • For each update, rewrite student number \(P\)’s score to \(V\)
   • Since student numbers start from 1 (1-indexed), convert to array index (0-indexed) by using P - 1

3. Counting: Traverse the final array Q and count the number of students with scores below \(K\)

Concrete Example

Sample input: \(N=3, M=2, K=50\), initial scores \([40, 60, 30]\) - Update 1: Change student 1’s score to 55 → \([55, 60, 30]\) - Update 2: Change student 3’s score to 70 → \([55, 60, 70]\) - Number of students below 50 points is 0

Complexity

• Time Complexity: \(O(N + M)\)

  • Reading initial scores: \(O(N)\)
  • \(M\) update operations: Each is \(O(1)\), so total is \(O(M)\)
  • Final counting: \(O(N)\)

• Space Complexity: \(O(N)\)

  • Only the array storing students’ scores

Implementation Notes

1. Index Conversion: While student numbers start from 1 (1-indexed) in the problem statement, Python arrays start from 0 (0-indexed), so adjustment like Q[P - 1] is necessary

2. When Number of Updates is 0: \(M = 0\) is possible, but since the for loop simply executes 0 times, no special handling is needed

3. How to Write the Count: In Python, using a generator expression like sum(1 for score in Q if score < K) is concise. len([score for score in Q if score < K]) gives the same result

Source Code

# Receive input
N, M, K = map(int, input().split())
Q = list(map(int, input().split()))

# Process M updates
for _ in range(M):
    P, V = map(int, input().split())
    Q[P - 1] = V  # Adjust since student numbers are 1-indexed

# Count the number of students with scores below K
count = sum(1 for score in Q if score < K)

# Output the result
print(count)

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