Overview

For \(N\) customers, compare the actual order \(T_i\) with what was communicated to the kitchen \(S_i\), and count the number of cases where they differ.

Analysis

This problem is very straightforward — you simply need to determine whether the two strings match for each customer.

• Key observation: Each customer is independent, and you just need to compare the pair of \(T_i\) and \(S_i\) for each one. There is no need to consider relationships with other customers’ orders.
• Is a naive approach sufficient?: Since we only perform one string comparison for each of the \(N\) customers, a naive one-by-one comparison is fast enough when \(N \leq 10^5\) and the maximum string length is \(20\). No special algorithm is needed.

Concrete Example

For instance, suppose \(N = 3\) with the following input:

Only \(i = 2\) is a mismatch, so the answer is 1.

Algorithm

1. Read \(N\).
2. Initialize a counter count to \(0\).
3. Loop \(N\) times, reading \(T_i\) and \(S_i\) on each line.
4. If \(T_i \neq S_i\), increment count by \(1\).
5. Output count at the end.

Complexity

• Time complexity: \(O(N \times L)\) (where \(L\) is the maximum string length; in this problem \(L \leq 20\))
  • String comparison takes \(O(L)\) for each customer, repeated \(N\) times. Since \(L\) can be treated as a constant, this is effectively \(O(N)\).
• Space complexity: \(O(L)\)
  • We only need to store \(T_i\) and \(S_i\) for each line; there is no need to store all the data at once.

Implementation Notes

• In Python, you can use input().split() to split a line by spaces and receive two strings at once.

• String comparison can be easily done with the != operator. Python’s string comparison is based on content (the actual characters in the string), so you can safely use it as-is.

  Source Code

N = int(input())
count = 0
for _ in range(N):
    T, S = input().split()
    if T != S:
        count += 1
print(count)

This editorial was generated by claude4.6opus-thinking.