Overview

When \(N\) people are standing in a line, this problem asks us to count the number of pairs where adjacent people have an “awkward handshake” (gloves of the same color) when shaking hands.

Analysis

Understanding the Problem

First, let’s organize the problem statement.

• When participant \(i\) and participant \(i+1\) shake hands, participant \(i\)’s right hand and participant \(i+1\)’s left hand touch each other
• If the touching gloves are the same color, it’s an “awkward handshake”
• We count the number of awkward handshakes for all pairs where \(i = 1, 2, \ldots, N-1\)

Thinking Through a Concrete Example

For example, suppose there are 3 participants wearing gloves as follows:

Participant	Left hand (\(L_i\))	Right hand (\(R_i\))
1	N	S
2	S	N
3	N	S

• Handshake between participant 1 and participant 2: Participant 1’s right hand (S) vs Participant 2’s left hand (S) → Same color → Awkward handshake
• Handshake between participant 2 and participant 3: Participant 2’s right hand (N) vs Participant 3’s left hand (N) → Same color → Awkward handshake

The answer is \(2\).

A Naive Approach Is Sufficient

In this problem, for each of the \(N-1\) adjacent pairs, we only need to compare the glove colors once. No special techniques are needed, and it can be solved in \(O(N)\) with a simple loop.

Algorithm

1. Read the glove information (left hand \(L_i\), right hand \(R_i\)) for \(N\) people and store them in an array
2. Initialize a counter count to \(0\)
3. For \(i = 0, 1, \ldots, N-2\) (0-indexed):
   • Compare participant \(i\)’s right hand color gloves[i][1] with participant \(i+1\)’s left hand color gloves[i+1][0]
   • If they are the same color, increment count by \(1\)
4. Output count

Complexity

• Time complexity: \(O(N)\)
  • \(O(N)\) for reading input
  • \(O(N-1) = O(N)\) for comparing pairs
• Space complexity: \(O(N)\)
  • For storing all participants’ glove information in an array

Implementation Notes

• In Python, you can read the left and right hand colors at once using input().split()
• Pay attention to array indexing: the code uses 0-indexed, so participant \(i\) (1-indexed) corresponds to index \(i-1\)
• Comparison can be done with simple string comparison == ('N' == 'N' is True, 'N' == 'S' is False)

Source Code

n = int(input())
gloves = []
for _ in range(n):
    l, r = input().split()
    gloves.append((l, r))

count = 0
for i in range(n - 1):
    # Compare participant i's right hand with participant i+1's left hand
    if gloves[i][1] == gloves[i + 1][0]:
        count += 1

print(count)

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