Imagine the snail starts on one grid square. That starting square is already part of its trail, because the problem says the snail makes the squares it touches slimy. In the example “move east 3 squares”, there are 4 slimy squares total, which includes the starting square.

Understanding the Problem

Every time the snail moves one square, we check whether the new square was already slimy.

• If it steps onto a square it has never touched before, nothing is added to the answer.
• If it steps onto a square that is already slimy, the snail has returned to an old square, so we add 1 to the answer.

So the problem is really asking: as the snail walks step by step, how many times does it revisit an old square?

Important Detail Students Often Miss

A movement like S2 does not mean one big jump. It means move 1 square south, then move 1 more square south.

You must process movements one square at a time, because the snail could enter an already slimy square in the middle of a longer movement. The movement format is a direction letter followed by a positive integer up to 20.

Core DSA Concepts

• Simulation: follow the snail exactly as described, updating its position and trail step by step.
• Hash set / visited set: remember which squares are already slimy using coordinate pairs such as (x, y).
• Coordinates on a grid: use x for east/west and y for north/south.
• Unit-step processing: break each command into single-square moves before checking whether the square was already visited.

Coordinate Updates

• North: y += 1
• South: y -= 1
• East: x += 1
• West: x -= 1

Big Idea

Start from the original square and add it to a visited set. For each movement command, repeat the movement one square at a time. After each single step, check whether the new coordinate is already in the set. If it is, increase the revisit count. Then add the coordinate to the set.

Python Solution

# CCC 2026 J4 - Snail Path
# Read the number of movements
M = int(input())

# Current position of the snail
x = 0
y = 0

# The starting square is already touched, so it is slimy
visited = set()
visited.add((x, y))

# Count how many times the snail enters a square
# that was already slimy
answer = 0

# Direction changes for one step
dx = {
    "E": 1,
    "W": -1,
    "N": 0,
    "S": 0
}

dy = {
    "E": 0,
    "W": 0,
    "N": 1,
    "S": -1
}

# Process each movement
for _ in range(M):
    move = input().strip()

    # First character is the direction
    direction = move[0]

    # The rest of the string is the distance
    distance = int(move[1:])

    # Move one square at a time
    for _ in range(distance):
        x += dx[direction]
        y += dy[direction]

        if (x, y) in visited:
            answer += 1

        visited.add((x, y))

print(answer)

Common Mistakes

• Treating S20 as one jump: that is wrong. The snail passes through 20 squares one by one, and any of those may already be slimy.
• Forgetting the starting square: the starting square must already be in visited, otherwise returning to start will not be counted correctly.
• Counting distinct repeated squares instead of total re-entries: the output is how many times the snail enters a slimy square, not how many slimy squares are revisited. A single square can be entered many times and should be counted many times. The statement explicitly notes that one slimy square can be entered multiple times.
• Using a list instead of a set: a set is much faster for checking whether a coordinate has been seen before.