Overview

This is a problem about two people moving on a number line, where we track the difference in their coordinates at each time step and count the number of times the difference becomes \(0\) (i.e., they are at the same coordinate).

Analysis

While it is possible to manage each person’s coordinate individually, the key insight is that we only need to focus on the difference (relative position) of their coordinates.

The difference in coordinates at time \(0\) is \(D = X - Y\). At each time step, Takahashi and Aoki each move, so we consider how the difference \(D\) changes.

• Takahashi moves R → \(D\) increases by \(+1\) (Takahashi moves away in the positive direction)
• Takahashi moves L → \(D\) decreases by \(-1\)
• Aoki moves R → \(D\) decreases by \(-1\) (when Aoki moves in the positive direction, the difference shrinks)
• Aoki moves L → \(D\) increases by \(+1\)
• S (stop) → no change

In other words, Takahashi’s movement affects the difference directly, while Aoki’s movement affects it in the opposite direction.

By updating this difference at each time step and counting the number of times \(D = 0\), we obtain the answer.

Concrete Example

For \(X = 3\), \(Y = 5\), \(A = \) RRL, \(B = \) LLS:

Time	Takahashi’s Move	Aoki’s Move	Change in \(D\)	\(D\)	Meet?
0	—	—	—	\(3 - 5 = -2\)	—
1	R (+1)	L (+1)	\(+2\)	\(0\)	✅
2	R (+1)	L (+1)	\(+2\)	\(2\)	❌
3	L (-1)	S (0)	\(-1\)	\(1\)	❌

The answer is 1 time.

Comparison with the Naive Approach

Managing the two coordinates separately also solves the problem in \(O(T)\), but tracking only the difference requires just one variable and is simpler. In this problem, \(T \leq 10^6\), so a linear scan in \(O(T)\) is more than sufficient.

Algorithm

1. Initialize the difference \(D = X - Y\).
2. For \(i = 0, 1, \ldots, T-1\):
   • If \(A[i]\) is R, add \(+1\) to \(D\); if L, add \(-1\).
   • If \(B[i]\) is R, add \(-1\) to \(D\); if L, add \(+1\) (opposite direction from Takahashi).
   • If \(D = 0\), increment the count by \(+1\).
3. Output the count.

Complexity

• Time complexity: \(O(T)\) — just a single scan through the strings
• Space complexity: \(O(T)\) — for storing the input strings (the difference tracking itself is \(O(1)\))

Implementation Notes

• Fast input: Since \(T\) can be as large as \(10^6\), in Python we use sys.stdin.buffer.read() to read all input at once, avoiding input bottlenecks.

• Sign direction of the difference: Be careful that Aoki’s movement affects the difference in the opposite direction. For example, when Aoki moves R (positive direction), the difference \(X - Y\) decreases. Getting this wrong will result in WA.

• Integer overflow: The initial values of \(X, Y\) can be up to \(\pm 10^9\), but since Python uses arbitrary-precision integers, there is no risk of overflow. In languages like C++, make sure to use long long.

  Source Code

import sys

def main():
    input_data = sys.stdin.buffer.read().decode()
    tokens = input_data.split()
    T = int(tokens[0])
    X = int(tokens[1])
    Y = int(tokens[2])
    A = tokens[3]
    B = tokens[4]
    
    diff = X - Y
    count = 0
    for i in range(T):
        a = A[i]
        b = B[i]
        if a == 'L':
            diff -= 1
        elif a == 'R':
            diff += 1
        if b == 'L':
            diff += 1
        elif b == 'R':
            diff -= 1
        if diff == 0:
            count += 1
    
    print(count)

main()

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