Mange the positions of left and right hands while inspecting the key to press for \(1,2,\dots,N\)-th press, and add the fatigue level of each move required.

In the sample code below, we prepare an array \(\text{pos}\) of length two, so that \(pos[0]\) and \(pos[1]\) manage the current position of left and right hands, respectively. Also, we set the initial positions of the hands to \(-1\), and do not add the fatigue level to the answer when moving a hand from \(-1\) to another key, so as to realize the action of “placing a hand at an appropriate position (the position of the first key to be pressed by the hand)” at the beginning of the performance.”

Sample code (C++):

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> pos = {-1, -1};  // 0: left, 1: right
    int ans = 0;
    for (int i = 0; i < n; i++) {
        int a;
        char s;
        cin >> a >> s;
        int hand = (s == 'R');
        if (pos[hand] != -1) ans += abs(pos[hand] - a);
        pos[hand] = a;
    }
    cout << ans << endl;
}

Sample code (Python) :

n = int(input())
pos = [-1, -1]   # 0: left, 1: right
ans = 0
for i in range(n):
    a, s = input().split()
    a = int(a)
    hand = (0 if s == 'L' else 1)
    if pos[hand] != -1:
        ans += abs(pos[hand] - a)
    pos[hand] = a
print(ans)