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Problem summary

People stand in a row facing left (L) or right (R). Each second, every adjacent RL pair shakes hands and both people turn, changing RL into LR simultaneously. Find the number of seconds until no handshake is possible and the total number of handshakes.

Constraints

String length is at most 100000.

Worked example

Original statement sample.

RLLRLRR

3 4

Solution analysis

1. Think of a right-facing person as a token moving right across left-facing people. Every R before an L forms one inversion, and they eventually shake hands exactly once. While scanning, count previous R characters; each L contributes that count to the total.
2. For the time, maintain seconds for the processed prefix. An L with r previous R characters needs at least r seconds to cross them. If earlier left-facing people are already moving, it may also need one second after their completion. Therefore update seconds=max(seconds+1,r) when r>0.
3. For RLLRLRR, the first two L characters have one R before them, contributing 1+1 handshakes and finishing after seconds 1 and 2. The third L has two R characters before it, contributes 2 more, and finishes at max(3,2)=3. Hence output 3 4.

Why it works

Each RL→LR exchange removes exactly one inversion and creates none, so infinitely many rounds are impossible. A left-facing token must wait for its preceding right-facing tokens and for the token ahead of it; these give the two terms in the completion-time recurrence. Once these constraints are met, every adjacent pair acts immediately, so the bound is attained.

Watch out. Do not update the string in place to simulate a simultaneous second: doing so can move a person multiple times. If there is no earlier R, an L creates no delay. Use 64-bit inversion totals.

An independently written complete program with standard headers.

#include<algorithm>
#include<iostream>
#include<string>
using namespace std;
int main(){
    string row;
    cin>>row;
    long long rightCount=0;
    long long handshakes=0;
    long long seconds=0;
    for(char facing:row){
        if(facing=='R'){
            rightCount++;
        }
        else if(rightCount>0){
            handshakes+=rightCount;
            // Account for both crossings and the token immediately ahead.
            seconds=max(seconds+1,rightCount);
        }
    }
    cout<<seconds<<' '<<handshakes<<'\n';
}