# Rock Paper Scissors ROPASCI Solution Codechef

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## Rock Paper Scissors ROPASCI Solution

There are NN players standing in a line, indexed 11 to NN from left to right. They all play a game of Rock, Paper, Scissors. Each player has already decided which move they want to play. You are given this information as a string SS of length NN, i.e,

• SiSi is equal to RR if player ii will play Rock.
• SiSi is equal to PP if player ii will play Paper.
• SiSi is equal to SS if player ii will play Scissors.

Let W(i,j)W(i,j) denote the move played by the winner if players i,i+1,…,ji,i+1,…,j compete in order from left to right. That is,

• First, players ii and i+1i+1 play a game
• The winner of this game plays against player i+2i+2
• The winner of the second game plays against player i+3i+3

⋮⋮

• The winner of the first j−i−1j−i−1 games plays against player jj, and the move played by the winner of this game is declared to be W(i,j)W(i,j).

If i=ji=j, then player ii is considered to be the winner and W(i,i)=SiW(i,i)=Si.

Your task is to find the value of W(i,N)W(i,N) for all ii from 11 to NN.

Note : If a person with index ii and index jj (i<ji<j) play against each other, then:

• If Si≠SjSi≠Sj, the winner is decided by classical rules, i.e, rock beats scissors, scissors beats paper, and paper beats rock.
• If Si=SjSi=Sj, the player with lower index (in this case, ii) wins.

### Input Format

• The first line of input contains a single integer TT, denoting the number of test cases. The description of TT test cases follows.
• The first line of each test case contains a single integer NN, the number of players.
• The second line of each test case contains the string SS of length NN, denoting the moves chosen by the players.

### Output Format

For each test case, print a single line containing a string of length NN, whose ii-th character is W(i,N)W(i,N).

### Constraints

• 1≤T≤1051≤T≤105
• 1≤N≤5⋅1051≤N≤5⋅105
• SiSi is either RR, PP or SS
• Sum of NN over all test cases doesn’t exceed 5⋅1055⋅105

• 1≤T≤10001≤T≤1000
• 1≤N≤50001≤N≤5000
• Sum of NN over all test cases doesn’t exceed 50005000

• Original constraints

```2
1
S
4
SSPR
```

```S
RRPR
```

### Explanation

Test Case 1. W(1,1)=SW(1,1)=S as there is only one player.

Test Case 2.

For W(1,4)W(1,4) the game is played as follows :

• Player 11 and 22 compete, player 11 wins.
• Player 11 and 33 compete, player 11 wins.
• Player 11 and 44 compete, player 44 wins.

Hence, we print W(1,4)=S4=RW(1,4)=S4=R

For W(3,4)W(3,4) the game is played as follows :

• Player 33 and 44 compete, player 33 wins.

Hence, we print W(3,4)=S3=P

## SOLUTION

Program Python: Rock Paper Scissors ROPASCI Solution in Python

```# cook your dish here
def match(a,b):
if a==b:
return a

if a == 'S' and b == 'R' or a == 'R' and b == 'S':
return 'R'
elif a == 'S' and b == 'P' or a == 'P' and b == 'S':
return 'S'
elif a == 'P' and b == 'R' or a == 'R' and b == 'P':
return 'P'
else:
return None

for _ in range(int(input())):
n = int(input())
s = input()

dp_r = [None]*(n+1)
dp_p = [None]*(n+1)
dp_s = [None]*(n+1)

ans = [None]*(n+1)

#base case
ans[n] = s[n-1]
dp_r[n] = match('R', s[n-1])
dp_p[n] = match('P', s[n-1])
dp_s[n] = match('S', s[n-1])

for i in range(n-1,1-1,-1):
#find dp_arrays and ans

#dp_r
r_res = match('R', s[i-1])

if r_res == 'R':
dp_r[i] = dp_r[i+1]
elif r_res == 'P':
dp_r[i] = dp_p[i+1]
elif r_res == 'S':
dp_r[i] = dp_s[i+1]

#dp_p
p_res = match('P', s[i-1])

if p_res == 'R':
dp_p[i] = dp_r[i+1]
elif p_res == 'P':
dp_p[i] = dp_p[i+1]
elif p_res == 'S':
dp_p[i] = dp_s[i+1]

#dp_s
s_res = match('S', s[i-1])

if s_res == 'R':
dp_s[i] = dp_r[i+1]
elif s_res == 'P':
dp_s[i] = dp_p[i+1]
elif s_res == 'S':
dp_s[i] = dp_s[i+1]

#ans
if s[i-1] == 'R':
ans[i] = dp_r[i+1]
elif s[i-1] == 'P':
ans[i] = dp_p[i+1]
elif s[i-1] == 'S':
ans[i] = dp_s[i+1]

print(''.join(ans[1:]))```

Program C++: Rock Paper Scissors ROPASCI Solution in C++

```#include <iostream>
#include<cassert>
#include<vector>
#include<string>
using namespace std;

char match(char a, char b)
{
if(a==b) return a;

else if(a=='R' && b=='P' || a=='P' && b=='R')
{
return 'P';
}
else if(a=='R' && b=='S' || a=='S' && b=='R')
{
return 'R';

}
else if(a=='P' && b=='S' || a=='S' && b=='P')
{
return 'S';
}
else
{
assert (false);
}

}
int main() {
int t;
cin>> t;

while(t--)
{
int n;
cin>>n;
string s;
cin>>s;

vector<char> dp_r(n+1);
vector<char> dp_p(n+1);
vector<char> dp_s(n+1);

vector<char> ans(n+1);

ans[n]=s[n-1];
dp_r[n]=match('R', s[n-1]);
dp_p[n]=match('P', s[n-1]);
dp_s[n]=match('S', s[n-1]);

for(int i =n-1;i>=1;i--)
{
char r_res = match('R', s[i-1]);
if(r_res=='R')
{
dp_r[i]= dp_r[i+1];
}
else if(r_res=='P')
{
dp_r[i]= dp_p[i+1];
}
else if(r_res=='S')
{
dp_r[i]= dp_s[i+1];
}

char p_res = match('P', s[i-1]);
if(p_res=='R')
{
dp_p[i]= dp_r[i+1];
}
else if(p_res=='P')
{
dp_p[i]= dp_p[i+1];
}
else if(p_res=='S')
{
dp_p[i]= dp_s[i+1];
}

char s_res = match('S', s[i-1]);
if(s_res=='R')
{
dp_s[i]= dp_r[i+1];
}
else if(s_res=='P')
{
dp_s[i]= dp_p[i+1];
}
else if(s_res=='S')
{
dp_s[i]= dp_s[i+1];
}

if(s[i-1] == 'R')
{
ans[i]=dp_r[i+1];
}

else if(s[i-1] == 'P')
{
ans[i]=dp_p[i+1];
}
else if(s[i-1] == 'S')
{
ans[i]=dp_s[i+1];
}
}

for(int i =1; i<=n;i++)
{
cout<<ans[i];
}
cout<<endl ;
}

}
```