Problem: An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.  You can do the following two kinds of operations any number of times in any order:    * Choose two stones (not necessarily adjacent) and swap them.   * Choose one stone and change its color (from red to white and vice versa).  According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?

Submission attempts (in order):
import itertools
from collections import deque,defaultdict,Counter
from itertools import accumulate
import bisect
from heapq import heappop,heappush,heapify
import math
from copy import deepcopy
import queue
import numpy as np

Mod = 1000000007
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1] 
for i in range(2, 10**5 + 1):
    fact.append((fact[-1] * i) % Mod)
    inv.append((-inv[Mod % i] * (Mod // i)) % Mod)
    factinv.append((factinv[-1] * inv[-1]) % Mod)
    
def cmb(n, r, p):
    if (r < 0) or (n < r):
        return 0
    r = min(r, n - r)
    return fact[n] * factinv[r] * factinv[n - r] % p
    
def sieve_of_eratosthenes(n):
    if not isinstance(n,int):
        raise TypeError("n is not int")
    if n<2:
        raise ValueError("n is not effective")
    prime = [1]*(n+1)
    for i in range(2,int(math.sqrt(n))+1):
        if prime[i] == 1:
            for j in range(2*i,n+1):
                if j%i == 0:
                    prime[j] = 0
    res = []
    for i in range(2,n+1):
        if prime[i] == 1:
            res.append(i)
    return res
 
 
class UnionFind:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]
    
    def findroot(self,x):
        if x == self.parent[x]:
            return x
        else:
            y = self.parent[x]
            y = self.findroot(self.parent[x])
            return y
    
    def union(self,x,y):
        px = self.findroot(x)
        py = self.findroot(y)
        if px < py:
            self.parent[y] = px
        else:
            self.parent[px] = py
 
    def same_group_or_no(self,x,y):
        return self.findroot(x) == self.findroot(y)
def pow_k(x, n):
 
    if n == 0:
        return 1
 
    K = 1
    while n > 1:
        if n % 2 != 0:
            K *= x
        x *= x
        n //= 2
 
    return K * x
def main():  #startline-------------------------------------------
    n = int(input())
    s = input()
    print(min(s[:n//2].count("W"),s[n//2+1:].count("R")))
        
                
if __name__ == "__main__":
    main() 

import itertools
from collections import deque,defaultdict,Counter
from itertools import accumulate
import bisect
from heapq import heappop,heappush,heapify
import math
from copy import deepcopy
import queue
import numpy as np

Mod = 1000000007
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1] 
for i in range(2, 10**5 + 1):
    fact.append((fact[-1] * i) % Mod)
    inv.append((-inv[Mod % i] * (Mod // i)) % Mod)
    factinv.append((factinv[-1] * inv[-1]) % Mod)
    
def cmb(n, r, p):
    if (r < 0) or (n < r):
        return 0
    r = min(r, n - r)
    return fact[n] * factinv[r] * factinv[n - r] % p
    
def sieve_of_eratosthenes(n):
    if not isinstance(n,int):
        raise TypeError("n is not int")
    if n<2:
        raise ValueError("n is not effective")
    prime = [1]*(n+1)
    for i in range(2,int(math.sqrt(n))+1):
        if prime[i] == 1:
            for j in range(2*i,n+1):
                if j%i == 0:
                    prime[j] = 0
    res = []
    for i in range(2,n+1):
        if prime[i] == 1:
            res.append(i)
    return res
 
 
class UnionFind:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]
    
    def findroot(self,x):
        if x == self.parent[x]:
            return x
        else:
            y = self.parent[x]
            y = self.findroot(self.parent[x])
            return y
    
    def union(self,x,y):
        px = self.findroot(x)
        py = self.findroot(y)
        if px < py:
            self.parent[y] = px
        else:
            self.parent[px] = py
 
    def same_group_or_no(self,x,y):
        return self.findroot(x) == self.findroot(y)
def pow_k(x, n):
 
    if n == 0:
        return 1
 
    K = 1
    while n > 1:
        if n % 2 != 0:
            K *= x
        x *= x
        n //= 2
 
    return K * x
def main():  #startline-------------------------------------------
    n = int(input())
    s = input()
    print(s.count("R")-s[:s.count("R")].count("R"))
        
                
if __name__ == "__main__":
    main() 

Verdicts: Wrong Answer

Accepted