Problem:
Score : 400 points

Problem Statement

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?

Constraints

2 \leq N \leq 200000

c_i is R or W.

Input

Input is given from Standard Input in the following format:

N
c_{1}c_{2}...c_{N}

Output

Print an integer representing the minimum number of operations needed.

Sample Input 1

4
WWRR

Sample Output 1

2

For example, the two operations below will achieve the objective.

Swap the 1-st and 3-rd stones from the left, resulting in RWWR.

Change the color of the 4-th stone from the left, resulting in RWWW.

Sample Input 2

2
RR

Sample Output 2

0

It can be the case that no operation is needed.

Sample Input 3

8
WRWWRWRR

Sample Output 3

3

Solution in Python (3.8.2) (judge verdict: Wrong Answer)

# D Alter Altar

n = int(input())
C = input()

# Rの数
num_R = C.count("R")

# 元々全部R or Wの場合
if len(C) == num_R or num_R == 0:
  print(0)
  exit()

# 一色に染めるときの回数で少ない方
num_all = min(len(C) - num_R, num_R)

# 動かす必要のないWの数
_C = C[::-1]
safe_W = 0
for _c in _C:
  if _c == "W":
    safe_W += 1
  else:
    break

# 右端のRの数
right_R = 0
sec_W = 0
frag = ""
for _c in _C:
  if _C == "R":
    right_R += 1
    frag = "right"
  elif frag == "right" and _C == "W":
    sec_W += 1
    frag = "sec"
  elif frag == "sec" and _C == "R":
    break
  else:
    break
  
#左端にRを固める作戦の回数
space = 0
for c in C:
  if c == "W":
    space += 1
  elif c == "R":
    break
if num_R <= space:
  num_left = min(len(C) - num_R - safe_W, num_R)
else:
  num_left = min(len(C) - num_R - safe_W, num_R - 1)
  if space >= right_R:
    num_left = min(num_left, len(C) - num_R - sec_W)

print(min(num_left, num_all))