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: Accepted)

#!/usr/bin/env python3
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(10000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]


def inside(y, x, H, W):
    return 0 <= y < H and 0 <= x < W


def ceil(a, b):
    return (a + b - 1) // b


def sum_of_arithmetic_progression(s, d, n):
    return n * (2 * s + (n - 1) * d) // 2


def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)


def lcm(a, b):
    g = gcd(a, b)
    return a / g * b


def solve():
    N = int(input())
    C = input()

    num_r = C.count("R")
    num_w = C.count("W")
    ans = min(num_r, num_w)

    left_r = 0
    left_w = 0
    for i in range(N):
        left_r += C[i] == 'R'
        left_w += C[i] == 'W'

        right_r = num_r - left_r
        right_w = num_w - left_w
        ans = min(ans, min(left_w, right_r) + abs(left_w - right_r))

    print(ans)


def main():
    solve()


if __name__ == '__main__':
    main()