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 Kotlin (1.3.71) (judge verdict: Accepted)

import java.util.*
import javax.print.attribute.IntegerSyntax
import kotlin.math.*

private fun readLn() = readLine()!! // string line
private fun readInt() = readLn().toInt() // single int
private fun readLong() = readLn().toLong() // single long
private fun readStrings() = readLn().split(" ") // list of strings
private fun readInts() = readStrings().map(String::toInt) // list of ints
private fun readLongs() = readStrings().map(String::toLong) // list of longs

fun main() {
    val n = readInt()
    val c0 = readLn().toCharArray()
    val c = c0.map {
        if (it == 'W') 0 else 1
    }.toIntArray()

//    val rCount: Int = c.sum()
//    val wCount: Int = c.size - rCount

    var r = 0

    var leftWhite = c.indexOfFirst { it == 0 }
    var rightRed = c.indexOfLast { it == 1 }

    while (true) {
        if (leftWhite == -1 || rightRed == -1 || leftWhite == n || leftWhite > rightRed) {
            println(r)
            break
        }
        c[leftWhite] = 1
        c[rightRed] = 0
        r++

        while (leftWhite < n && c[leftWhite] != 0) {
            leftWhite++
        }
        while (rightRed >= 0 && c[rightRed] != 1) {
            rightRed--
        }
    }
}