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: Wrong Answer)

import java.io.IOException
import java.lang.NumberFormatException
import java.lang.StringBuilder
import kotlin.math.max

object BMain{
    fun solve(){
        val n = nextInt()
        val c = next()
        val numw = c.count { it == 'W' }
        val numr = n - numw
        var tnumw = 0
        var tnumr = 0
        for(i in 0 until n){
            if(c[i] == 'R'){
                tnumr++
            }else{
                tnumw++
            }
            if(tnumw > numr - tnumr) break
        }
        println(max(tnumw, numr - tnumr))
    }





    // Scanner based on uwi-san's Java code
    val stream = System.`in`
    val buffer = ByteArray(1024)
    var ptr = 0
    var buflen = 0

    fun hasNextByte(): Boolean{
        if(ptr < buflen){
            return true
        }else{
            ptr = 0
            try{
                buflen = stream.read(buffer)
            }catch (e: IOException){
                e.printStackTrace()
            }
            if(buflen <= 0){
                return false
            }
        }
        return true
    }

    fun readByte() =
        if(hasNextByte()) buffer[ptr++]
        else -1

    fun isPrintableChar(c: Byte) = c in 33..126

    fun hasNext(): Boolean{
        while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++
        return hasNextByte()
    }

    fun next(): String{
        if(!hasNext()) throw NoSuchElementException()
        val sb = StringBuilder()
        var b = readByte()
        while(isPrintableChar(b)){
            sb.appendCodePoint(b.toInt())
            b = readByte()
        }
        return sb.toString()
    }

    fun nextLong(): Long{
        if(!hasNext()) throw NoSuchElementException()
        var n = 0L
        var minus = false
        var b = readByte()
        if(b == '-'.toByte()){
            minus = true
            b = readByte()
        }
        if(b !in '0'.toByte()..'9'.toByte())
            throw NumberFormatException()
        while(true){
            if(b in '0'.toByte()..'9'.toByte()){
                n *= 10
                n += b - '0'.toByte()
            }else if(!isPrintableChar(b)){
                return if(minus) -n else n
            }else{
                throw NumberFormatException()
            }
            b = readByte()
        }
    }

    fun nextInt(): Int{
        val n1 = nextLong()
        if(n1 !in Int.MIN_VALUE..Int.MAX_VALUE)
            throw NumberFormatException()
        return n1.toInt()
    }

    fun NextDouble() = next().toDouble()
}

fun main(){
    BMain.solve()
}