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 kotlin.math.abs
import kotlin.math.min

fun main() {
    val n = readInt()
    val colors = readLn()
    val white = colors.count { it == 'W' }
    val whiteSuffix = IntArray(n)
    for (i in n - 1 downTo 0) {
        if (colors[i] == 'W') whiteSuffix[i] = 1
        if (i < n - 1) whiteSuffix[i] += whiteSuffix[i + 1]
    }

    var best = Int.MAX_VALUE
    for (i in n downTo 0) {
        val d = n - i
        val change = abs(white - d)
        val swaps = if (i < n) d - whiteSuffix[i] else d
        best = min(best, change + swaps)
    }

    println(best)
}

// Input Reader
private fun readLn() = readLine()!!

private fun readInt() = readLn().toInt()