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 C (GCC 9.2.1) (judge verdict: Accepted)

#include <float.h>
#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

// 内部定数
#define D_STONE_MAX		200005									// 最大石数

// 内部変数
static FILE *szpFpI;											// 入力
static char sc1Stone[D_STONE_MAX];								// 石
static int siSCnt;												// 石数
static int si1SumLW[D_STONE_MAX];								// 白石合計[左位置]
static int si1SumRR[D_STONE_MAX];								// 赤石合計[右位置]

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpA;
	static int siTNo;
#endif

// １行出力
int
fOutLine(
	char *pcpLine				// <I> １行
)
{
	char lc1Buf[1024];

#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, pcpLine)) {
		siRes = -1;
	}
#else
	printf("%s", pcpLine);
#endif

	return 0;
}

// 最小値 - 取得
int
fGetMinI(
	int piVal1					// <I> 値１
	, int piVal2				// <I> 値２
)
{
	if (piVal1 < piVal2) {
		return piVal1;
	}
	else {
		return piVal2;
	}
}

// 実行メイン
int
fMain(
)
{
	int i;
	char lc1Buf[1024];

	// 石数 - 取得
	fgets(lc1Buf, sizeof(lc1Buf), szpFpI);
	sscanf(lc1Buf, "%d", &siSCnt);

	// 石 - 取得
	fgets(&sc1Stone[1], sizeof(sc1Stone) - 1, szpFpI);

	// 石合計 - セット
	si1SumLW[0] = 0;
	for (i = 1; i <= siSCnt; i++) {
		si1SumLW[i] = si1SumLW[i - 1];
		if (sc1Stone[i] == 'W') {
			si1SumLW[i]++;
		}
	}
	si1SumRR[siSCnt + 1] = 0;
	for (i = siSCnt; i >= 1; i--) {
		si1SumRR[i] = si1SumRR[i + 1];
		if (sc1Stone[i] == 'R') {
			si1SumRR[i]++;
		}
	}

	// 最小の操作回数 - 取得
	int liMin = INT_MAX;
	for (i = 0; i <= siSCnt; i++) {

		// 石交換回数
		int liCCnt = fGetMinI(si1SumLW[i], si1SumRR[i + 1]);

		// 変色回数 - 加算
		int liSum = liCCnt;
		liSum += si1SumLW[i] - liCCnt;
		liSum += si1SumRR[i + 1] - liCCnt;

		// 最小の操作回数 - 更新
		liMin = fGetMinI(liMin, liSum);
	}

	return liMin;
}

// １回実行
int
fOne(
)
{
	int liRet;
	char lc1Buf[1024];

	// 入力 - セット
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", siTNo);
	szpFpI = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", siTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#else
	szpFpI = stdin;
#endif

	// 実行メイン
	liRet = fMain();

	// 出力
	if (liRet == 0) {
		sprintf(lc1Buf, "Yes\n");
	}
	else {
		sprintf(lc1Buf, "No\n");
	}
	sprintf(lc1Buf, "%d\n", liRet);
	fOutLine(lc1Buf);

	// 残データ有無
#ifdef D_TEST
	lc1Buf[0] = '\0';
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, "")) {
		siRes = -1;
	}
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpI);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", siTNo);
	}
	else {
		printf("NG %d\n", siTNo);
	}
#endif

	return 0;
}

// プログラム開始
int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		siTNo = i;
		fOne();
	}
#else
	fOne();
#endif

	return 0;
}