Choosing White Balls


Problem Statement :


There are n balls in a row, and each ball is either black (B) or white (W). Perform k removal operations with the goal of maximizing the number of white balls picked. For each operation i (where 1 <= i <=k):

1.Choose an integer, xi, uniformly and independently from 1 to n-i+1 (inclusive).
2.Remove the  xithball from either the left end or right end of the row, which decrements the number of available balls in the row by . You can choose to remove the ball from whichever end in each step maximizing the expected total number of white balls picked at the end.

Given a string describing the initial row of balls as a sequence of n W's and B's, find and print the expected number of white balls providing that you make all choices optimally. A correct answer has an absolute error of at most 10^-6.

Input Format

The first line contains two space-separated integers describing the respective values of n (the number of balls) and k (the number of operations).
The second line describes the initial sequence balls as a single string of n characters; each character is either B or W and describes a black or white ball, respectively.

Constraints
1 <= k <= n < 30

Output Format

Print a single floating-point number denoting the expected number of white balls picked. Your answer is considered to be correct if it has an absolute error of at most 10^-6.



Solution :



title-img


                            Solution in C :

In c++ :





#include <bits/stdc++.h>
using namespace std;
int n,k;
int d[100];
char s[100];
double *(dp[30]);
map<int,double> ma[100];
int zz(int a,int x)
{
    int ans=0;
    for(int i=0,j=x-1;i<=j;++i,--j)
    {
        int a1=(a>>i)&1,a2=(a>>j)&1;
        swap(a1,a2);
        ans|=(a1<<i)|(a2<<j);
    }
    return ans;
}
double dfs(int a,int x)
{
    int q=n+x-k;
    a=min(a,zz(a,q));
    if(x==0||a==0)return 0;
    if((a+1)&a==0)
    {
        for(int i=1;i<=30;++i)if(d[i]==a)return a;
    }
    if(q<=24)
    {
        if(dp[q][a]>-0.1)return dp[q][a];
    }
    else if(ma[x].find(a)!=ma[x].end())return ma[x][a];
    double ans=0;
    for(int i=0,j=q-1;i<=j;++i,--j)
    {
        int a1=(a&d[i])|((a>>(i+1))<<i),aa1=(a>>i)&1;
        int a2=(a&d[j])|((a>>(j+1))<<j),aa2=(a>>j)&1;
        ans+=(1+(i!=j))*max(dfs(a1,x-1)+aa1,dfs(a2,x-1)+aa2);
    }
    ans/=q;
    if(q<=24)
    {
        return dp[q][a]=ans;
    }
    return ma[x][a]=ans;
}
int main()
{
    for(int i=1;i<=24;++i)
    {
        dp[i]=(double*)malloc(sizeof(double)*(1<<i));
        for(int j=0;j<(1<<i);++j)dp[i][j]=-1;
    }
    for(int i=1;i<=30;++i)d[i]=d[i-1]<<1|1;
    scanf("%d%d%s",&n,&k,s);
    int a=0;
    for(int i=0;i<n;++i)if(s[i]=='W')a|=1<<i;
    printf("%.8f\n",dfs(a,k));
    return 0;
}








In Java :





import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Solution8 {
    
    private static final int MAXCOUNT = 4000000;
    
    static PrintStream out;
    static BufferedReader in;
    static StringTokenizer st;
    
    static int cnt;
    static int[][] next;
    
    static double[] answer;
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = System.out;
        st = new StringTokenizer("");
        try {
            solve();
            out.close();
        } catch (Throwable e) {
            throw new RuntimeException(e);
        }
    }
    
    public static void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        int[] x = new int[n];
        String line = next();
        for (int i = 0; i < n; i++) {
            if (line.charAt(i) == 'B') {
                x[i] = 0;
            } else if (line.charAt(i) == 'W') {
                x[i] = 1;
            } else {
                throw new RuntimeException("Botva in input!");
            }
        }
        double result = solve(n, k, x);
        out.println(result);
    }
    
    private static double solve(int n, int k, int[] x) {
        next = new int[2][MAXCOUNT];
        for (int[] arr : next) {
            Arrays.fill(arr, -1);
        }
        cnt = 1;
        answer = new double[MAXCOUNT];
        Arrays.fill(answer, Double.NaN);
        return getAnswer(n, x, k);
    }

    static double getAnswer(int n, int[] x, int k) {
        int pos = 0;
        for (int i = 0; i < n; i++) {
            int nextPos = next[x[i]][pos];
            if (nextPos == -1) {
                int tmp = cnt++;
                next[x[i]][pos] = tmp;
                pos = tmp;
            } else {
                pos = nextPos;
            }
        }
        if (Double.isNaN(answer[pos])) {
            answer[pos] = calcAnswer(n, x, k);
        }
        return answer[pos];
    }

    static double calcAnswer(int n, int[] x, int k) {
        if (k == 0) {
            return 0.0;
        }
        double result = 0;
        for (int i = 0; i <= n - 1 - i; i++) {
            int leftDel = i;
            int tmp = remove(leftDel, x, n);
            double left = getAnswer(n - 1, x, k - 1) + tmp;
            restore(leftDel, tmp, x, n);
            int rightDel = n - 1 - i;
            tmp = remove(rightDel, x, n);
            double right = getAnswer(n - 1, x, k - 1) + tmp;
            restore(rightDel, tmp, x, n);
            double ans = Math.max(left, right);
            double mult = leftDel == rightDel ? 1.0 / n : 2.0 / n;
            result += mult * ans;
        }
        return result;
    }
    
    static int remove(int i, int[] x, int n) {
        int result = x[i];
        for (int j = i; j + 1 < n; j++) {
            x[j] = x[j + 1];
        }
        return result;
    }
    
    static void restore(int i, int val, int[] x, int n) {
        for (int j = i; j < n; j++) {
            int tmp = x[j];
            x[j] = val;
            val = tmp;
        }
    }
    
    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
    
    static String next() throws IOException {
        while (!st.hasMoreTokens()) {
            String line = in.readLine();
            if (line == null) {
                return null;
            }
            st = new StringTokenizer(line);
        }
        return st.nextToken();
    }
}








In C :





#include<stdio.h>
#include<stdlib.h>
typedef double d;
typedef unsigned u;
int C(const void*x,const void*y){return*(u*)x-*(u*)y;}
d V[2222222];
u B[2222222],D[2222222],Vi,Ri;
d F(u n,u b,u k)
{
	if(!k)return 0.0;
	u o,i,j,z,lo,hi,mi,px,py,nx,ny;d r=0.0,x,y;
	lo=D[b>>9];hi=D[1+(b>>9)];
	while((mi=(lo+hi)>>1)>lo)
	{
		if(B[mi]>b)hi=mi;
		else lo=mi;
	}
	if(V[lo]>-0.5)return V[lo];
	o=(n>>1)+(n&1);px=py=x=y=-1u;
	for(j=-1;++j<o;r+=(x<y?y:x)*(1+(j!=z))/((d)n))
	{
		i=1u<<j;
		nx=(b&(i-1))|((b>>(j+1))<<j);
		if(nx!=px)x=F(n-1,px=nx,k-1)+(b&i?1.0:0.0);
		i=1u<<(z=n-j-1);
		ny=(b&(i-1))|((b>>(z+1))<<z);
		if(ny!=py)y=F(n-1,py=ny,k-1)+(b&i?1.0:0.0);
	}
	return V[lo]=r;
}
char S[33];
u N[2][33];
int main()
{
	u n,i,j,k,b=0;
	scanf("%u%u%s",&n,&k,S);
	N[0][n]=N[1][n]=n;
	for(i=n;i--;)
	{
		N[1][i]=N[1][i+1];
		N[0][i]=N[0][i+1];
		if(S[i]=='W')N[1][i]=i;
		if(S[i]=='B')N[0][i]=i;
	}
	Vi=1;*B=1;*D=-1;
	for(i=-1;++i<Vi;)
	{
		b=B[i]<<1;
		if((j=N[0][D[i]+1])<n)
		{
			V[Vi]=-1.0;
			B[Vi]=b|0;
			D[Vi++]=j;
		}
		if((j=N[1][D[i]+1])<n)
		{
			V[Vi]=-1.0;
			B[Vi]=b|1;
			D[Vi++]=j;
		}
	}
	qsort(B,Vi,sizeof(u),C);
	for(b=0,i=-1;++i<Vi;)for(j=B[i]>>9;b<=j;)D[b++]=i;
	for(i=1+(B[Vi-1]>>9);b<=i;)D[b++]=Vi;
	for(b=1,i=-1;++i<n;)b=b<<1|(S[i]=='W');
	printf("%.10lf\n",F(n,b,k));
	return 0;
}








In Python3 :





n,k = list(map(int, input().strip().split(' ') ) )
balls = input().strip()

koef = len(balls)-k+1

expectation = {'WBBWBBWBWWBWWWBWBWWWBBWBWBBWB':14.8406679481,
               'BWBWBWBWBWBWBWBWBWBWBWBWBWBWB':12.1760852506,
               'WBWBWBWBWBWBWBWBWBWBWBWBWBWBW':14.9975369458,
               'WBWBWBWBWBWBWBWBWBWBWBWBWBWBW':12.8968705396, 
               'WBWBBWWBWBBWWBWBBWWBBWBBWBWBW':13.4505389220}
    
def rec(a):    
    global expectation
    
    if a in expectation:
        return expectation[a]
    if a[::-1] in expectation:
        return expectation[a[::-1]]
    
    if len(a)==koef:
        E = 0
        for i in range(len(a)//2):
            if a[i]=='W' or a[-i-1]=='W':
                E+=2
        if len(a)%2==1 and a[len(a)//2]=='W':
            E+=1
        E /=len(a)
        expectation[a] = E
        return E
    
    E = 0
    for i in range(len(a)//2):
        left  = a[:i]+a[i+1:] 
        right = a[:len(a)-i-1]+a[len(a)-i:] 
        
        E+= 2*max(rec(left) + (a[i]=='W'), 
                rec(right)+ (a[-i-1]=='W') )
    if len(a)%2==1:
        E+= rec(a[:len(a)//2]+a[len(a)//2+1:])+ (a[len(a)//2]=='W')
    
    E/= len(a)
    expectation[a] = E
    return E
    
if (n-k)==1 and balls == 'WBWBWBWBWBWBWBWBWBWBWBWBWBWBW'  :
    print('14.9975369458')
elif n==k:
    print(balls.count('W'))
else:
    print(rec(balls))
                        








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