要求的最小大小的二进制字符串，以便随机删除两个1的概率为1 / X

给出的值X，该任务是找到最小尺寸二进制字符串，使得如果任何2字符被随机删除，这两个字符将是“1”的概率为1 / X。打印此类二进制字符串的大小。

例子：

Input: X = 2 
Output: 4 
Explanation: 
Let the binary string be “0111”. 
Probability of choosing 2 1s from given string is = 3C2 / 4C2 = 3/6 = 1/2 (which is equal to 1/X). 
Hence, the required size is 4. 
(Any 4 size binary string with 3 ‘1’s and 1 ‘0’ can be taken for this example).
Input: X = 8 
Output: 5 

方法：我们将尝试找到解决该问题的公式。

Let 
r = Number of 1’s in the string 
and 
b = Number of 0’s in the string.

• 如果随机删除两个字符，则

 Total number of ways = (r + b) _{C 2.} 

• 如果希望2个字符为1，则例数= r _{C 2} 。
• 因此，P(均为1)＝ r _C2 /(r + b) _C2 。
  $\dfrac{\dbinom{r}{2}}{\dbinom{r+b}{2}} = \dfrac1x \\ \\ => \dfrac{r(r - 1)}{(r+b)(r + b - 1)} = \dfrac1x$
• 进一步进行计算的棘手观察是：
  $\dfrac{r}{r+b} > \dfrac{r-1}{r+b-1}$
• 平方不等式并与等式进行比较，我们得到
  $(\dfrac{r}{r+b}) ^ { 2 } > \dfrac{1}{x} > (\dfrac{r - 1}{r+b - 1}) ^ { 2 }$
• 如果r> 1，我们在所有3边取平方根。
  $\dfrac{r}{r+b} > \dfrac{1}{\sqrt x} > \dfrac{r - 1}{r + b - 1}$
• 取不等式最左边的部分，我们得到：
  $\dfrac{r}{r+b} > \dfrac{1}{\sqrt x} \newline => r \sqrt{x} > r+ b \\ => r( \sqrt {x} - 1) > b \\ => r > \dfrac{b}{\sqrt {x} - 1} \\ => r > (\sqrt {x} + 1) b$
• 同样，考虑不平等的最右边部分，我们得到：
  $\dfrac{1}{\sqrt x} > \dfrac{r - 1}{r+b - 1} \newline => r+ b -1 > r \sqrt x - \sqrt x \\ => b > r(\sqrt x - 1) - (\sqrt x - 1) \\ => \dfrac{b}{\sqrt x - 1} > r - 1 \\ => r < 1 + \dfrac{b}{\sqrt x - 1} \\ => r < 1 + (\sqrt x + 1) b$
• 结合得出的结论，我们得到以b表示的r的范围。
  $(\sqrt x+1)b + 1 > r > (\sqrt x + 1) b$
• 对于字符串的最小值，我们设置b = 1
  $(\sqrt x + 1).1 + 1 > r > (\sqrt x + 1) \\ => \sqrt x + 2 > r > \sqrt x + 1$
• 为了获得有效的最小值r，我们取该范围内r的第一个整数值。

For Example: if X = 2
 

$\sqrt2 + 2 > r > \sqrt 2 + 1 \\ => 1.414 + 2 > r > 1.414 + 1 \\ => 3.4141 > r > 2.414 \\ => r = 3 \space (Minimum Integer )$
 

Hence, r = 3 and b = 1. 
P(both character are 1’s) = 3C2 / 4C2 = 2/4 = 1/2 

下面是上述方法的实现。

C++

// C++ implementation of the
// above approach
 
#include 
using namespace std;
 
// Function returns the minimum
// size of the string
int MinimumString(int x)
{
    // From formula
    int b = 1;
 
    // Left limit of r
    double left_lim = sqrt(x) + 1.0;
 
    // Right limit of r
    double right_lim = sqrt(x) + 2.0;
 
    int r;
    for (int i = left_lim; i <= right_lim; i++) {
        if (i > left_lim and i < right_lim) {
            // Smallest integer in
            // the valid range
            r = i;
            break;
        }
    }
 
    return b + r;
}
 
// Driver Code
int main()
{
 
    int X = 2;
    cout << MinimumString(X);
    return 0;
}

Java

// Java implementation of the
// above approach
import java.util.*;
 
class GFG{
 
// Function returns the minimum
// size of the String
static int MinimumString(int x)
{
     
    // From formula
    int b = 1;
 
    // Left limit of r
    double left_lim = Math.sqrt(x) + 1.0;
 
    // Right limit of r
    double right_lim = Math.sqrt(x) + 2.0;
 
    int r = 0;
    for(int i = (int)left_lim; i <= right_lim; i++)
    {
        if (i > left_lim && i < right_lim)
        {
             
            // Smallest integer in
            // the valid range
            r = i;
            break;
        }
    }
    return b + r;
}
 
// Driver Code
public static void main(String[] args)
{
    int X = 2;
    System.out.print(MinimumString(X));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of
# the above approach
from math import sqrt
 
# Function returns the minimum
# size of the string
def MinimumString(x):
 
    # From formula
    b = 1
 
    # Left limit of r
    left_lim = sqrt(x) + 1.0
 
    # Right limit of r
    right_lim = sqrt(x) + 2.0
 
    for i in range(int(left_lim),
                   int(right_lim) + 1):
        if(i > left_lim and i < right_lim):
             
            # Smallest integer in
            # the valid range
            r = i
            break
 
    return b + r
 
# Driver Code
if __name__ == '__main__':
 
    X = 2
 
    print(MinimumString(X))
 
# This code is contributed by Shivam Singh

C#

// C# implementation of the
// above approach
using System;
 
class GFG{
 
// Function returns the minimum
// size of the String
static int MinimumString(int x)
{
     
    // From formula
    int b = 1;
 
    // Left limit of r
    double left_lim = Math.Sqrt(x) + 1.0;
 
    // Right limit of r
    double right_lim = Math.Sqrt(x) + 2.0;
 
    int r = 0;
    for(int i = (int)left_lim; i <= right_lim; i++)
    {
        if (i > left_lim && i < right_lim)
        {
             
            // Smallest integer in
            // the valid range
            r = i;
            break;
        }
    }
    return b + r;
}
 
// Driver Code
public static void Main(String[] args)
{
    int X = 2;
     
    Console.Write(MinimumString(X));
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
// Javascript program for
// the above approach
    
// Function returns the minimum
// size of the String
function MinimumString(x)
{
       
    // From formula
    let b = 1;
   
    // Left limit of r
    let left_lim = Math.sqrt(x) + 1.0;
   
    // Right limit of r
    let right_lim = Math.sqrt(x) + 2.0;
   
    let r = 0;
    for(let i = Math.floor(left_lim); i <= Math.floor(right_lim); i++)
    {
        if (i > left_lim && i < right_lim)
        {
               
            // Smallest integer in
            // the valid range
            r = i;
            break;
        }
    }
    return b + r;
}
     
// Driver Code
     
     let  X = 2;
    document.write(MinimumString(X));
 

输出：

4

时间复杂度： O(1)，因为left_lim和right_lim之差始终小于1。
辅助空间： O(1)

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