Minimum size binary string required such that probability of deleting two 1’s at random is 1/X

Given a value X, the task is to find a minimum size binary string, such that if any 2 characters are deleted at random, the probability that both the characters will be ‘1’ is 1/X. Print the size of such binary string.
Example:

Input: X = 2 
Output:
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:

Approach: We will try to find a formula to solve this problem. 
 

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

  • If two characters are deleted at random, then 
     

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



  • If 2 characters are desired to be 1’s, Favourable number of cases = r C 2
     
  • Hence, P(both are 1’s) = rC2 / (r + b)C2.
    \dfrac{\dbinom{r}{2}}{\dbinom{r+b}{2}} = \dfrac1x \\ \\ => \dfrac{r(r - 1)}{(r+b)(r + b - 1)} = \dfrac1x
  • A tricky observation to further proceed our calculation is:
    \dfrac{r}{r+b} > \dfrac{r-1}{r+b-1}
  • Squaring the inequality and comparing with the equality, we get
    (\dfrac{r}{r+b}) ^ { 2 } > \dfrac{1}{x} > (\dfrac{r - 1}{r+b - 1}) ^ { 2 }
  • If r > 1, we take square root on all 3 sides.
    \dfrac{r}{r+b} > \dfrac{1}{\sqrt x} > \dfrac{r - 1}{r + b - 1}
  • Taking the leftmost part of the inequality, we get: 
    \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

  • Similarly, taking the rightmost part of the inequality, we get: 
     \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
  • Combining the derived conclusions, we get the range of r in terms of b. 
    (\sqrt x+1)b + 1 > r > (\sqrt x + 1) b
  • For the minimum value of string, we set b = 1 
    (\sqrt x + 1).1 + 1 > r > (\sqrt x + 1) \\ => \sqrt x + 2 > r > \sqrt x + 1
  • In order to get a valid minimum r, we take the first integer value of r in this range.

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 

Below is the implementation of the above approach.

C++

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// C++ implementation of the
// above approach
  
#include <bits/stdc++.h>
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;
}

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Python3

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# 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

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Output: 

4

 

Time Complexity: O(1), as the difference between left_lim and right_lim will be always less than 1. 
Auxiliary Space: O(1)
 

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