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

• Last Updated : 10 May, 2021

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 • A tricky observation to further proceed our calculation is: • Squaring the inequality and comparing with the equality, we get • If r > 1, we take square root on all 3 sides. • Taking the leftmost part of the inequality, we get: • Similarly, taking the rightmost part of the inequality, we get: • Combining the derived conclusions, we get the range of r in terms of b. • For the minimum value of string, we set b = 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 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++

 // 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

 

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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