# 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 stringint 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 Codeint main(){     int X = 2;    cout << MinimumString(X);    return 0;}

## Java

 // Java implementation of the// above approachimport java.util.*; class GFG{ // Function returns the minimum// size of the Stringstatic 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 Codepublic 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 approachfrom math import sqrt # Function returns the minimum# size of the stringdef 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 Codeif __name__ == '__main__':     X = 2     print(MinimumString(X)) # This code is contributed by Shivam Singh

## C#

 // C# implementation of the// above approachusing System; class GFG{ // Function returns the minimum// size of the Stringstatic 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 Codepublic 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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