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.
Input: X = 2
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
Approach: We will try to find a formula to solve this problem.
r = Number of 1’s in the string
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.
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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