Given N three-digit numbers, your task is to find bit score of all N numbers and then print the number of pairs possible based on these calculated bit score.

1. Rule for calculating bit score from three digit number:

From the 3-digit number,

• extract largest digit and multiply by 11 then
• extract smallest digit multiply by 7 then
• add both the result for getting bit pairs.

Note: - Bit score should be of 2-digits, if above results in a 3-digit bit score, simply ignore most significant digit.

Consider following examples:

Say, number is 286

Largest digit is 8 and smallest digit is 2

So, 8*11+2*7 =102 so ignore most significant bit , So bit score = 02.

Say, Number is 123

Largest digit is 3 and smallest digit is 1

So, 3*11+7*1=40, so bit score is 40.

2. Rules for making pairs from above calculated bit scores

Condition for making pairs are

• Both bit scores should be in either odd position or even position to be eligible to form a pair.
• Pairs can be only made if most significant digit are same and at most two pair can be made for a given significant digit.

Constraints
N<=500

Input Format
First line contains an integer N, denoting the count of numbers.

Second line contains N 3-digit integers delimited by space

Output
One integer value denoting the number of bit pairs.

Test Case

Explanation
Example 1

Input

8 234 567 321 345 123 110 767 111

Output

3 Explanation

After getting the most and least significant digits of the numbers and applying the formula given in Rule 1 we get the bit scores of the numbers as:

58 12 40 76 40 11 19 18

No. of pair possible are 3:

40 appears twice at odd-indices 3 and 5 respectively. Hence, this is one pair.

12, 11, 18 are at even-indices. Hence, two pairs are possible from these three-bit scores.

Hence total pairs possible is 3.

Answer rating: 100% (QA)

Here is the Java code for the problem package comcheggbitscores import jav... View the full answer