Digit Pair

Problem Statement:

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.

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, 811+27 =102 so ignore most significant bit , So bit score = 02.

Say, Number is 123

Largest digit is 3 and smallest digit is 1

So, 311+71=40, so bit score is 40.

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.

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.

Solution

#include<iostream>
using namespace std;
int bit_score(int n)
{
int a, b, c, largest, smallest;
int score;
a = n%10;	n/=10;
b = n%10;	n/=10;
c = n%10;	n/=10;
largest = (a>b)?a:b;
largest = (c>largest)?c:largest;
smallest = (a<b)?a:b;
smallest = (c<smallest)?c:smallest;
score = largest*11 + smallest*7;
score = score % 100;
return score;
}
int findPairs (int score_array[], int N)	
{
 int sig_dig[10], i, pairs = 0, msb;
 for(i=0; i<10; i++)	
 {
 sig_dig[i] = 0;
 }
 for(i=0; i<N; i=i+2)	
 {
   msb = score_array[i] / 10;
   for(int j =i+2; j<N; j=j+2)
   {
      if(msb == score_array[j]/10)
      {
        if(sig_dig[msb] < 2)	
        {
	   sig_dig[msb]++;
        }
      }
    }
   }
   for(i=1; i<N; i=i+2)	
   {
     msb = score_array[i] / 10;
     for(int j =i+2; j<N; j=j+2)
     {
         if(msb == score_array[j]/10)
         {
                if(sig_dig[msb] < 2)	
                {
	         sig_dig[msb]++;
	        }
            }
        }
	}
	for(i=0; i<10; i++)	
        {
	   pairs = pairs + sig_dig[i];
	}
	return pairs;
}
int main()	
{
int N, i;
int ip_array[501];
int score_array[501];
int pairs;
cin>>N;
for(i=0; i<N; i++)	
{
  cin>>ip_array[i];
}
for(i=0; i<N; i++)	
{
score_array[i] = bit_score(ip_array[i]);
}
pairs = findPairs(score_array, N);
cout<<pairs;
return 0;
}