csfx-py/hacktober2020

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Digit_pair.py

Digit Pairs
Problem Description
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, 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.
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.
   Timeout
   1
   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

def dig(n):
    s,l=9,0
    while(n):
        r=n%10
        if(s>r):
            s=r
        if(l<r):
            l=r
        n=n//10
    return [s,l]
  
def pair(t):
    if(t==2):
        return 1
    if(t>2):
        return 2
    return 0
      
n=int(input())
array=list(map(int,input().split()))
even=[]
odd=[]
for i in range(n):
    t=dig(array[i])
    s,l=t[0],t[1]
    l*=11
    s*=7
    array[i]=(l+s)%100
    
even=[0]*10
odd=[0]*10
for i in range(n):
    t=array[i]//10
    if (i+1)%2==0:
        even[t]+=1
    else:
        odd[t]+=1

c=[0]*10
for i in range(10):
    if(even[i]<=1) and odd[i]<=1:
        continue
    c[i]+=pair(even[i])+pair(odd[i])
    c[i]=min(2,c[i])
        
    #print(p,odd[i])
print(sum(c))