Program to find maximum size of any sequence of given array where every pair is nice in Python

Python Server Side Programming Programming

Suppose we have a sequence nums of size n. We have to find the maximum size of subsequence of nums in which every pair (p, q) is a nice pair? A pait is said to be nice pair if and only if it holds at least one of these conditions: 1. The parity of the number of distinct prime divisors of p is equal to that of b. For example, the value 18 has two distinct prime divisors: 2 and 3. 2. The parity of the sum of all positive divisors of p is same as q.

So, if the input is like nums = [2,3,6,8], then the output will be 3

To solve this, we will follow these steps −

n := size of nums
Define three empty lists cnt, total, result
for each i in nums, do
- count := 0, tot := 0
- prime := a new list
- for each j in nums, do
  - if (j mod k for all k in range 2 to j) is true, then
    - insert j at the end of prime
- for each j in prime, do
  - if i mod j is 0, then
    - count := count + 1
- if count is even, then
  - insert 'odd' at the end of cnt
- otherwise,
  - insert 'even' at the end of cnt
- for j in range 1 to i, do
  - if i mod j is same as 0, then
    - tot := tot + j
- if tot is odd, then
  - insert 'odd' at the end of total
- otherwise,
  - insert 'even' at the end of total
for i in range 0 to n-2, do
- for j in range i+1 to n - 1, do
  - if cnt[i] is same as cnt[j] or total[i] is same as total[j], then
    - insert nums[i] at the end of result
    - if j is same as n-2, then
      - insert nums[j] at the end of result
result := a new list from a new set from result
return size of result

Example

Let us see the following implementation to get better understanding −

def solve(nums):
   n = len(nums)
   cnt = []
   total = []
   result = []
   for i in nums:
      count = 0
      tot = 0

      prime = []
      for j in nums:
         if all(j % k for k in range(2, j)) == True:
            prime.append(j)

      for j in prime:
         if i % j == 0:
            count += 1
      if count % 2:
         cnt.append('odd')
      else:
         cnt.append('even')

      for j in range(1,i+1):
         if i % j == 0:
            tot += j

      if tot % 2:
         total.append('odd')
      else:
         total.append('even')

   for i in range(n-1):
      for j in range(i+1, n):

         if cnt[i] == cnt[j] or total[i] == total[j]:
            result.append(nums[i])

            if j == n-1:
               result.append(nums[j])

   result = list(set(result))
   return len(result)

nums = [2,3,6,8]
print(solve(nums))

Input

15, 3, 8

Output

Arnab Chakraborty

Published on 23-Oct-2021 08:41:00

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