I was trying to figure out this problem, and I did. However, my code is so abhorrently ugly that I want to tear out my eyeballs when I look at it:

with open("gymnastics.in", "r") as fin:
  rounds, cows = [int(i) for i in fin.readline().split()]
  nums = [tuple(map(int, line.split())) for line in fin]

def populatePairs(cows):
  pairs = []
  for i in range(cows):
      for j in range(cows):
          if i != j:
  return pairs

def consistentPairs(pairs, nums):
  results = []

  for i in pairs:
      asdf = True
      for num in nums:
          for j in num:
              if j == i[1]:
                  asdf = False
              if j == i[0]:
          if not asdf:
      if asdf:

  return results

pairs = populatePairs(cows)

with open("gymnastics.out", "w+") as fout:
  print(len(consistentPairs(pairs, nums)), file=fout)

I feel like that there should definitely be a better solution that is faster than \$O(n^3)\$, and without the triple nested for-loop with the if-statements trailing behind them, but I cannot, for the love of god, think of a better solution.

Problem synopsis:

Given an \$n\$ by \$m\$ grid of points, find the number of pairs in which number is consistently placed before the other.



3 4
4 1 2 3
4 1 3 2
4 2 1 3

Output: 4

Explanation: The consistent pairs of cows are (1,4), (2,4), (3,4), and (3,1), in which case 4 is consistently greater than all of them, and 1 is always greater than 3.

  • \$\begingroup\$ Do you have test cases with this? \$\endgroup\$ – Mast Oct 25 '20 at 17:12
  • \$\begingroup\$ @Mast yes. i'm going to edit the question with the test case included \$\endgroup\$ – 12 rhombi in grid w no corners Oct 25 '20 at 22:28


  • Use the variable names from the problem specification instead of inventing your own (just make them lower-case since that's preferred in Python code). Makes it easier to see the connections. Unless yours are much better.
  • I guess nums is plural so num is a single number, but you can iterate it? Bad name. And wth does asdf mean?
  • Python prefers snake_case for function names.
  • To be honest, the lack of any explanation of your method, the name asdf and the highly convoluted code made me give up reading it. But here's my solution, simply counting occurring pairs and then the result is the number of pairs that appeared K times:
from itertools import combinations
from collections import Counter

ctr = Counter()
with open('gymnastics.in') as f:
    k, _ = map(int, next(f).split())
    for session in f:
        cows = session.split()
        ctr.update(combinations(cows, 2))

with open('gymnastics.out', 'w') as f:
    print(list(ctr.values()).count(k), file=f)
  • \$\begingroup\$ respect for the brutal honesty; i was incredibly tired when i was writing the code so i didnt put even an iota of thought into writing the variable/function names. \$\endgroup\$ – 12 rhombi in grid w no corners Oct 26 '20 at 0:10
  • 1
    \$\begingroup\$ WAIT UR SOLUTION IS SO SMART WTF. jesus christ, mr. superb rain, you are incredibly cool \$\endgroup\$ – 12 rhombi in grid w no corners Oct 26 '20 at 0:15

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