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Discussion on: Daily Challenge #29 - Xs and Os

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mariocsee profile image
Mario See

A little Python here.

def count_xo(s):
  count_x, count_o = 0, 0
  for c in s:
    if c.lower() == "x":
      count_x += 1
    if c.lower() == "o":
      count_o += 1
  return count_x == count_o

This method uses a for loop to go through each character. It converts each character to lower case for a case insensitive comparison with "x" and "o" and increments the respective counters. The return line checks for count equality to return a boolean.

The time complexity in big O notation is O(n) or linear, where n is the size of input string s, and the space complexity is O(1) or constant since it just uses two variables to keep track of counts.

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colinb profile image
Colin Bounouar

Is there a reason why you prefer a for loop over a single call to lower on the str, and then returning lowered.count('x') == lowered.count('o') ?