re: Daily Challenge #29 - Xs and Os VIEW POST

FULL DISCUSSION
 

This can be done in linear time and linear space in terms of complexity by keeping track of the number of letter in a Map object as you traverse the string. Do note that the letter should be case insensitive which that actually gives us a hint that we can make each of the character to be always lowercased before using it as a cache key.

Pseudocode:

a = 'xoxo'
m = Map()
loop for each n in a {
l = n.toLowerCase()

m.set(l, 1) if not m.has(l)
else m.set(l, 1 + m.get(l))
}

return True if not m.has('x') and not m.has('o')

return m.get('x') == m.get('o');

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