1.3 Given two strings, write a method to decide if one is a permutation of the other.

isPermutation.py

# use a dict(hashmap) to count the appearance of characters of the first string, then minus the number of that in the other string.
# once the hash value is less than zero, return false
# time complexity: O(n)
# space complexity: O(1) for certain set of characters, say ascii

class Solution:
  def isPermutation(self, str1, str2):
    if len(str1) != len(str2):
      return False

    myDict = {}
    for key in str1:
      if myDict.has_key(key):
        myDict[key] += 1
      else:
        myDict[key] = 1
        
    for key in str2:
      if not myDict.has_key(key):
        return False
      myDict[key] -= 1
      if myDict[key] < 0:
        return False

    return True

Line 16, myDict[key] = 1.
should be myDict[key] = 0, right?

In addition, may be you can use list instead of dict, such as:
letters = []
for i in range(256):
letters.append(0)

    for char in str1:
        letters[ord(char)] += 1

    for char2 in str2:
        letters[ord(char2)] -= 1
        if letters[ord(char2)] < 0:
            return False
    else:
        return True