check permutation v1

my implementation and solution

Chapter 1 Arrays and strings/1.2. check_permutation.md

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+# 1.2. Check permutation
+
+## given two strings, decide if one is a permutation of the other
+
+## First idea
+
+if one is a permutation of the other, they need to have the same length:
+
+```python
+if len(s1) != len(s2):
+    return False
+```
+
+first sort, then check: O(nlogn + n) = O(nlogn).
+
+Can it be done in O(1)? BCR? don't think so, even doing it manually takes O(n).
+
+> example: s1 = "ane", s2= "ena"
+
+## array to keep count of characters
+
+array of ints with count of each character?
+
+```python
+alphabet = 26
+def check_permutation(s1, s2):
+    if len(s1) != len(s2):
+        return False
+    checker = [0 for _ in range(alphabet)]
+    for char in s1:
+        checker[ord(char)] += 1
+    for char in s2:
+        checker[ord(char)] -= 1
+        if checker[ord(char)] < 0:
+            return False
+    return True
+```
+
+## optimization from this
+
+how many characters do we admit? 128 for starters in ASCII.
+
+```python
+alphabet = 128
+...
+if ord(char) >= 128:
+    sys.exit(1)
+```
+
+> Tests:
+
+* empty strings: correct
+* one character strings: correct
+* different length strings: correct
+* same length strings, True: correct
+* same length strings, False: correct
+
+This takes O(128 + n + n) = O(n) time and O(128) space. We can't beat BCR but maybe less space?
+
+## Use hash tables
+
+substitute array of 128 by hash table  of 128 in worst case, but generally less. O(1). no need to limit string size to 128.
+
+```python
+def check_permutation(s1, s2):
+    if len(s1) != len(s2):
+        return False
+    checker = dict()
+
+    for char in s1:
+        ascii_val = ord(char)
+        if ascii_val not in checker:
+            checker[ascii_val] = 1
+        else:
+            checker[ascii_val] += 1
+
+    for char in s2:
+        ascii_val = ord(char)
+        if ascii_val not in checker:
+            return False
+        checker[ascii_val] -= 1
+        if checker[ascii_val] < 0:
+            return False
+    return True
+```
+
+O(n) time, O(1) space.
+
+## Forgot about detail (remembered reading the solution)
+
+I don't check that the dictionary is empty at the end. I should delete the key when the count reaches 0, and check that it's an empty hash table at the end. Tests:
+
+> 'aane' vs 'ane'
+
+There'll be count of 1 in checker[0] at the end. But it'll be discarded because of different length.
+
+> 'aane' vs 'anee'
+
+It passes the test, but why? Ah! If both strings are the same length, the first string might have a bigger count of one character, but the second will have a bigger count of same/another character, so this character (`e` in this case), will make the count negative, thus returning False.
+
+## Solution
+
+Check if permutation is case sensitive:
+
+> is 'God' a permutation of 'dog'?
+
+Also check if whitespace if significant:
+
+> 'god   ' is a permutation of 'dog'?
+
+The solution assumes that the comparison is case sensitive and whitespace is significant (like my solution). It also checks that the lengths are different.
+
+### First solution
+
+If two strings are permutations, they have the same characters in different orders. Sort them, and compare. O(nlogn)
+
+### Check count of characters
+
+The definition of a permutation is: two words with the same character counts. Create an array that operates as a hash table, mapping each character to its frequency. Increment through the first string, decrement through the second, and at the end that the array is all zeroes. Terminate early if a value turns negative.
+
+Solved it right!

Chapter 1 Arrays and strings/1.2. check_permutation.py

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+import unittest
+
+def check_permutation(s1, s2):
+    if len(s1) != len(s2):
+        return False
+    checker = dict()
+
+    for char in s1:
+        ascii_val = ord(char)
+        if ascii_val not in checker:
+            checker[ascii_val] = 1
+        else:
+            checker[ascii_val] += 1
+
+    for char in s2:
+        ascii_val = ord(char)
+        if ascii_val not in checker:
+            return False
+        checker[ascii_val] -= 1
+        if checker[ascii_val] == 0:
+            del checker[ascii_val]
+        if checker[ascii_val] < 0:
+            return False
+    if checker == {}:
+        return True
+    else:
+        return False
+
+class Test(unittest.TestCase):
+    dataT = [('abcd', 'dbca'), ('dfgh', 'dfgh'), ('', '')]
+    dataF = [('aane', 'anee'), ('ab', 'ca'), ('ane', 'aane')]
+
+    def test(self):
+        # true check
+        for test_string in self.dataT:
+            s1, s2 = test_string
+            res = check_permutation(s1, s2)
+            self.assertTrue(res)
+        # false check
+        for test_string in self.dataF:
+            s1, s2 = test_string
+            res = check_permutation(s1, s2)
+            self.assertFalse(res)
+
+if __name__ == "__main__":
+    unittest.main()