1.5. levenshtein distance

officially called 'One Away'

Chapter 1 Arrays and strings/1.5. levenshtein.md

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+# 1.4. Levenshtein distance
+
+## Given two strings, check if they are one edit (or zero edits) away. Edits can be insertion, substitution or deletion
+
+## First idea
+
+Difference in length must be at most 1
+
+```python
+if abs(len(s1) - len(s2)) > 1:
+    return False
+if s1 == s2:
+    return True
+```
+
+BCR is O(n) time, O(1) space.
+
+This 'edit away' is Levenshtein distance, task is to check if Levenshtein distance <= 1.
+
+Can we do `s1 - s2`? Or `s1.difference(s2)`? Not in python.
+
+## Second idea
+
+From the Internet, this is O(1) time:
+
+```python
+def levenshtein(s1, s2):
+    return len("".join(s1.rsplit(s2))) <= 1
+```
+
+But this splits the first word with the second word, it doesn't work if there's a gap/addition in the middle of the string.
+
+## Third idea
+
+Keep a boolean `flag` that must be at most 1. Iterate through the smallest string, check for differences, if difference: flag += 1, if flag == 2, return False.
+
+### If equal lengths
+
+Same length, match s1[i] vs s2[i]. if difference: flag += 1, if flag == 2, return False. if iteration done, return True
+
+### Unequal lengths
+
+Iterate through the smallest string, match s1[i] vs s2[i]. If difference, the smallest string must have a character deletion. flag = 1, and keep matching s1[i] vs s2[i+1] (where s1 is the small string). If the smallest string is iterated and no errors, return True
+
+> Tests:
+
+* ("", "") -> True, correct
+* ("an", "a") -> True, correct
+* ("a", "an") -> True, correct
+* ("ane", "ae") -> True, correct
+* ("ane", "ne") -> True, correct
+* ("asdfgh", "asdgh") -> True, correct
+* ("asdf", "asgf") -> True, correct
+* ("asdfg", "adfgh") -> False, corrrect
+* ("an", "ne") -> False, correct
+
+Tests passed, O(min(len(s1), len(s2)))) time, O(1) space. Possibly code can be compressed but I don't know how.
+
+## Solution
+
+First solution, 3 different cases for equal lengths, len1 < len2, len2 < len1. Seeing this, my code:
+
+```python
+elif len1 < len2:  # s1 shorter than s2
+    for i in range(len1):
+        if not flag:
+            if s1[i] != s2[i]:
+                if flag:
+                    return False
+                flag = True
+        else: # there's already one difference
+            if s1[i] != s2[i + 1]:
+                return False
+    return True
+
+elif len2 < len1:  # s2 shorter than s1
+    for i in range(len2):
+        if not flag:
+            if s1[i] != s2[i]:
+                if flag:
+                    return False
+                flag = True
+        else:  # there's already one difference
+            if s1[i + 1] != s2[i]:
+                return False
+    return True
+```
+
+can be merged into one:
+
+```python
+else:
+    index1 = 0
+    index2 = 0
+    while index1 < len1 and index2 < len2:
+        if s1[index1] != s2[index2]: # difference found
+            if flag:
+                return False
+            flag = True
+
+            ## need to check for shorter string here...
+```
+
+The solution creates 2 new strings, short and long (using O(2n) space), and updates indexes accordingly (see java file).
+
+What's better, create 2 new strings (O(n) space) and make code more compact, or use just 3 variables and 10 more lines of code? Don't know.

Chapter 1 Arrays and strings/1.5. levenshtein.py

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+import unittest
+
+def levenshtein(s1, s2):
+    if abs(len(s1) - len(s2)) > 1:
+        return False
+    if s1 == s2:
+        return True
+    
+    len1 = len(s1)
+    len2 = len(s2)
+    flag = False
+
+    if len1 == len2:  # equal lengths
+        for i in range(len1):
+            if s1[i] != s2[i]:
+                if flag:
+                    return False
+                flag = True
+        return True
+    
+    elif len1 < len2:  # s1 shorter than s2
+        for i in range(len1):
+            if not flag:
+                if s1[i] != s2[i]:
+                    if flag:
+                        return False
+                    flag = True
+            else: # there's already one difference
+                if s1[i] != s2[i + 1]:
+                    return False
+        return True
+
+    elif len2 < len1:  # s2 shorter than s1
+        for i in range(len2):
+            if not flag:
+                if s1[i] != s2[i]:
+                    if flag:
+                        return False
+                    flag = True
+            else:  # there's already one difference
+                if s1[i + 1] != s2[i]:
+                    return False
+        return True
+
+
+class Test(unittest.TestCase):
+    dataT = [("", ""), ("an", "a"), ("a", "an"), ("ane", "ae"),
+             ("ane", "ne"), ("asdfgh", "asdgh"), ("asdf", "asgf")]
+    dataF = [("asdfg", "adfgh"), ("an", "ne"), ("asdf", "as")]
+
+    def test_unique(self):
+        for test in self.dataT:
+            res = levenshtein(*test)
+            self.assertTrue(res)
+
+        for test in self.dataF:
+            res = levenshtein(*test)
+            self.assertFalse(res)
+
+        return
+
+if __name__ == "__main__":
+    unittest.main()