1.9. string rotation

Chapter 1 Arrays and strings/1.9. string_rotation.java

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+public class Question {
+	public static boolean isSubstring(String big, String small) {
+		if (big.indexOf(small) >= 0) {
+			return true;
+		} else {
+			return false;
+		}
+	}
+	
+	public static boolean isRotation(String s1, String s2) {
+	    int len = s1.length();
+	    /* check that s1 and s2 are equal length and not empty */
+	    if (len == s2.length() && len > 0) { 
+	    	/* concatenate s1 and s1 within new buffer */
+	    	String s1s1 = s1 + s1;
+	    	return isSubstring(s1s1, s2);
+	    }
+	    return false;
+	}
+	
+	public static void main(String[] args) {
+		String[][] pairs = {{"apple", "pleap"}, {"waterbottle", "erbottlewat"}, {"camera", "macera"}};
+		for (String[] pair : pairs) {
+			String word1 = pair[0];
+			String word2 = pair[1];
+			boolean is_rotation = isRotation(word1, word2);
+			System.out.println(word1 + ", " + word2 + ": " + is_rotation);
+		}
+	}
+
+}

Chapter 1 Arrays and strings/1.9. string_rotation.md

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+# 1.9. String rotation
+
+## Given s1 and s2, calculate if s2 is a rotation of s1 using only one call to isSubstring, a function that checks if a string is a substring of another
+
+> example: 'waterbottle' is a rotation of 'erbottlewat'
+
+## First idea
+
+Need to have the same lengths and same count of characters.
+
+```python
+from collections import Counter
+if len(s1) != len(s2) or Counter(s1) != Counter(s2):
+    return 0
+if len(s1) < 3 or s1 == s2:
+    return 1
+```
+
+Find s1[0] in s2, index `i`. Check that s2, from i to the end is equal to the substring of s1, and same with the first part.
+
+```python
+indexes = [i for i, e in enumerate(s2) if e == s1[0]]
+for idx in indexes:
+    offset = len(s1) - idx
+    if s2[idx:] == s1[:offset] and s2[:idx] == s1[offset:]:
+        return 1
+return 0
+```
+
+Finds all occurences of s1[0] in s2, checks if 'wat' == 'wat', and then 'erbottle' == 'erbottle'. Takes O(n) time, O(1) space. No calls to isSubstring, either 2 calls or none. I don't see the point of just one.
+
+## Hints and solution
+
+Let s1 be divided into two parts, x and y. s1 = xy and s2 = yx. If we concatenate s2 with itself, s2 = yxxy. In which case s1 is a substring of s2s2. But this takes O(n) time in `isSubstring` and O(n) space.
+
+```python
+return isSubstring(s1, s2 + s2)
+```

Chapter 1 Arrays and strings/1.9. string_rotation.py

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+import unittest
+from collections import Counter
+
+def rotate_matrix(s1, s2):
+
+    if len(s1) != len(s2) or Counter(s1) != Counter(s2):
+        return 0
+    if len(s1) < 3 or s1 == s2:
+        return 1
+
+    indexes = [i for i, e in enumerate(s2) if e == s1[0]]
+    for idx in indexes:
+        offset = len(s1) - idx
+        if s2[idx:] == s1[:offset] and s2[:idx] == s1[offset:]:
+            return 1
+    return 0
+
+
+class Test(unittest.TestCase):
+    '''Test Cases'''
+    dataT = [('', ''),
+             ('a', 'a'),
+             ('ab', 'ba'),
+             ('waterbottle', 'waterbottle'),
+             ('erbottlewat', 'waterbottle'),
+             ('qwerqwer', 'erqwerqw')]
+
+    dataF = [('', 'a'),
+            ('a', 'b'),
+            ('qwerq', 'wreqq')]
+
+    def test_rotate_matrix(self):
+        for test in self.dataT:
+            res = rotate_matrix(*test)
+            self.assertTrue(res)
+
+        for test in self.dataF:
+            res = rotate_matrix(*test)
+            self.assertFalse(res)
+
+if __name__ == "__main__":
+    unittest.main()