Create maximum-number-of-occurrences-of-a-substring.py

Author: kamyu104

Python/maximum-number-of-occurrences-of-a-substring.py

@@ -0,0 +1,54 @@
+# Time:  O(n)
+# Space: O(n)
+
+import collections
+
+
+# rolling hash (Rabin-Karp Algorithm)
+class Solution(object):
+    def maxFreq(self, s, maxLetters, minSize, maxSize):
+        """
+        :type s: str
+        :type maxLetters: int
+        :type minSize: int
+        :type maxSize: int
+        :rtype: int
+        """
+        M, p = 10**9+7, 113
+        power, rolling_hash = pow(p, minSize-1, M), 0
+
+        left = 0
+        lookup, count = collections.defaultdict(int), collections.defaultdict(int)
+        for right in xrange(len(s)):
+            count[s[right]] += 1
+            if right-left+1 > minSize:
+                count[s[left]] -= 1
+                rolling_hash = (rolling_hash - ord(s[left])*power) % M
+                if count[s[left]] == 0:
+                    count.pop(s[left])
+                left += 1
+            rolling_hash = (rolling_hash*p + ord(s[right])) % M
+            if right-left+1 == minSize and len(count) <= maxLetters:
+                lookup[rolling_hash] += 1
+        return max(lookup.values() or [0])
+
+
+# Time:  O(m * n), m = 26
+# Space: O(m * n)
+class Solution2(object):
+    def maxFreq(self, s, maxLetters, minSize, maxSize):
+        """
+        :type s: str
+        :type maxLetters: int
+        :type minSize: int
+        :type maxSize: int
+        :rtype: int
+        """
+        lookup = {}
+        for right in xrange(minSize-1, len(s)):
+            word = s[right-minSize+1:right+1]
+            if word in lookup:
+                lookup[word] += 1
+            elif len(collections.Counter(word)) <= maxLetters:
+                lookup[word] = 1
+        return max(lookup.values() or [0])