added 0014_longest_common_prefix.py

Author: nyierr

problems/easy/0014_longest_common_prefix.py

@@ -0,0 +1,92 @@
+"""
+Write a function to find the longest common prefix string amongst an array of strings.
+
+If there is no common prefix, return an empty string "".
+
+Example 1:
+  Input: strs = ["flower","flow","flight"]
+  Output: "fl"
+
+Example 2:
+  Input: strs = ["dog","racecar","car"]
+  Output: ""
+  Explanation: There is no common prefix among the input strings. 
+
+Constraints:
+  * 1 <= strs.length <= 200
+  * 0 <= strs[i].length <= 200
+  * strs[i] consists of only lower-case English letters.
+"""
+
+class Solution:
+    # O(S) where S is the total number of characters    
+    #def longestCommonPrefix(self, strs: List[str]) -> str: 
+    #    prefix = strs[0]
+    #    
+    #    for s in strs[1:]:
+    #        if len(prefix) == 0 or len(s) == 0:
+    #            prefix = ""
+    #            break
+                
+    #        no_change = True
+    #        max_i = min(len(s), len(prefix))
+            
+    #        for i in range(max_i):
+    #            if s[i] != prefix[i]:
+    #                prefix = prefix[:i]
+    #                no_change = False
+    #                break
+                    
+    #        if no_change and i == len(s) - 1:
+    #            prefix = s[:i+1]
+                
+    #    return prefix
+    
+    # O(S) where S is the total number of characters
+    # Similar to the previous one but using recursion
+    #def longestCommonPrefix(self, strs: List[str]) -> str: 
+    #    if len(strs) == 1:
+    #        return strs[0]
+        
+    #    mid = len(strs) // 2
+    #    prefix = self.longestCommonPrefix(strs[:mid])
+    #    check = self.longestCommonPrefix(strs[mid:])
+        
+    #    if prefix == "" or check == "":
+    #        return ""
+        
+    #    matched = True
+    #    max_i = min(len(prefix), len(check))
+    #    for i in range(max_i):
+    #        if prefix[i] != check[i]:
+    #            prefix = prefix[:i]
+    #            matched = False
+    #            break
+                
+    #    if matched and i == len(check) - 1:
+    #        prefix = check[:i+1]
+    #        
+    #    return prefix
+
+    # O(S*log m) Binary-search-like approach
+    def longestCommonPrefix(self, strs: List[str]) -> str: 
+        r = len(strs[0])
+        for s in strs[1:]:
+            r = min(r, len(s))
+
+        l = 1
+        while l <= r:
+            mid = (l + r) // 2
+            if self.isShared(strs, mid):
+                l = mid + 1
+            else:
+                r = mid - 1
+        
+        return strs[0][:(l+r) // 2]
+            
+    def isShared(self, strs, mid):
+        prefix = strs[0][:mid]
+        for s in strs[1:]:
+            if prefix != s[:mid]:
+                return False
+        return True