Merge pull request #922 from dsrao711/issue_921

Python solution for longest common prefix -- DSA450

Author: smv1999

DSA 450 GFG/LongestCommonPrefix.py

@@ -0,0 +1,66 @@
+# Link : https://leetcode.com/problems/longest-common-prefix/submissions/ 
+
+# Approach : 
+
+# Example
+# Input: strs = ["flower","flow","flight"]
+# Output: "fl"
+
+# We are supposed to return the longest common prefix  
+
+# Variables : 
+# prefix => longest common substring 
+
+# 1. Assume the longest common substring to be the first element of the list . prefix = strs[0]
+# 2. Iterate through the list of words and keep comparing each char of the word with the prefix 
+# 3. If the char of word != char of prefix , reduce the prefix to last matched char
+# 4. If the word is shorter than the prefix , and all charachters are matched , then prefix = word
+
+
+# Working
+# prefix = "flower"
+
+# i = 1
+# prefix = "flower" , word = "flow"
+
+# f = f , pass
+# l = l , pass 
+# o = o , pass
+# w = w , pass
+# i = 4 , 4 >= len("flow") , prefix = "flow"  , break
+
+
+
+# i = 2 
+# prefix = "flow" , word = "flight"
+# f = f , pass
+# l = l , pass
+# o != i  => prefix = prefix[:i] = "fl" , break
+
+
+# o/p = "fl"
+
+
+class Solution(object):
+    def longestCommonPrefix(self, strs):
+        """
+        :type strs: List[str]
+        :rtype: str
+        """
+        n = len(strs)
+        prefix = strs[0]
+        
+        for word in strs:
+          for i in range(0 , len(prefix)):
+            if(i >= len(word)):
+              prefix = word
+              break
+              
+            elif(prefix[i] != word[i]):
+              prefix = prefix[:i]
+              break
+              
+        return prefix
+            
+            
+# TC : O(mn)