-

 LARGEST COMMON PREFIX

14. Longest Common Prefix
Easy
12.7K
3.8K
Companies

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.
My_CODE:
class Solution: def longestCommonPrefix(self, strs: List[str]) -> str: a="" for i in range(len(strs[0])): b=strs[0][i] for j in range(1,len(strs)):
#try-except:if next letter index is not present in next string eg['ab',a'] try: if b!=strs[j][i]: return a except IndexError: return a a+=b return a

OTHER CODE:
(1)
def longestCommonPrefix(self, strs): prefix="" if len(strs)==0: return prefix for i in range(len(min(strs))): c=strs[0][i] if all(a[i]==c for a in strs): prefix+=c else: break return prefix


Comments

Post a Comment

Popular posts from this blog

-
Image
from tkinter import * from PIL import ImageTk , Image from sqlite3 import * from tkinter import ttk from tkinter import messagebox import time def create ():     global BILL_N , BILL_A , BILL_A_DUE , BILL_UNITS     conn = connect ( 'database.db' ) #     c = conn . cursor ()         # c.execute('''CREATE TABLE K_BILL (account_number integer PRIMARY KEY,     #  name text,     #  units integer,     #  amount_due integer,         #  amount integer)''')             # c.execute(''' INSERT INTO K_Bill VALUES (4902233658724, 'SAMIULLAH', 800 , 20267,800*13)''')     # c.execute(''' INSERT INTO K_Bill VALUES (3769186408275, 'Ali', 25000, 25267,436*13)''')     # c.execute(''' INSERT INTO K_Bill VALUES (9863867106183, 'Faez', 22000, 22567,658*13)''')     c . execute ( ''' UPDATE K_BILL set units=436 wh...
Read more
-
Image
from tkinter import * from PIL import ImageTk , Image from sqlite3 import * from tkinter import ttk from tkinter import messagebox import time # Save updated record def update_record ():     # Grab record number     selected = my_tree . focus ()     # Save new data         my_tree . item ( selected , text = "" , values =( upd_id , items_entry . get (), address_entry . get (), Email_entry . get (), Ph_No_entry . get (), Price_entry . get ()))     #update database     conn = connect ( 'database.db' )     c = conn . cursor ()     c . execute ( '''UPDATE CHECKOUT SET items=?, address=?, email=?, ph_No=?, price=? WHERE id=?''' ,               (( items_entry . get (), address_entry . get (), Email_entry . get (), Ph_No_entry . get (), Price_entry . get (), id ,)))         conn . commit ()     conn . close ()     ...
Read more
-
20. Valid Parentheses Easy 18.3K 1K Companies Given a string  s  containing just the characters  '(' ,  ')' ,  '{' ,  '}' ,  '['  and  ']' , determine if the input string is valid. An input string is valid if: Open brackets must be closed by the same type of brackets. Open brackets must be closed in the correct order. Every close bracket has a corresponding open bracket of the same type.   Example 1: Input: s = "()" Output: true Example 2: Input: s = "()[]{}" Output: true Example 3: Input: s = "(]" Output: false   Constraints: 1 <= s.length <= 10 4 s  consists of parentheses only  '()[]{}' . My code: class Solution: def isValid(self, s: str) -> bool: a='([{' b=')]}' c=[] for i in s: if i in a and s.count(i)!=s.count(b[a.find(i)]):#checks if no.of opening element==no of closing element print(i,b[...
Read more