-

 ROMAN TO INTEGER

12. Integer to Roman
Medium
5.3K
5K
Companies

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, 2 is written as II in Roman numeral, just two one's added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

  • I can be placed before V (5) and X (10) to make 4 and 9. 
  • X can be placed before L (50) and C (100) to make 40 and 90. 
  • C can be placed before D (500) and M (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral.

 

Example 1:

Input: num = 3
Output: "III"
Explanation: 3 is represented as 3 ones.

Example 2:

Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.

Example 3:

Input: num = 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

 

Constraints:

  • 1 <= num <= 3999

CODE:

r={'I':1,'V':5,'X':10,'L':50,'C':100,'D':500,'M':1000}

s='III'
a=0
count=0
assert 1<=len(s)<=15,"Invalid length"
for i in s:
    if i not in r:
        raise ValueError("Invalid numeral")
    
while count<=len(s)-1:
    #instead of so many ifs: #at start# s = s.replace("IV", "IIII").replace("IX", "VIIII").replace("XL", "XXXX").replace("XC", "LXXXX").replace("CD", "CCCC").replace("CM", "DCCCC")
   
if count==s.find("IV"):
        a+=4
        count+=2
        continue
    if count==s.find("IX"):
        a+=9
        count+=2
        continue
    if count==s.find("XL"):
        a+=40
        count+=2
        continue
    if count==s.find("XC"):
        a+=90
        count+=2
        continue
    if count==s.find("CD"):
        a+=400
        count+=2
        continue
    if count==s.find("CM"):
        a+=900
        count+=2
        continue
    else:
         
         a+=r[s[count]]
         count+=1
         
print(a)

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