2022-08-15 Daily-Challenge
Today I have done leetcode's August LeetCoding Challenge with cpp
.
August LeetCoding Challenge 15
Description
Roman to Integer
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 ones 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 beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(1000) to make 400 and 900.
Given a roman numeral, convert it to an integer.
Example 1:
Input: s = "III"
Output: 3
Explanation: III = 3.
Example 2:
Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.
Example 3:
Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
Constraints:
1 <= s.length <= 15
s
contains only the characters('I', 'V', 'X', 'L', 'C', 'D', 'M')
.- It is guaranteed that
s
is a valid roman numeral in the range[1, 3999]
.
Solution
unordered_map<char, int> mp = {
{'M', 1000},
{'D', 500},
{'C', 100},
{'L', 50},
{'X', 10},
{'V', 5},
{'I', 1}
};
class Solution {
public:
int romanToInt(string s) {
int len = s.length();
int pos = 0;
int answer = 0;
while(pos < len) {
if(pos < len - 1 && mp[s[pos]] < mp[s[pos + 1]]) {
answer += mp[s[pos + 1]] - mp[s[pos]];
pos += 2;
} else {
answer += mp[s[pos]];
pos += 1;
}
}
return answer;
}
};
// Accepted
// 3999/3999 cases passed (12 ms)
// Your runtime beats 71.04 % of cpp submissions
// Your memory usage beats 86.74 % of cpp submissions (6 MB)