For numbers from 100 to 900, using the symbols "C" and "D". The same rules work for numbers from 10 to 90, using "X" and "L".For numbers from 1 to 9, the symbols are "I" and "V", and the coding works like this. "IIX isn't allowed, this would be "VIII". When one number is followed by a larger number, it means subtraction. When a number is followed by the same or smaller number, it means addition. An additional symbol for thousands covers all the relevant bases. The rules for Roman Numerals involve using four pairs of symbols for ones and five, tens and fifties, hundreds and five hundreds.From the given number, pick successive digits, using %10 and /10 to gather the digits from right to left. 1: We iterate the Roman Numeral string backwards.You can use following guidelines to develop solution for it: We are using a constant number of variables, so space complexity = O(1).Write a program to convert a given number into equivalent Roman number (store its value as a string). In both approaches, we are running a single loop and doing O(1) operations at each iteration. We need to take care of the boundary condition when we reach the last character.This would be the scenario when string symbols will appear in descending order or three characters being repeated in successions. Otherwise, we add the integer value of the current character from the overall output.If yes, then we subtract the integer value of the current character from the overall output. Access each roman character using a loop and check the current character is less than the next character or not.Or in other words, there will be at max two Roman characters will be in increasing order. Return output Another solution idea similar to the above approachįrom the Roman string of a number, we can observe one thing: there is only one smaller number before a larger number. Solution code C++ĭef integerValue (c ) : if c = 'I' : return 1 if c = 'V' : return 5 if c = 'X' : return 10 if c = 'L' : return 50 if c = 'C' : return 100 if c = 'D' : return 500 if c = 'M' : return 1000 return - 1Ĭurr = integerValue (S ) if i + 1 = next : Step 3: By end of the loop, we return the value stored in the variable output. Boundary condition: before calculating the value of next, we need to check: do we present at the last index? If yes, then there is no need to calculate next and we just add the curr to the output i.e.Note: we are also incrementing i in the loop. Here (next - curr) is the combined integer value of the two roman characters, so we also increment the value of i by 1. If (curr So we add the curr value to the output i.e. If (curr >= next): This is a situation of case 1.We are using a helper function integerValues(char c) to get the integer value of a roman character. Inside the loop: We declare two variables curr and next to track the integer value of two consecutive roman characters at index i and i + 1. Step 2: Now we scan the input string using a loop. Step 1: We declare variable output to store the integer value of the given roman string. Suppose function romanToInteger(String S, int n) is converting a roman string S of size n to its integer value. One solution idea would be to scan the input string and incrementally generate the integer value based on the appearance of case 1 and case 2 .
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |