Roman to Integer

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class Solution:
    def romanToInt(self, s: str) -> int:
        translations = {
            "I": 1,
            "V": 5,
            "X": 10,
            "L": 50,
            "C": 100,
            "D": 500,
            "M": 1000
        }
        number = 0
        s = s.replace("IV", "IIII").replace("IX", "VIIII")
        s = s.replace("XL", "XXXX").replace("XC", "LXXXX")
        s = s.replace("CD", "CCCC").replace("CM", "DCCCC")
        for char in s:
            number += translations[char]
        return number

Problem Classification

This is a Conversion Problem.

  1. String Manipulation: The problem involves converting a string representation of Roman numerals into an integer. Thus, it deals with the processing and manipulation of string data.

  2. Mathematics: Since we are dealing with numerical conversion, this problem falls into the category of mathematics. Specifically, it relates to number theory and arithmetic as it requires understanding of the Roman numeral system and its conversion to decimal.

  3. Data Structures: This problem involves the use of a specific data structure - the dictionary or map, to maintain a mapping between Roman numerals and their integer equivalents. This brings it under the scope of problems dealing with data structures.

  4. Algorithm: The problem requires a specific order of operations (algorithm) to accurately convert the Roman numeral string to its integer equivalent. This means it falls under the category of algorithmic problems.

This is a multi-disciplinary problem that combines elements of string manipulation, mathematics, data structures, and algorithms.

Language Agnostic Coding Drills

This involves understanding of dictionaries, string manipulation and iteration and mathematical addition:

  1. Understand and use dictionaries in Python: Dictionaries are key-value pairs in Python. Here, the dictionary ’translations’ is used to map Roman numerals to their corresponding integer values. The dictionary method is quite powerful in Python, and can be used in various other scenarios where a mapping is needed.

  2. String manipulation: This problem requires the coder to understand and implement various string manipulation techniques. This includes replacing parts of a string with other strings, which is done using the replace() method. This method is called on a string object and requires two arguments, the string to replace and the string to replace it with.

  3. Loops and iteration: Understanding how to iterate through a string or any other iterable object in Python is crucial for this problem. The coder needs to use a for loop to iterate through the modified string and add up the corresponding integer values.

  4. Basic mathematics (Addition): Addition of integers is an integral part of this problem. This involves incrementing the value of the ’number’ variable by the corresponding integer value of the Roman numeral character.

In terms of a step-by-step approach to the problem:

  1. The first step is to create a dictionary that maps Roman numerals to their corresponding integer values.

  2. The input string is then manipulated to replace subtractive notations like IV, IX, etc. with their equivalent additive notations. This simplifies the calculation process.

  3. The program then iterates over the string, and for each character, it finds the corresponding integer value from the ’translations’ dictionary.

  4. The integer values are added together to get the final output, which is the integer representation of the Roman numeral string.

Targeted Drills in Python

  1. Understanding and using dictionaries in Python:
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# create a dictionary
translations = {
    "I": 1,
    "V": 5,
    "X": 10,
    "L": 50,
    "C": 100,
    "D": 500,
    "M": 1000
}
# access a dictionary
print(translations["I"])  # Output: 1
  1. String manipulation:
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# replace parts of a string with another string
s = "IV"
s = s.replace("IV", "IIII")
print(s)  # Output: "IIII"
  1. Loops and iteration:
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# iterate through a string
s = "IIII"
for char in s:
    print(char)
# Output: "I"
#         "I"
#         "I"
#         "I"
  1. Basic mathematics (Addition):
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# initialize a number
number = 0
# add to a number
number += 1
print(number)  # Output: 1

Now, let’s combine all the drills to solve the problem:

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class Solution:
    def romanToInt(self, s: str) -> int:
        # create and use a dictionary
        translations = {
            "I": 1,
            "V": 5,
            "X": 10,
            "L": 50,
            "C": 100,
            "D": 500,
            "M": 1000
        }

        # initialize a number
        number = 0
        
        # string manipulation
        s = s.replace("IV", "IIII").replace("IX", "VIIII")
        s = s.replace("XL", "XXXX").replace("XC", "LXXXX")
        s = s.replace("CD", "CCCC").replace("CM", "DCCCC")
        
        # loop through the string and add to the number
        for char in s:
            number += translations[char]
            
        return number

# create an object of the Solution class
solution = Solution()
print(solution.romanToInt("III"))  # Output: 3
print(solution.romanToInt("IV"))  # Output: 4
print(solution.romanToInt("IX"))  # Output: 9
print(solution.romanToInt("LVIII"))  # Output: 58
print(solution.romanToInt("MCMXCIV"))  # Output: 1994

Q&A

Time Limit Exceeded, is it because I used a switch case?

`int romanToInt(char * s){

int num = 0; int i = 0;

while(s[i] != ‘\0’);{ switch (s[i]){ case ‘M’: if(s[i-1] == ‘C’){ num += 900; }else{ num += 1000; }

    case 'D':
        if(s[i-1] == 'C'){
            num += 400;
        }else{
            num += 500;
        }

    case 'C':
        if(s[i+1] == 'M' || s[i+1] == 'D'){
            break;
        }else{
            if(s[i-1] == 'X'){
                num+=90;
            }else{
                num+= 100;
            }
        }
    
    case 'L':
        if(s[i-1] == 'X'){
            num += 40;
        }else{
            num += 50;
        }

    case 'X':
        if(s[i+1] == 'L' || s[i+1] == 'C'){
            break;
        }else{
            if(s[i-1] == 'I'){
                num += 9;
            }else{
                num += 10;
            }
        }


    case 'V':
        if(s[i-1] == 'I'){
            num += 4;
        }else{
            num += 5;
        }


    case 'I':
        if(s[i+1] == 'V' || s[i+1] == 'X'){
            break;
        }else{
            num += 1;
        }
}
i++;

} return num; }`

Yes, the code runs into an infinite loop, and that’s the cause for the Time Limit Exceeded error. The reason is that you have a semicolon ; right after your while condition, which is making the while loop run indefinitely without any statements inside it. The code block following this semicolon is then executed as a separate, single-time block, and it’s not inside the loop as you intended. This leads to an infinite loop because there’s no operation inside the while loop that changes the loop’s condition (s[i] != '\0'), so it just keeps running.

The correct way to write the while loop is:

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while(s[i] != '\0') {
    // your switch case and code goes here
}

By fixing this, you should avoid the Time Limit Exceeded error.

However, there is another logical error in your code related to how you’re handling the roman numeral conversion. The switch-case construct doesn’t have an automatic break once a case is executed. This means if a case like ‘M’ or ‘D’ is hit, the code continues to execute the next cases (‘C’, ‘L’, ‘X’, ‘V’, ‘I’) as well, leading to incorrect results. To avoid this, you need to put a break statement at the end of each case:

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switch (s[i]){
    case 'M':
        // some code here
        break;
    case 'D':
        // some code here
        break;
    // similar for other cases
}

By adding these break statements, you ensure that the switch-case construct exits after executing a matching case.

Why is this not being accepted ? s=input()

d={“I”:1,“V”:5,“X”:10,“L”:50,“C”:100,“D”:500,“M”:1000} total=0 prev_value=0

for char in reversed(s) : if 1 <= len(s) <= 15 and (char in d): current_value = d[char] if current_value<prev_value : total-=current_value #LIX XIL

 else :
     total+=current_value
     prev_value=current_value

else : quit() if total in range(1,4000) : print(total)

can any one explain why the third test case is failing for my code class Solution { public: int romanToInt(string s) { int sum1=0; int sum2=0; int sum; int count=0; for(int k=0;s[k]!=’\0’;k++) { count++; } for(int i=0;i<count;i++) { if(s[i]==‘I’){ s[i]=1; } if(s[i]==‘V’){ s[i]=5; } if(s[i]==‘X’){ s[i]=10; } if(s[i]==‘L’){ s[i]=50; } if(s[i]==‘C’){ s[i]=100; } if(s[i]==‘D’){ s[i]=500; } if(s[i]==‘M’){ s[i]=1000; } } for(int j=0;j<count-1;j++){ if(s[j]<s[j+1]) { sum1=sum1+s[j]; } if(s[j]>=s[j+1]) { sum2=sum2+s[j]; } } sum2=sum2+s[count-1]; sum=sum2-sum1; return sum; } };

The issue in your code stems from the way you are trying to replace the Roman numeral characters (like ‘I’, ‘V’, ‘X’, etc.) with their integer counterparts. You are treating these characters as integers directly in the following statements:

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if(s[i]=='I'){
    s[i]=1;
}
// similar for other Roman numerals

In C++, this doesn’t work as you intend it to. It doesn’t convert the characters in the string to the corresponding integers, instead, it replaces the characters with the ASCII character that corresponds to the integer value (1, 5, 10, etc.). This is because strings in C++ are made up of characters, not integers.

Later on, when you try to do the operations s[j]<s[j+1] and s[j]>=s[j+1], you’re not comparing the integer values you think you are, but rather the ASCII character values that you’ve mistakenly inserted into the string.

Here is the corrected code using a std::map to map the Roman numerals to their integer counterparts and a separate integer vector to hold the converted values:

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class Solution {
public:
    int romanToInt(string s) {
        std::map<char, int> romanToInt = {{'I', 1}, {'V', 5}, {'X', 10}, {'L', 50}, {'C', 100}, {'D', 500}, {'M', 1000}};
        std::vector<int> intValues;
        for (char c : s) {
            intValues.push_back(romanToInt[c]);
        }
        int sum = 0;
        for (int i = 0; i < intValues.size() - 1; ++i) {
            if (intValues[i] < intValues[i + 1]) {
                sum -= intValues[i];
            } else {
                sum += intValues[i];
            }
        }
        sum += intValues.back();  // add the last value
        return sum;
    }
};

This code correctly converts the Roman numerals to integers and stores them in a separate integer vector. It then uses this vector to calculate the sum according to the rules of Roman numerals.