grid47 Exploring patterns and algorithms
Nov 5, 2024
5 min read
Solution to LeetCode 12: Integer to Roman Problem
You are given a positive integer and need to convert it into a Roman numeral. Roman numerals are constructed by combining symbols from the set: I, V, X, L, C, D, M, where each symbol has a specific value. The conversion rules are based on subtractive notation for values like 4 (IV), 9 (IX), and so on, to avoid using symbols multiple times where not allowed.
Problem
Approach
Steps
Complexity
Input: The input is a single integer `num` (1 <= num <= 3999), which represents the number to be converted to a Roman numeral.
Example: Input: num = 1987
Constraints:
• 1 <= num <= 3999
Output: Return the Roman numeral representation of the given number.
Example: Output: 'MCMLXXXVII'
Constraints:
• The result should be a string representing the Roman numeral equivalent of the input number.
Goal: The goal is to break down the number into its decimal place values and convert them into Roman numerals according to the Roman numeral conversion rules.
Steps:
• Create a list of values for Roman numerals starting from the largest to the smallest.
• For each value, check if it can be subtracted from the current number and add the corresponding Roman numeral symbol to the result.
• If a subtractive form like 4 (IV) or 9 (IX) is needed, use those instead of repeating smaller symbols.
Goal: The solution should handle numbers efficiently up to 3999 and provide the correct Roman numeral string.
Steps:
• The number must be between 1 and 3999, inclusive.
Assumptions:
• The input number will always be a valid positive integer within the specified range.
• Input: Input: num = 1987
• Explanation: 1987 is split into 1000 (M), 900 (CM), 80 (LXXX), and 7 (VII). Therefore, the Roman numeral is 'MCMLXXXVII'.
• Input: Input: num = 58
• Explanation: 58 is split into 50 (L) and 8 (VIII). Therefore, the Roman numeral is 'LVIII'.
Approach: To convert the number to a Roman numeral, we use a greedy approach that subtracts the largest possible Roman numeral values from the number at each step, appending the corresponding symbols to the result string.
Observations:
• We should process the Roman numeral symbols from largest to smallest to maximize the efficiency of the conversion.
• Start with the largest value (1000, M) and work downwards. This avoids adding unnecessary symbols.
Steps:
• Define a list of Roman numeral symbols and their corresponding values in descending order.
• Iterate over the list of Roman numeral values. For each value, subtract it from the number if it is less than or equal to the remaining number, and append the corresponding symbol to the result string.
• Handle the subtractive cases (like 4, 9, etc.) as needed.
Empty Inputs:
• There are no empty inputs since the number will always be between 1 and 3999.
Large Inputs:
• The solution should work efficiently for the largest input (3999).
Special Values:
• Ensure that the subtractive notation is correctly applied, e.g., 4 (IV), 9 (IX), 40 (XL), etc.
Constraints:
• The solution should process the number in O(1) time since the number of possible Roman numeral symbols is constant.
Initialize a vector `chrs` to store the corresponding Roman numerals for each decimal value.
4 : Variable Initialization
int idx =0;
Initialize an index `idx` to keep track of the current position in the `nums` and `chrs` vectors.
5 : Variable Initialization
string res ="";
Initialize an empty string `res` to store the resulting Roman numeral.
6 : Loop Iteration
while(num >0)
Start a loop that continues while `num` is greater than 0.
7 : Loop Iteration
while(num >= nums[idx])
Start an inner loop that continues while `num` is greater than or equal to the current decimal value.
8 : Mathematical Operations
num -= nums[idx];
Subtract the current decimal value from `num`.
9 : String Manipulation
res += chrs[idx];
Append the corresponding Roman numeral to the `res` string.
10 : Index Update
idx++;
Increment the index `idx` to move to the next decimal value and its corresponding Roman numeral.
11 : Loop Iteration
}
End of the outer loop.
12 : Return Value
return res;
Return the final Roman numeral string `res`.
Best Case: O(1)
Average Case: O(1)
Worst Case: O(1)
Description: Since the number of Roman numeral symbols is fixed and there are only a few steps involved in the conversion, the time complexity is O(1).
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is O(1) because we are only using a fixed number of variables to store the result and intermediate values.
