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This approach iterates through numbers from 1 to n and applies conditional logic using modulus operations to determine if a number should be represented as "Fizz", "Buzz", or "FizzBuzz". If none of these conditions are met, the number itself is returned as a string.
Time Complexity: O(n) as we iterate through each number from 1 to n once.
Space Complexity: O(n) for the output array.
1#include <vector>
2#include <string>
3
4std::vector<std::string> fizzBuzz(int n) {
5 std::vector<std::string> result;
6 for (int i = 1; i <= n; ++i) {
7 if (i % 3 == 0 && i % 5 == 0) {
8 result.push_back("FizzBuzz");
9 } else if (i % 3 == 0) {
10 result.push_back("Fizz");
11 } else if (i % 5 == 0) {
12 result.push_back("Buzz");
13 } else {
14 result.push_back(std::to_string(i));
15 }
16 }
17 return result;
18}
19This C++ solution utilizes the vector class to dynamically build the result array. Strings are appended based on conditional checks for divisibility, making use of push_back to add elements.
This approach uses a hash map to store possible outputs, simplifying conditional checks. By mapping integers to their respective Fizz or Buzz values, we consolidate decision logic, reducing redundancy in the code.
Time Complexity: O(n), since each element is evaluated once.
Space Complexity: O(n) due to storage for result strings.
1using System;
using System.Collections.Generic;
public class Solution {
public IList<string> FizzBuzz(int n) {
string[] values = { "", "Fizz", "Buzz", "FizzBuzz" };
IList<string> result = new List<string>();
for (int i = 1; i <= n; i++) {
int index = (i % 15 == 0) ? 3 : (i % 3 == 0) ? 1 : (i % 5 == 0) ? 2 : 0;
result.Add(index == 0 ? i.ToString() : values[index]);
}
return result;
}
}
This C# solution also uses an array to manage potential outputs. Through index calculation based on divisibility, choice logic is centralized in an optimal and less repetitive manner.