Here, this tutorial will help you to understand about Fizz Buzz solution of leetcode 412 problem with algorithm.
You are given an integer value lets say n. You have to return a string array result where:
result[i] == "FizzBuzz"ifiis divisible by3and5.result[i] == "Fizz"ifiis divisible by3.result[i== “Buzz” ifiis divisible by5.result[i] == i(as a string) if none of the above conditions are true.
Example 1:
Input: num = 4 result = ["1", "2", "Fizz", "4"] num = 5 result = ["1","2","Fizz","4","Buzz"]
Algorithm for Fizz Buzz problem:
- Iterate over number from 1 to n
- For each number, If It is divisible by 3 and 5 then store “
FizzBuzz” to string array. - For each number, If It is divisible by 3 then store “
Fizz” to string array. - For each number, If It is divisible by 5 then store “
Buzz” to string array. - Store number as string in array If none of the above condition is true.
- For each number, If It is divisible by 3 and 5 then store “
Fizz Buzz solution code in Go
package main
import (
"fmt"
"strconv"
)
func fizzBuzz(n int) []string {
str := []string{}
for i := 1; i <= n; i++ {
if i % 3 == 0 && i % 5 == 0 {
str = append(str, "FizzBuzz")
} else if i % 3 == 0 {
str = append(str, "Fizz")
} else if i % 5 == 0 {
str = append(str, "Buzz")
} else {
str = append(str, strconv.Itoa(i))
}
}
return str
}
func main() {
str := fizzBuzz(4)
for _, val := range(str) {
fmt.Println(val)
}
}
Fizz Buzz problem code in C++
#include <iostream>
#include <vector>
using namespace std;
vector<string> fizzBuzz(int n) {
vector<string> v;
for (int i = 1; i <= n; i++) {
if (i % 3 == 0 && i % 5 == 0) {
v.push_back("FizzBuzz");
} else if (i % 3 == 0) {
v.push_back("Fizz");
} else if (i % 5 == 0) {
v.push_back("Buzz");
} else {
v.push_back(to_string(i));
}
}
return v;
}
int main() {
vector<string> v = fizzBuzz(4);
for(auto it : v) {
cout<<it<<" ";
}
}
To learn more about C++ pls refer given below link:
https://www.techieindoor.com/category/golang/
References: