Given an integer n, return a string array answer (1-indexed) where:
answer[i] == "FizzBuzz " if i is divisible by 3 and 5.answer[i] == "Fizz " if i is divisible by 3.answer[i] == "Buzz " if i is divisible by 5.answer[i] == i (as a string) if none of the above conditions are true.Example 1:
Input: n = 3 Output: ["1","2","Fizz"]
Example 2:
Input: n = 5 Output: ["1","2","Fizz","4","Buzz"]
Example 3:
Input: n = 15 Output: ["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14","FizzBuzz"]
Constraints:
1 <= n <= 104program fizzbuzz
implicit none
integer, parameter :: n = 15
character(len=10), dimension(n) :: answer
integer :: i
do i = 1, n
if (mod(i, 3) == 0 .and. mod(i, 5) == 0) then
answer(i) = "FizzBuzz "
else if (mod(i, 3) == 0) then
answer(i) = "Fizz "
else if (mod(i, 5) == 0) then
answer(i) = "Buzz "
else
write (answer(i), '(I10)') i
end if
end do
print *, answer
end program fizzbuzz1 2Fizz 4Buzz Fizz 7 8Fizz Buzz 11Fizz 13 14FizzBuzz
module fizzbuzz
contains
function fizzbuzz(n) result(answer)
integer, intent(in) :: n
character(len=10) :: answer(n)
integer :: i
do i = 1, n
if (mod(i, 3) == 0 .and. mod(i, 5) == 0) then
answer(i) = "FizzBuzz "
else if (mod(i, 3) == 0) then
answer(i) = "Fizz "
else if (mod(i, 5) == 0) then
answer(i) = "Buzz "
else
answer(i) = trim(adjustl(str(i))) // " "
end if
end do
end function fizzbuzz
end module fizzbuzz
program main
use fizzbuzz
implicit none
integer :: n
do n = 1, 10
write (*, *) fizzbuzz(n)
end do
end program main
temp.f95:5:17:
5 | function fizzbuzz(n) result(answer)
| 1
Error: MODULE attribute of ‘fizzbuzz’ conflicts with PROCEDURE attribute at (1)
temp.f95:6:28:
6 | integer, intent(in) :: n
| 1
Error: Unexpected data declaration statement in CONTAINS section at (1)
temp.f95:7:34:
7 | character(len=10) :: answer(n)
| 1
Error: Explicit shaped array with nonconstant bounds at (1)
temp.f95:9:16:
9 | integer :: i
| 1
Error: Unexpected data declaration statement in CONTAINS section at (1)
temp.f95:11:15:
11 | do i = 1, n
| 1
Error: Unexpected DO statement in CONTAINS section at (1)
temp.f95:12:53:
12 | if (mod(i, 3) == 0 .and. mod(i, 5) == 0) then
| 1
Error: Unexpected block IF statement in CONTAINS section at (1)
temp.f95:13:35:
13 | answer(i) = "FizzBuzz "
| 1
Error: Unexpected assignment statement in CONTAINS section at (1)
temp.f95:14:37:
14 | else if (mod(i, 3) == 0) then
| 1
Error: Unexpected ELSE IF statement in CONTAINS section at (1)
temp.f95:15:31:
15 | answer(i) = "Fizz "
| 1
Error: Unexpected assignment statement in CONTAINS section at (1)
temp.f95:16:37:
16 | else if (mod(i, 5) == 0) then
| 1
Error: Unexpected ELSE IF statement in CONTAINS section at (1)
temp.f95:17:31:
17 | answer(i) = "Buzz "
| 1
Error: Unexpected assignment statement in CONTAINS section at (1)
temp.f95:18:12:
18 | else
| 1
Error: Unexpected ELSE statement in CONTAINS section at (1)
temp.f95:19:52:
19 | answer(i) = trim(adjustl(str(i))) // " "
| 1
Error: Unexpected assignment statement in CONTAINS section at (1)
temp.f95:20:11:
20 | end if
| 1
Error: Expecting END MODULE statement at (1)
temp.f95:21:7:
21 | end do
| 1
Error: Expecting END MODULE statement at (1)
temp.f95:22:3:
22 | end function fizzbuzz
| 1
Error: Expecting END MODULE statement at (1)
temp.f95:27:9:
27 | use fizzbuzz
| 1
Fatal Error: Cannot open module file ‘fizzbuzz.mod’ for reading at (1): No such file or directory
compilation terminated.
def fizzBuzz(n):
result = []
for i in range(1, n + 1):
if i % 15 == 0:
result.append("FizzBuzz")
elif i % 3 == 0:
result.append("Fizz")
elif i % 5 == 0:
result.append("Buzz")
else:
result.append(str(i))
return result
The algorithm for this problem simply consists of iterating over the integers from 1 to n, checking the divisibility rules for FizzBuzz, and inserting the appropriate string into the array for each index. For each number i in the range [1,n], we check if it's divisible by 3 and 5 (i % 15 == 0), just 3 (i % 3 == 0), just 5 (i % 5 == 0), or none of the above. Based on the conditions met, we append the corresponding string to the array (i.e., "FizzBuzz", "Fizz", "Buzz", or the number itself as a string).
The code implementations in each language follow their respective syntax, but the core logic is the same across all languages.
#include <vector>
#include <string>
std::vector<std::string> fizzBuzz(int n) {
std::vector<std::string> result(n);
for(int i = 1; i <= n; ++i) {
if (i % 15 == 0) result[i - 1] = "FizzBuzz";
else if (i % 3 == 0) result[i - 1] = "Fizz";
else if (i % 5 == 0) result[i - 1] = "Buzz";
else result[i - 1] = std::to_string(i);
}
return result;
}
The algorithm for this problem simply consists of iterating over the integers from 1 to n, checking the divisibility rules for FizzBuzz, and inserting the appropriate string into the array for each index. For each number i in the range [1,n], we check if it's divisible by 3 and 5 (i % 15 == 0), just 3 (i % 3 == 0), just 5 (i % 5 == 0), or none of the above. Based on the conditions met, we append the corresponding string to the array (i.e., "FizzBuzz", "Fizz", "Buzz", or the number itself as a string).
The code implementations in each language follow their respective syntax, but the core logic is the same across all languages.