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412. Fizz Buzz

Description

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 <= 104

Solutions

Solution 1: Simulation

Thinking

The output is a direct transcription of \(1..n\) under a few divisibility rules, so a single pass is enough.

Test divisibility by \(15\) before \(3\) or \(5\), then fall back to the decimal representation. Checking \(15\) first prevents writing only one of the two words when both apply.

We iterate through each integer from 1 to \(n\). For each integer, we check whether it is a multiple of both 3 and 5, or just a multiple of 3, or just a multiple of 5. Based on the check result, we add the corresponding string to the answer array.

The time complexity is \(O(n)\), where \(n\) is the integer given in the problem. Ignoring the space consumption of the answer array, the space complexity is \(O(1)\).

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class Solution:
    def fizzBuzz(self, n: int) -> List[str]:
        ans = []
        for i in range(1, n + 1):
            if i % 15 == 0:
                ans.append('FizzBuzz')
            elif i % 3 == 0:
                ans.append('Fizz')
            elif i % 5 == 0:
                ans.append('Buzz')
            else:
                ans.append(str(i))
        return ans

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class Solution {
    public List<String> fizzBuzz(int n) {
        List<String> ans = new ArrayList<>();
        for (int i = 1; i <= n; ++i) {
            String s = "";
            if (i % 3 == 0) {
                s += "Fizz";
            }
            if (i % 5 == 0) {
                s += "Buzz";
            }
            if (s.length() == 0) {
                s += i;
            }
            ans.add(s);
        }
        return ans;
    }
}

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class Solution {
public:
    vector<string> fizzBuzz(int n) {
        vector<string> ans;
        for (int i = 1; i <= n; ++i) {
            string s = "";
            if (i % 3 == 0) {
                s += "Fizz";
            }
            if (i % 5 == 0) {
                s += "Buzz";
            }
            if (s.empty()) {
                s = to_string(i);
            }
            ans.push_back(s);
        }
        return ans;
    }
};

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func fizzBuzz(n int) []string {
    ans := make([]string, 0, n)
    for i := 1; i < n+1; i++ {
        switch {
        case i%15 == 0:
            ans = append(ans, "FizzBuzz")
        case i%3 == 0:
            ans = append(ans, "Fizz")
        case i%5 == 0:
            ans = append(ans, "Buzz")
        default:
            ans = append(ans, strconv.Itoa(i))
        }
    }
    return ans
}

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/**
 * @param {number} n
 * @return {string[]}
 */
var fizzBuzz = function (n) {
    const ans = [];
    for (let i = 1; i <= n; ++i) {
        if (i % 15 === 0) {
            ans.push('FizzBuzz');
        } else if (i % 3 === 0) {
            ans.push('Fizz');
        } else if (i % 5 === 0) {
            ans.push('Buzz');
        } else {
            ans.push(`${i}`);
        }
    }
    return ans;
};

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class Solution {
    /**
     * @param Integer $n
     * @return String[]
     */
    function fizzBuzz($n) {
        $ans = [];
        for ($i = 1; $i <= $n; ++$i) {
            $s = '';
            if ($i % 3 == 0) {
                $s .= 'Fizz';
            }
            if ($i % 5 == 0) {
                $s .= 'Buzz';
            }
            if (strlen($s) == 0) {
                $s .= $i;
            }
            $ans[] = $s;
        }
        return $ans;
    }
}

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