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Fibonacci Number⚓︎

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Description⚓︎

The Fibonacci numbers, commonly denoted F(n) form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. That is,

  • F(0) = 0, F(1) = 1
  • F(n) = F(n - 1) + F(n - 2), for n > 1.

Given n, calculate F(n).

Example 1:

  • Input: n = 2
  • Output: 1
  • Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.

Example 2:

  • Input: n = 3
  • Output: 2
  • Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.

Example 3:

  • Input: n = 4
  • Output: 3
  • Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.

Constraints:

  • 0 <= n <= 30

Solution⚓︎

Way 1 (Memorization: Top-Down)⚓︎

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class Solution {
private:
    int Fibonacci(int i, vector<int>& dp) {
        if (i <= 1) return i;
        if (dp[i] != -1) return dp[i];
        int res = Fibonacci(i - 1, dp) + Fibonacci(i - 2, dp);
        dp[i] = res;
        return res;
    }
public:
    int fib(int n) {
        vector<int> dp(n + 1, -1);
        return Fibonacci(n, dp);
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(n)\).

Way 2 (Bottom-Up)⚓︎

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class Solution {
public:
    int fib(int n) {
        if (n <= 1) return n;
        vector<int> dp(n + 1);
        dp[0] = 0;
        dp[1] = 1;
        for (int i = 2; i <= n; i++)
            dp[i] = dp[i - 1] + dp[i - 2];
        return dp[n];
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(n)\).

Way 3⚓︎

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class Solution {
public:
    int fib(int n) {
        if (n <= 1) return n;
        int dp[2];
        dp[0] = 0;
        dp[1] = 1;
        for (int i = 2; i <= n; i++) {
            int sum = dp[0] + dp[1];
            dp[0] = dp[1];
            dp[1] = sum;
        }
        return dp[1];
    }
};
  • Time complexity: \(O(n)\);
  • Space complexity: \(O(1)\).