509. Fibonacci Number

Recursion

/*
 * @lc app=leetcode id=509 lang=cpp
 *
 * [509] Fibonacci Number
 *
 * https://leetcode.com/problems/fibonacci-number/description/
 *
 * algorithms
 * Easy (72.00%)
 * Likes:    8563
 * Dislikes: 381
 * Total Accepted:    2.3M
 * Total Submissions: 3.1M
 * Testcase Example:  '2'
 *
 * 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
 *
 *
 */

// @lc code=start
class Solution {
public:
    int fib(int n)
    {
        if(n < 2) return n;
        return fib(n - 1) + fib(n - 2);
    }
};
// @lc code=end

• T: $O(2^N)$
• S: $O(N)$

DP

class Solution {
public:
    int fib(int n)
    {
        if(n < 2) return n;
        vector<int> dp(n + 1);
        dp[0] = 0, dp[1] = 1;
        for(int i = 2; i < n + 1; i++)
        {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
};

• T: $O(N)$
• S: $O(N)$

DP improved

class Solution {
public:
    int fib(int n)
    {
        if(n < 2) return n;
        int first = 1, second = 1;
        for(int i = 2; i < n; i++)
        {
            int temp = first;
            first = first + second;
            second = temp;
        }
        return first;
    }
};

• T: $O(N)$
• S: $O(1)$