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
My Solution
We can solve this problem with dynamic programming. Here we will use memoization. We have a helper that takes as arguments a number n and a integer array we will use as our cache. For the base cases, if n is 0 we return 0 and if n is 1 we return 1. If the cache has stored the result for n, we just return that. Otherwise we recursively call dp(n-1, cache) + dp(n-2, cache) as shown by the recurrence relation and store the result in cache and return it.
Time complexity is O(n) because there are n subproblems that we must solve before we get to the final answer. Space complexity is O(n) also due to the space required for the cache.
class Solution {
public int fib(int n) {
int[] cache = new int[n+1];
return dp(n, cache);
}
public int dp(int n, int[] cache) {
if (n == 1) {
return 1;
}
if (n == 0) {
return 0;
}
if (cache[n] != 0) {
return cache[n];
}
cache[n] = dp(n-1, cache) + dp(n-2, cache);
return cache[n];
}
}