23(March) Fibonacci series up to Nth term

23. Fibonacci Series up to Nth Term

The problem can be found at the following link: Question Link

Problem Description

Given an integer n, return the Fibonacci series up to the nth term. Since the terms can become very large, return the terms modulo (10^9 + 7).

Example:

Input:

n = 5

Output:

0 1 1 2 3 5

Explanation: 0 1 1 2 3 5 is the Fibonacci series up to the 5th term.

My Approach

1. Initialization:

   • Create a vector ans to store the Fibonacci series up to the (n)th term.

   • Initialize the first two terms of the series, ans[0] and ans[1], as 0 and 1 respectively.

2. Fibonacci Calculation:

   • Iterate from i = 2 to n.

   • Calculate the (i)th term of the Fibonacci series as the sum of the previous two terms: (ans[i] = (ans[i - 1] + ans[i - 2]) % MOD), where MOD is (10^9 + 7).

3. Return:

   • Return the vector ans containing the Fibonacci series up to the (n)th term.

Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), as we iterate through the series up to the (n)th term.

• Expected Auxiliary Space Complexity: O(n), as we use a vector of size (n + 1) to store the Fibonacci series.

Code (C++)

class Solution {
public:
    const int MOD = 1e9 + 7;

    vector<int> Series(int n) {
        vector<int> ans(n + 1);
        ans[0] = 0;
        ans[1] = 1;

        for (int i = 2; i <= n; ++i) {
            ans[i] = (ans[i - 1] + ans[i - 2]) % MOD;
        }

        return ans;
    }
};

Contribution and Support

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