23. Fibonacci series up to Nth term

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

My Approach

• Initialize an empty vector vec to store the Fibonacci series.

• Initialize variables n1 and n2 with initial Fibonacci values.

• If n is 0, push n1 into vec and return vec.

• If n is 1, push both n1 and n2 into vec and return vec.

• For i starting from 2 up to n, calculate the next Fibonacci number by adding the last two elements of vec, take modulo mod, and push it into vec.

• Return the vec vector containing the Fibonacci series.

Time and Auxiliary Space Complexity

• Time Complexity: The time complexity of this approach is O(N), because it iterates N times to generate the Fibonacci series.

• Auxiliary Space Complexity: The auxiliary space complexity is O(N), where N is the user input.

Code (C++)

class Solution {
  public:
    const int mod=1e9+7;
    vector<int> Series(int n)
    {
        vector<int>vec;
        int n1=0, n2=1;
        vec.push_back(n1);
        if (n==0)
            return vec;
        vec.push_back(n2);
        if (n==1)
            return vec;
        for (int i=2;i<=n;i++)
        {
            int num=(vec[i-1]+vec[i-2])%mod;
            vec.push_back(num);
        }
        return vec;
    }
};

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