23. Rohan's Love for Matrix

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

My Approach

• Initialize two variables prev1 and prev2 to store the values of the first and second elements of the matrix respectively.

• Iterate from 2 to n to calculate the nth power of the matrix.

• Inside the loop, calculate the next element curr of the matrix as the sum of the previous two elements (prev1 and prev2) modulo 1000000007.

• Update prev1 and prev2 with the values of prev2 and curr respectively.

• After the loop, return the value of prev2, which represents the first element of the resulting matrix an.

Time and Auxiliary Space Complexity

• Time Complexity : O(N)

• Auxiliary Space Complexity : O(1)

Code (C++)

class Solution {
  public:
    int firstElement(int n)
    {
        int prev1=0, prev2=1;
        for (int i=2;i<=n;i++)
        {
            int curr=(prev1+prev2)%1000000007;
            prev1=prev2;
            prev2=curr;
        }
        return prev2;
    }
};

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