The problem can be found at the following link: Question Link
Simple go with the flow of question statement.
- If n <= 2, return a vector containing n ones.
- Otherwise, initialize a vector 'out' of size n and set the first two elements to 1 (since the first two Fibonacci numbers are always 1).
- Use a loop to calculate the remaining Fibonacci numbers and store them in the 'out' vector.
- Time Complexity: The loop runs for
n
iterations to calculate the Fibonacci numbers, which isO(n)
. - Auxiliary Space Complexity: I use a vector of size
n
to store the Fibonacci numbers, which isO(n)
.
class Solution {
public:
vector<long long> printFibb(int n)
{
if(n <= 2)
return vector<long long>(n, 1);
vector<long long> out(n);
out[0] = out[1] = 1;
for(int i = 2; i < n; ++i) {
out[i] = out[i - 1] + out[i - 2];
}
return out;
}
};
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