Q14. 5. TIme and Space
Find the time complexity of the given code.
void pat(){
for (int i = 2; i <=n; i = pow(i, c)) {
//Print statement - O(1) operation
}
//Here fun is sqrt or cuberoot or any other constant root
for (int i = n; i > 1; i = fun(i)) {
//Print statement - O(1) operation
}
Why does the loop run logN times?
first iteration i is 2
in second iteration i will be 2^k
in third iteration i will be (2^k)^k = 2^(k^2)
in fourth iteration i will be 2^(k^3)
in i+1th iteration i will be 2^(k^(i))
for loop to stop
2^(k^(i) ) > n
take log base 2 both side
k^(i) > log2(n)
take log base k both side
i > log k log2(n)
so this option is correct->LogLog N
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