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In the book Programming Interviews Exposed it says that the complexity of the program below is O(N), but I don't understand how this is possible. Can someone explain why this is?
int var = 2;
for (int i = 0; i < N; i++) {
for (int j = i+1; j < N; j *= 2) {
var += var;
}
}
315
Every time the outer loop executes, the inner loop executes M times. As a result, the statements in the inner loop execute a total of N * M times. Thus, the total complexity for the two loops is O(N2).
the size in the for loop is increasing every time the for loop is being executed, starting at 1,2,...,n-1 and the while loop runs n-1 times. That means that the time complexity is O(n^3)?
The loop may, at first glance, look like O(n^2) but the inner loop maximally creates 5 iterations for every iteration of the outer loop. So we receive only 5 * n iterations in total, not something like n^2 . The complexity thus is O(n) .
These are typically used for working with two dimensions such as printing stars in rows and columns as shown below. When a loop is nested inside another loop, the inner loop runs many times inside the outer loop. In each iteration of the outer loop, the inner loop will be re-started.
You need a bit of math to see that. The inner loop iterates Θ(1 + log [N/(i+1)]) times (the 1 + is necessary since for i >= N/2, [N/(i+1)] = 1 and the logarithm is 0, yet the loop iterates once). j takes the values (i+1)*2^k until it is at least as large as N, and
(i+1)*2^k >= N <=> 2^k >= N/(i+1) <=> k >= log_2 (N/(i+1))
using mathematical division. So the update j *= 2 is called ceiling(log_2 (N/(i+1))) times and the condition is checked 1 + ceiling(log_2 (N/(i+1))) times. Thus we can write the total work
N-1 N
∑ (1 + log (N/(i+1)) = N + N*log N - ∑ log j
i=0 j=1
= N + N*log N - log N!
Now, Stirling's formula tells us
log N! = N*log N - N + O(log N)
so we find the total work done is indeed O(N).
179
Outer loop runs n times. Now it all depends on the inner loop.
The inner loop now is the tricky one.
Lets follow:
i=0 --> j=1 ---> log(n) iterations
...
...
i=(n/2)-1 --> j=n/2 ---> 1 iteration.
i=(n/2) --> j=(n/2)+1 --->1 iteration.
i > (n/2) ---> 1 iteration
(n/2)-1 >= i > (n/4) ---> 2 iterations
(n/4) >= i > (n/8) ---> 3 iterations
(n/8) >= i > (n/16) ---> 4 iterations
(n/16) >= i > (n/32) ---> 5 iterations
(n/2)*1 + (n/4)*2 + (n/8)*3 + (n/16)*4 + ... + [n/(2^i)]*i
N-1
n*∑ [i/(2^i)] =< 2*n
i=0
--> O(n)
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