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    <p>Let </p> <p><strong>L3</strong> = log to the base 3 <strong>L2</strong> = Log to the base 2</p> <p>Then the correct answer is <strong>O(L3(L2(n))</strong> and NOT O(L2(L2(n)).</p> <p>Start with <strong>x = x * 2</strong>. x will increase exponentially till it reaches n, thus making the time complexity O(L2(n))</p> <p>Now consider <strong>x = x * x</strong>. x increases faster than the above. In every iteration the value of x jumps to the square of its previous value. Doing some simple math, here is what we get:</p> <p>For x = 2 n = 4, iterations taken = 1 n = 16, iterations taken = 2 n = 256, iterations taken = 3 n = 65536, iterations taken = 4</p> <p>Thus, the time complexity is <strong>O(L2(L2(n))</strong>. You can verify this by putting values above values for n.</p> <p>Now coming to your problem, <strong>x = x * x * x</strong>. This will increase even faster than x = x * x. Here is the table:</p> <p>For x = 2 n = 8, iterations taken = 1 n = 512, iterations taken = 2 n = (512*512*512), iterations taken = 3 and so on</p> <p>If you look at this carefully, this turns out to be <strong>O(L3(L2(n))</strong>. L2(n) will get you the power of two, but since you are taking cube of x in every iteration, you will have to take log to the base 3 of it to find out the correct number of iteration taken.</p> <p>So I think the correct answer is <strong>O(log-to-base-3(log-to-base-2(n))</strong></p> <p>Generalizing this, if <strong>x = x * x * x * x * .. (k times)</strong>, then the time complexity is <strong>O(log-to-base-k(log-to-base-2(n)</strong></p>
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    1. COthe base of the logarithm is irrelevant as `log_a(x) = log_b(x)/log_b(a)` and `O(c·x) = O(x)` for `c = const`
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    2. COYeah, I realize that now. It is kinda confusing. It gives an impression that log_4(x) would run twice as fast as log_2(x).
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