JAVA: What is the Big-O cost of the following code (n is a large positive integer)?...

Question

Question

JAVA: What is the Big-O cost of the following code (n is a large positive integer)?

int count = 0;
for (int i = 1; i < n; i *= 2) {
   for (int k = i; k <= i + 8; k ++ ) {
       count ++;
   }
}

engineering Computer-Science

Answer 1

Answer #1

For Outer Loop :There are n iterations, however, instead of simply incrementing, 'i' is increased by 2*itself each run. Thus the loop is log(n).

For Inner Loop :

for (int p = 0; p < log(N); p++)

for (int j = 0; j < pow(2, p) + 8 ; j++)

--> N*log2

total = N*log2*log(n)

Answer 2

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