The code segment below prints some of the elements in an array with size n. Select an expression to

complete the code segment so that the resulting algorithm has O(log n) running time.

for

System. out. println(array[j]);

int j = 1; j < array.length; j =j *2

Explanation:

j=j*2 loops have O(logn) so the running time of algorithm is O(logn).

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i guess the correct option is letter a.

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