Consider the following function
Answer : B
Explanation :
Loop A runs Log N times if you watch closely.
[ J becomes 2, 4, 8, 16, ...., 2^X. It stops when 2^X = N. That is X = Log N.]
Loop B runs for n/2 times. Each time Loop A runs it increments K by n/2.
So, value of K is approx, n/2 times the statement inside A runs.
=> n/2 * Number of times Loop B runs * Number of times Loop A runs
=> n/2 * n/2 * Log N
=> O(n^2 Log N)
Topics to revise :
C language
References :
None as of now
Explanation :
Loop A runs Log N times if you watch closely.
[ J becomes 2, 4, 8, 16, ...., 2^X. It stops when 2^X = N. That is X = Log N.]
Loop B runs for n/2 times. Each time Loop A runs it increments K by n/2.
So, value of K is approx, n/2 times the statement inside A runs.
=> n/2 * Number of times Loop B runs * Number of times Loop A runs
=> n/2 * n/2 * Log N
=> O(n^2 Log N)
Topics to revise :
C language
References :
None as of now
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