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There are three possible operations on queue- Enqueue, Dequeue and MultiDequeue. MultiDequeue is calling Dequeue multiple times based on a global variable <math>k</math>. Since, the queue is initially empty, whatever be the order of these operations, there cannot be more no. of Dequeue operations than Enqueue operations. Hence, the total no. operations will be <math>n</math> only. | There are three possible operations on queue- Enqueue, Dequeue and MultiDequeue. MultiDequeue is calling Dequeue multiple times based on a global variable <math>k</math>. Since, the queue is initially empty, whatever be the order of these operations, there cannot be more no. of Dequeue operations than Enqueue operations. Hence, the total no. operations will be <math>n</math> only. | ||
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− | [[Category:Algorithms & Data Structures]] | + | [[Category:Algorithms & Data Structures Questions from GATE]] |
Consider the following operation along with Enqueue and Dequeue operations on queues, where k is a global parameter. <syntaxhighlight lang="c"> MultiDequeue(Q){ m = k while (Q is not empty) and (m > 0) { Dequeue(Q) m = m – 1 } } </syntaxhighlight> What is the worst case time complexity of a sequence of $n$ queue operations on an initially empty queue?
(A) $Θ(n)$
(B) $Θ(n + k)$
(C) $Θ(nk)$
(D) $Θ(n^2)$
There are three possible operations on queue- Enqueue, Dequeue and MultiDequeue. MultiDequeue is calling Dequeue multiple times based on a global variable <math>k</math>. Since, the queue is initially empty, whatever be the order of these operations, there cannot be more no. of Dequeue operations than Enqueue operations. Hence, the total no. operations will be <math>n</math> only.
Consider the following operation along with Enqueue and Dequeue operations on queues, where k is a global parameter. <syntaxhighlight lang="c"> MultiDequeue(Q){ m = k while (Q is not empty) and (m > 0) { Dequeue(Q) m = m – 1 } } </syntaxhighlight> What is the worst case time complexity of a sequence of $n$ queue operations on an initially empty queue?
(A) $Θ(n)$
(B) $Θ(n + k)$
(C) $Θ(nk)$
(D) $Θ(n^2)$
There are three possible operations on queue- Enqueue, Dequeue and MultiDequeue. MultiDequeue is calling Dequeue multiple times based on a global variable <math>k</math>. Since, the queue is initially empty, whatever be the order of these operations, there cannot be more no. of Dequeue operations than Enqueue operations. Hence, the total no. operations will be <math>n</math> only.