1
GATE CSE 2004
MCQ (Single Correct Answer)
+2
-0.6
The elements 32, 15, 20, 30, 12, 25, 16, are inserted one by one in the given
order into a max Heap. The resultant max Heap is
2
GATE CSE 1999
MCQ (Single Correct Answer)
+2
-0.6
If T1 = O(1), give the correct matching for the following pairs:
(N) Tn = Tn/2 + n
(O) Tn = Tn/2 + nlog n
(P) Tn = Tn - 1 + log n
(V) Tn = O(nlogn)
(W) Tn = O(n2)
(X) Tn = O(log2n)
List - I
(M) Tn = Tn - 1 + n(N) Tn = Tn/2 + n
(O) Tn = Tn/2 + nlog n
(P) Tn = Tn - 1 + log n
List - II
(U) Tn= O(n)(V) Tn = O(nlogn)
(W) Tn = O(n2)
(X) Tn = O(log2n)
3
GATE CSE 1996
MCQ (Single Correct Answer)
+2
-0.6
The minimum number of interchanges needed to convert the array
89, 19, 40, 17, 12, 10, 2, 5, 7, 11, 6, 9, 70
into a heap with the maximum element at the root is4
GATE CSE 1996
MCQ (Single Correct Answer)
+2
-0.6
The average number of key comparisons done on a successful sequential search
in list of length n is
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