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GATE | GATE-IT-2004 | Question 53

  • Last Updated : 28 Jun, 2021

An array of integers of size n can be converted into a heap by adjusting the heaps rooted at each internal node of the complete binary tree starting at the node ⌊(n – 1) /2⌋, and doing this adjustment up to the root node (root node is at index 0) in the order ⌊(n – 1)/2⌋, ⌊(n – 3)/ 2⌋, ….., 0. The time required to construct a heap in this manner is
(A) O(log n)
(B) O(n)
(C) O (n log log n)
(D) O(n log n)


Answer: (B)

Explanation: The above statement is actually algorithm for building a Heap of an input array A.

BUILD-HEAP(A) 
    heapsize := size(A); 
    for i := floor(heapsize/2) downto 1 
        do HEAPIFY(A, i); 
    end for 
END

Upper bound of time complexity is O(n) for above algo

 

See- https://www.geeksforgeeks.org/g-fact-85/

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