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Asked: 2026-05-29T05:05:59+00:00

I need to determine the complexity of the pseudocode I wrote while root ≠

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I need to determine the complexity of the pseudocode I wrote

while root ≠ null
    while hasChild(root)
        push(parentTree) ← root
        root ← pop(getChilds(root))
        ...
    is parentTree isEmpty
        root ← null
    else    
        root ← pop(parentTree)

How can I know the number of execution (for each line) in a worst case scenario ?

I am not able to determine it, because I actually does not know the first two lines. After, it’s easy, but I don’t know the count for the two first lines…

It’s a tree implementation using a stack, and root is the root node, as you see.

By the way, it’s the first time I write pseudo code, so I am not sure i wrote it in a good way. If it’s not correct, I can rewrite it.

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1 Answer

  1. Editorial Team

    prima facie analysis leads me to think the runtime is O(logn*logn)

    Reasoning:
    Outer while loop executes at most clogn times (where c is a constant). This is due to the fact that it relies on the ‘root’ variable, which in turn relies on the ‘pop parenttree’
    parent tree only gets populated with the ‘original’ root’s grandchildren, iteratively. At most it will have all the children down one path in the tree. Its known that the length of a single path down a tree is logn

    Inner while loop also executes at most d logn times (d is constant), if the … does not execute in O(1) then it would execute in dlogn+X, and the overall runtime would be O(logn*(logn+X)), likely simplifying to O(Xlogn)

    Assuming the is is an if, the if/else statements run in O(1)

    Outer*Inner = O(clogn*dlogn)

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