The order of growth execution time of the push operation when using the LinkedStack class, assuming a stack size of N is; O(1)
<h3>Understanding Computer Programming Language</h3>
Let Top be the position of last element inserted in Array.
For Push operation, the process is as follows;
if(Top == last index of Array) {
Printf(“Stack Overflow”)
} else {
Top = Top + 1
a[Top] = element you want to insert
}
Read more about Computer Programming at; brainly.com/question/22654163
Answer:
Explanation:
The minimum depth occurs for the path that always takes the smaller portion of the
split, i.e., the nodes that takes α proportion of work from the parent node. The first
node in the path(after the root) gets α proportion of the work(the size of data
processed by this node is αn), the second one get (2)
so on. The recursion bottoms
out when the size of data becomes 1. Assume the recursion ends at level h, we have
(ℎ) = 1
h = log 1/ = lg(1/)/ lg = − lg / lg
Maximum depth m is similar with minimum depth
(1 − )() = 1
m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )
Answer: ANSWER is B I'm pretty sure
Explanation: hope this help
s :)
