to be able to pop i elements out of a stack we must first check if the stack actually has i or more elements and return false if it doesnt. If it does, we will count i elements and make top point to the i+1 th element from the top of the stack
bool multipop(i, node* top)
{
node * temp = ⊤ int count = 1;
while(count<i)
{
if(temp->next == NULL)
return false;
else
{
temp = temp->next; count ++;
}
}
top = &temp;
return true;
}
this process takes O(i) time.
(a) Prove that n log^3 n is O(n^2). Prove that n^3 is not O(n^2 log n)....
1. Prove that log2(n) is O(n) 2. Prove that log(n!) is O(n log(n))
Using a recurrence relation, prove that the time complexity of the binary search is O(log n). You can use ^ operator to represent exponentiation operation. For example, 2^n represents 2 raised to the power of n.
What is the computational complexity of deleting an item from a heap? O(1) O(log n) O(n) O(n log n) O(n^2) (n squared)
poin (a) 20n-O(n) (c) n=o(log n) (e) log n!= 0(n log nioo) (b) 3(2) 2: 100
1) Which of the following are in O (n): a) n + lg n b) n + 2n c) n + n2 d) 1000 n + 4500 lg n + 54 n 2) What is a stack? Give an ADT. 3) How would you implement the push method for a Stack implemented as a Linked List? 4) How would you implement the pop method for a Stack implemented as a Linked List?
Prove this using the definition R7: log(n*) is O(log n) for any fixed x > 0
1. For each of the following, prove using the definition of O): (a) 7n + log(n) = O(n) (b) n2 + 4n + 7 =0(na) (c) n! = ((n") (d) 21 = 0(221)
O(log(log(N))) < O(log(N)) a. True b. False O(N ) < O(log(N)) a. True b. False O( N5) < O(N2 - 3N + 2) a. True b. False O(2N) < O(N2) a. True b. False
What is the time complexity of this code? I'm unsure if it is O(log(n)) or O(n). I think that the while loop is logn but the for loop that comes after runs the same number of times as the while loop. string toBinary(int num) { string binary = "", temp = ""; while (num != 0) { temp += to_string(num%2); num /= 2; for (int i = temp.size() - 1; i >= 0; i--) { binary += temp[i]; return binary;
n)2" log log(n)O(n)? I don't How does =n. VIn) T n understand how VITn) 2" log 7 -)? I know we can take out the T, because 1) Vn) T* n it's in our natural logarithm. It's a constant factor. but how does (n) show up in the denominator after it used to be in the numerator? I need to know how the expression (1) right on the left is equal to the expression (1) on the n)2" log log(n)O(n)?...