Answer:
B.O(n).
Explanation:
Since the time complexity of visiting a node is O(1) in iterative implementation.So the time complexity of visiting every single node in binary tree is O(n).We can use level order traversal of a binary tree using a queue.Which can visit every node in O(n) time.Level order traversal do it in a single loop without doing any extra traversal.
Answer:
2(101101) - 2(1100) = 200002
Explanation:
the text is everywhere don't know what to answer
Answer:
C. The experience of the park with The Gates was so different that some people saw the park in a new way.
Explanation:
The couple, Christo Javacheff and Jeanne-Claude Denat based in New York were two top contemporary artists who formed collaborative arts.
On the 3rd of January, 2005 they both made an art work known as The Gates, in Central Park.
The experience of the park with "The Gates" was so different that some people saw the park in a new way.
Answer:
The solution code is written in Python 3:
- def modifyList(listNumber):
- posCount = 0
- negCount = 0
-
- for x in listNumber:
- if x > 0:
- posCount += 1
- else:
- negCount += 1
-
- if(posCount == len(listNumber)):
- listNumber.append(max(listNumber))
-
- if(negCount == len(listNumber)):
- listNumber.append(min(listNumber))
-
- print(listNumber)
-
- modifyList([-1,-99,-81])
- modifyList([1,99,8])
- modifyList([-1,99,-81])
Explanation:
The key step to solve this problem is to define two variables, posCount and negCount, to track the number of positive value and negative value from the input list (Line 2 - 3).
To track the posCount and negCount, we can traverse through the for-loop and create if else statement to check if the current number x is bigger than 0 then increment posCount by 1 otherwise increment negCount (Line 5- 9).
If all number in the list are positive, the posCount should be equal to the length of the input list and the same rule is applied to negCount. If one of them happens, the listNumber will append either the maximum number (Line 11 -12) or append the minimum number (Line 14-15).
If both posCount and negCount are not equal to the list length, the block of code Line 11 -15 will be skipped.
At last we can print the listNumber (Line 17).
If we test our function using the three sets of input list, we shall get the following results:
[-1, -99, -81, -99]
[1, 99, 8, 99]
[-1, 99, -81]
