Consider a min heap containing all the integers from 1 to 1023 exactly once. If root is at depth 0, the maximum depth at which 9 can appear is?
The answer to the question is 8.
But, considering that a min heap is a nearly complete BT with-
1) for d <- 0 to h-1, all levels have 2^d nodes.
2) for d <- h, nodes are filled from left.
Source:http://homepages.math.uic.edu/~leon/cs-mcs401-s08/handouts/nearly_complete.pdf
What mistake is in answer being 4,as the level order traversal would be {1,2 3,4 5 6,7 8 9...}
The min-heap requires to put elements which are greater than their parent node.
Considering the question, one can put 1 as root, then 2 as its left child and any element greater than 9 (say 512) as its right child.For 2, one can continue in this way by putting 3 as left child and, say 513 as its right child. The final min heap obtained will be -
1
/ \
/ \
2 512
/ \ /\
/ \ / \
3 513 514 515
/\ /\ /\ /\
/ \
4 516 . . . . . .
/ / \ /\ /\ /\ /\ /\ /\
5 . . .. .. .. .. .. ...
/ /\ /\ /\/\
6 . . . . ...........................
/
7 .......................................................
/
8 ......................................................
/
9
The dots denote filled levels and can be replaced by elements from [517,758], as the levels must be filled.
The depth of 9 is 8
Related
Here are the combinations of different BSTs with same element, the arrangement looks different depending on the sequence of addition of nodes to the tree structure:
1 1 1 1 1
\ \ \ \ \
2 2 3 4 4
\ \ / \ / /
3 4 2 4 3 2
\ / / \
4 3 2 3
2 2 3 3 4 4 4
/ \ / \ / \ / \ / / /
1 3 1 4 2 4 1 4 3 2 3
\ / / \ / / \ /
4 3 1 2 2 1 3 1
/ \
1 2
4 4
/ /
1 1
\ \
2 3
\ /
3 2
Their in-order traversal will be same, so how do we distinguish them? Especially when there are more than one sequence of adding the nodes and they all generate the same structure, example 2,1,4,5,2,4,1,3,2,4,3,1.
You mentioned that all your examples have the same in-order traversal (1234), meaning inorder is not enough information to derive the tree structure. However, both post-order and pre-order sequences are sufficient for deriving the tree structure of a binary search tree.
For example, given the pre-order 2-1-3-4, the only binary search tree that satisfies this pre-order is your row 2 col 1 example. Compare that to pre-order 2-1-4-3 which would be your row 2 col 2 example.
For all of your examples, here are their pre-order notations
[1234] [1243] [1324] [1432] [1423]
[2134] [2143] [3214] [3124] [4321] [4213] [4312]
[4123] [4132]
As you can see, they are all distinct. You can repeat the same process for their post-order traversal, and you should get all distinct results.
Does a binary search beat an exponential search in any way, except in space complexity?
Both these algorithms search for a value in an ordered list of elements, but they address different issues. Exponential search is explicitly designed for unbounded lists whereas binary search deals with bounded lists.
The idea behind exponential search is very simple: Search for a bound, and then perform a binary search.
Example
Let's take an example. A = [1, 3, 7, 8, 10, 11, 12, 15, 19, 21, 22, 23, 29, 31, 37]. This list can be seen as a binary tree (although there is no need to build the tree):
15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
Binary search
A binary search for e = 27 (for example) will undergo the following steps
b0) Let T, R be the tree and its root respectively
15 (R)
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
b1) Compare e to R: e > 15. Let T, R be T right subtree and its root respectively
15
____/ \____
/ \
__8__ _23_(R)
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
b2) Compare e to R: e > 23. Let T, R be T right subtree and its root respectively
15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
b3) Compare e to R: e < 31. Let T, R be T left subtree and its root respectively
15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31___
/ \ / \ / \ / \
1 7 10 12 19 22 29 (R) 37
b4) Compare e to R: e <> 29: the element is not in the list, since T has no subtree.
Exponential search
An exponential search for e = 27 (for example) will undergo the following steps
Let T, R be the leftmost subtree (ie the leaf 1) and its root (1) respectively
15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31
/ \ / \ / \ / \
(R) 1 7 10 12 19 22 29 37
e1) Compare e to R: e > 1. Let R be the parent of R and T be the tree having R as root
15
____/ \____
/ \
__8__ _23__
/ \ / \
(R) 3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e2) Compare e to R: e > 3. Let R be the parent of R and T be the tree having R as root:
15
____/ \____
/ \
(R)_8__ _23__
/ \ / \
3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e3) Compare e to R: e > 8. Let R be the parent of R and T be the tree having R as root:
(R) 15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e4) Compare e to R: e > 15. R has no parent. Let T be the right subtree of T and R be its root:
15
____/ \____
/ \
__8__ _23_(R)
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e5..7) See steps b2..4)
Time complexity
For the sake of demonstration, let N = 2^n be the size of A and let indices start from 1. If N is not a power of two, the results are almost the same.
Let 0 <= i <= n be the minimum so that A[2^(i-1)] < e <= A[2^i] (let A[2^-1] = -inf). Note that this kind of interval may not be unique if you have duplicate values, hence the "minimum".
Exponential search
You need i + 1 iterations to find i. (In the example, you are jumping from child to parent repeatedly until you find a parent greater than e or there is no more parent)
Then you use a binary search on the selected interval. The size of this interval is 2^i - 2^(i-1) = 2^(i-1).
The cost of a binary search in an array of size 2^k is variable: you might find the value in the first iteration, or after k iterations (There are sophisticated analysis depending on the distribution of the elements, but basically, it's between 1 and k iterations and you can't know it in advance)
Let j_i, 1 <= j_i <= i - 1 be the number of iterations needed for the binary search in our case (The size of this interval is 2^(i-1)).
Binary search
Let i be the minimum so that A[2^(i-1)] < e <= A[2^i]. Because of the assumption that N = 2^n, the binary search will meet this interval:
We start with the root A[2^(n-1)]. If e > A[2^(n-1)], i = n because R = A[2^(n-1)] < e < A[2^n]. Else, we have e <= A[2^(n-1)]. If e > A[2^(n-2)], then i = n-1, else we continue until we find i.
You need n - i + 1 steps to find i using a binary search:
if i = n, you know it at the first iteration (e > R) else, you select the left subtree
if i = n-1, you need two iterations
and so on: if i = 0, you'll need n iterations.
Then you'll need j_i iterations as shown above to complete the search.
Comparison
As you see, the j_i iterations are common to both algorithms. The question is: Is i + 1 < n - i + 1? i.e. Is i < n - i or 2i < n? If yes, the exponential search will be faster than the binary search. If no, the binary search will be faster than the exponential search (or equally fast)
Let's get some distance: 2i < n is equivalent to (2^i)^2 < 2^n or 2^i < sqrt(2^n). While 2^i < sqrt(N), the exponential search is faster. As soon as 2^i > sqrt(N), the binary search is faster. Remember that the index of e is lower or equal than 2^i because e <= A[2^i].
In simple words, if you have N elements and if e is in the firstsqrt(N) elements, then exponential search will be faster, else binary search will be faster.
It depends on the distribution, but N - sqrt(N) > sqrt(N) if N > 4, and thus the binary search is likely to be faster than the exponential search unless you know that the element will be among the first ones or the list is ridiculously short.
If 2^n < N < 2^(n+1)
I won't go into details, but this does not change the general conclusion.
If the value is beyond the last power of two, the cost of exponential to find the bound is already n+2, more than the binary search (less than or equal to 2^(n+1)). Then you have a binary search to perform, maybe in a small interval, but binary search is already the winner.
Else you add the value A[N] to the list until you have 2^(n+1) value. This won't change anything for exponential search, and this will slow down the binary search. But this slow binary search remains faster if e is not in the firstsqrt(2^(n+1)) values.
Space complexity
That's an interesting question which I don't talk about, size of the pointer and things like that. If you are performing an exponential search and consuming elements as they arrive (imagine timestamps), you don't need to store the whole list at once. You just have to store one element (the first), then one element (the second), then two elements (the third and the fourth), then four elements, ... then 2^(i-1) elements. If i is small, then you won't need to store a large list as in a regular binary search.
Implementation
Implementation is really not a problem here. See the Wikipedia pages for information: Binary search algorithm and Exponential search.
Applications and how to choose among the two
Use the exponential search only when the sequence is unbounded or when you know the value is likely to be among the first ones. Unbounded: I like the example of timestamps: they are strictly growing. You can imagine a server with stored timestamps. You can ask for n timestamps and you are looking for a specific timestamp. Ask 1, then 2, then 4, then 8,... timestamps and perform the binary search when one timestamps exceeds the value you are looking for.
In other cases, use the binary search.
Remark: the idea behind the first part of the exponential search has some applications:
Guess an integer number when the upper limit is unbounded: Try 1, 2, 4, 8, 16,... and narrow the guess when you exceed the number (this is exponential search);
Find a bridge to cross a river by a foggy day: Make 100 steps left. If you didn't find the bridge, return to the initial point and make 200 steps right. If you still didn't find the bridge, return to the initial point and make 400 steps left. Repeat until you find the bridge (or swim);
Comput a congestion window in the TCP slow start: Double the quantity of data sent until there is a congestion. The TCP congestion algorithms are in general more careful and perform something similar to a linear search in the second part of the algorithm, because exceeding tries have a cost here.
I wonder if there is any relation between data_array data position to tree_array data position.
int data[N];
int tree[M]; // lets M = 2^X-1, where X = nearest ceiling power of 2 to N;
void build_segment_tree();
I wonder if I can say n'th value of data[] is mapped with i'th value of tree[]. is there any mathematical resolution?
You certainly can. For example segment tree is used for it's capapbility to store
segment information.
Now you will see that if you want to create a segment tree out of N elements then
you will need ceil(log_2(N))+1 levels. And in the last level you will find all the
1 length-range or the single elements.
These elements will be precisely in the position (1-index) 2^ceil(log_2(N)) to 2^ceil(log_2(N))+N-1.
[1-8]
/ \
[1-4] [5-8]
/ \ / \
[1-2][3-4] [5-6][7-8]
/\ /\ /\ /\
[1][2] [3][4] [5][6] [7][8]
1-11
/ \
1-6 7-11
1-3 4-6 7-9 10-11
1-2 3 4-5 6 7-8 9 10 11
1 2 4 5 7 8
This answer is for only valid for segment tree of power of 2 elements.
But for other elements the elements are not necessarily organized.
So the answer will be false for N those are not power of 2.
On that case you can't find any formualitve rule.
1
/ \
2 7
/ \ / \
3 4 5 6
1
/ / \ \
3 4 5 6
I'm looking for an algorithm to reduce a tree by cutting out nodes at a specific level and save all the leaves. This example is very simple, so the function my have to be recursive - imagine cutting a tree at level 3 and eliminating nodes at levels 4,5,6,7 - but preserving the leaves by inserting them all at level 4.
I have a graph shaped as a triangle.
8
/ \
1 4
/ \ / \
4 2 0
/ \ / \ / \
9 1 9 4
In the above graph the longest path is {8, 4, 2, 9}
My current algorithm calculates the max number of the adjacent nodes and add it to the list, then calculates the sum of that list. This works in the above graph but won't work in situations such as this scenario:
8
/ \
0 1
/ \ / \
4 0 4
/ \ / \ / \
9 99 3 4
My algorithm will mistakenly go through {8,1,4,4} where the correct longest path is {8,0,4,99}
The only solution I can think of is Backtracking. Where I have to go through all the paths and calculate the max path, which will be insanely slow in a huge graph. This about a 100k nodes graph.
My question is can I do better than this?
Start at the top.
For each node, pick the maximum of its parents (the nodes above connected to it) and add its own value.
Then, in the last row, pick the maximum.
This will just give you the value of the longest path, but you could easily get the actual path by simply starting at the value picked at the bottom and moving upwards, always picking the greater parent.
The running time would be linear in the number of nodes.
Example:
Original:
First example: Second example:
8 8
/ \ / \
1 4 0 1
/ \ / \ / \ / \
4 2 0 4 0 4
/ \ / \ / \ / \ / \ / \
9 1 9 4 9 99 3 4
Output:
First example: Second example:
8 8
/ \ / \
9 12 8 9
/ \ / \ / \ / \
13 14 12 12 9 13
/ \ / \ / \ / \ / \ / \
22 15 23 16 21 111 16 17
Then you'd pick 23 for the first and 111 for the second.
To get the path, we'd have 23-14-12-8, which corresponds to 9-2-4-8, for the first, and 111-12-8-8, which corresponds to 99-4-0-8, for the second.
I'm of course assuming we have a tree, as stated. For general graphs, this problem is quite a lot more difficult - NP-hard, to be exact.
You do not need backtracking here - you can use breadth-first search to propagate the max for the path that you have found so far to the corresponding node, level by level.
Start at the root, and set its max to its own value.
Go through nodes level-by-level
For each node check the max stored in its parent. There may be one or two of these parents. Pick the max of two max-es, add the value of the node itself, and store it in the current node
When the path through the graph is complete, the result would look like this:
Max graph:
8
/ \
8 9
/ \ / \
12 9 13
/ \ / \ / \
21 111 16 17
To recover a path, find the max value in the bottom layer. This is the final node of your path. You can reconstruct the path from the max graph and the original by starting at the max (111), subtracting the value (99), looking for the result (111-99=12) in the max graph, and continuing to that node until you reach the top:
111 - 99 = 12 -- Take 99
12 - 4 = 8 -- Take 4
8 - 0 = 8 -- Take 0
8 is the root -- Take 8
This gives you the max path in reverse. Note that this may not be the unique path (think of a graph filled with equal values to see how there may be multiple max paths). In this case, however, any path that you would recover will satisfy the max path requirement.