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I have a histogram with integer heights and constant width 1. I want to maximize the rectangular area under a histogram.
e.g.:
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The answer for this would be 6, 3 * 2, using col1 and col2.
O(n^2) brute force is clear to me, I would like an O(n log n) algorithm. I'm trying to think dynamic programming along the lines of maximum increasing subsequence O(n log n) algo, but am not going forward. Should I use divide and conquer algorithm?
PS: People with enough reputation are requested to remove the divide-and-conquer tag if there is no such solution.
After mho's comments: I mean the area of largest rectangle that fits entirely. (Thanks j_random_hacker for clarifying :) ).
The above answers have given the best O(n) solution in code, however, their explanations are quite tough to comprehend. The O(n) algorithm using a stack seemed magic to me at first, but right now it makes every sense to me. OK, let me explain it.
First observation:
To find the maximal rectangle, if for every bar x, we know the first smaller bar on its each side, let's say l and r, we are certain that height[x] * (r - l - 1) is the best shot we can get by using height of bar x. In the figure below, 1 and 2 are the first smaller of 5.
OK, let's assume we can do this in O(1) time for each bar, then we can solve this problem in O(n)! by scanning each bar.
Then, the question comes: for every bar, can we really find the first smaller bar on its left and on its right in O(1) time? That seems impossible right? ... It is possible, by using a increasing stack.
Why using an increasing stack can keep track of the first smaller on its left and right?
Maybe by telling you that an increasing stack can do the job is not convincing at all, so I will walk you through this.
Firstly, to keep the stack increasing, we need one operation:
while x < stack.top():
stack.pop()
stack.push(x)
Then you can check that in the increasing stack (as depicted below), for stack[x], stack[x-1] is the first smaller on its left, then a new element that can pop stack[x] out is the first smaller on its right.
Still can't believe stack[x-1] is the first smaller on the left on stack[x]?
I will prove it by contradiction.
First of all, stack[x-1] < stack[x] is for sure. But let's assume stack[x-1] is not the first smaller on the left of stack[x].
So where is the first smaller fs?
If fs < stack[x-1]:
stack[x-1] will be popped out by fs,
else fs >= stack[x-1]:
fs shall be pushed into stack,
Either case will result fs lie between stack[x-1] and stack[x], which is contradicting to the fact that there is no item between stack[x-1] and stack[x].
Therefore stack[x-1] must be the first smaller.
Summary:
Increasing stack can keep track of the first smaller on left and right for each element. By using this property, the maximal rectangle in histogram can be solved by using a stack in O(n).
Congratulations! This is really a tough problem, I'm glad my prosaic explanation didn't stop you from finishing. Attached is my proved solution as your reward :)
def largestRectangleArea(A):
ans = 0
A = [-1] + A
A.append(-1)
n = len(A)
stack = [0] # store index
for i in range(n):
while A[i] < A[stack[-1]]:
h = A[stack.pop()]
area = h*(i-stack[-1]-1)
ans = max(ans, area)
stack.append(i)
return ans
There are three ways to solve this problem in addition to the brute force approach. I will write down all of them. The java codes have passed tests in an online judge site called leetcode: http://www.leetcode.com/onlinejudge#question_84. so I am confident codes are correct.
Solution 1: dynamic programming + n*n matrix as cache
time: O(n^2), space: O(n^2)
Basic idea: use the n*n matrix dp[i][j] to cache the minimal height between bar[i] and bar[j]. Start filling the matrix from rectangles of width 1.
public int solution1(int[] height) {
int n = height.length;
if(n == 0) return 0;
int[][] dp = new int[n][n];
int max = Integer.MIN_VALUE;
for(int width = 1; width <= n; width++){
for(int l = 0; l+width-1 < n; l++){
int r = l + width - 1;
if(width == 1){
dp[l][l] = height[l];
max = Math.max(max, dp[l][l]);
} else {
dp[l][r] = Math.min(dp[l][r-1], height[r]);
max = Math.max(max, dp[l][r] * width);
}
}
}
return max;
}
Solution 2: dynamic programming + 2 arrays as cache.
time: O(n^2), space: O(n)
Basic idea: this solution is like solution 1, but saves some space. The idea is that in solution 1 we build the matrix from row 1 to row n. But in each iteration, only the previous row contributes to the building of the current row. So we use two arrays as previous row and current row by turns.
public int Solution2(int[] height) {
int n = height.length;
if(n == 0) return 0;
int max = Integer.MIN_VALUE;
// dp[0] and dp[1] take turns to be the "previous" line.
int[][] dp = new int[2][n];
for(int width = 1; width <= n; width++){
for(int l = 0; l+width-1 < n; l++){
if(width == 1){
dp[width%2][l] = height[l];
} else {
dp[width%2][l] = Math.min(dp[1-width%2][l], height[l+width-1]);
}
max = Math.max(max, dp[width%2][l] * width);
}
}
return max;
}
Solution 3: use stack.
time: O(n), space:O(n)
This solution is tricky and I learnt how to do this from explanation without graphs and explanation with graphs. I suggest you read the two links before reading my explanation below. It's hard to explain without graphs so my explanations might be hard to follow.
Following are my explanations:
For each bar, we must be able to find the biggest rectangle containing this bar. So the biggest one of these n rectangles is what we want.
To get the biggest rectangle for a certain bar (let's say bar[i], the (i+1)th bar), we just need to find out the biggest interval
that contains this bar. What we know is that all the bars in this interval must be at least the same height with bar[i]. So if we figure out how many
consecutive same-height-or-higher bars are there on the immediate left of bar[i], and how many consecutive same-height-or-higher bars are there on the immediate right of the bar[i], we
will know the length of the interval, which is the width of the biggest rectangle for bar[i].
To count the number of consecutive same-height-or-higher bars on the immediate left of bar[i], we only need to find the closest bar on the left that is shorter
than the bar[i], because all the bars between this bar and bar[i] will be consecutive same-height-or-higher bars.
We use a stack to dynamicly keep track of all the left bars that are shorter than a certain bar. In other words, if we iterate from the first bar to bar[i], when we just arrive at the bar[i] and haven't updated the stack,
the stack should store all the bars that are no higher than bar[i-1], including bar[i-1] itself. We compare bar[i]'s height with every bar in the stack until we find one that is shorter than bar[i], which is the cloest shorter bar.
If the bar[i] is higher than all the bars in the stack, it means all bars on the left of bar[i] are higher than bar[i].
We can do the same thing on the right side of the i-th bar. Then we know for bar[i] how many bars are there in the interval.
public int solution3(int[] height) {
int n = height.length;
if(n == 0) return 0;
Stack<Integer> left = new Stack<Integer>();
Stack<Integer> right = new Stack<Integer>();
int[] width = new int[n];// widths of intervals.
Arrays.fill(width, 1);// all intervals should at least be 1 unit wide.
for(int i = 0; i < n; i++){
// count # of consecutive higher bars on the left of the (i+1)th bar
while(!left.isEmpty() && height[i] <= height[left.peek()]){
// while there are bars stored in the stack, we check the bar on the top of the stack.
left.pop();
}
if(left.isEmpty()){
// all elements on the left are larger than height[i].
width[i] += i;
} else {
// bar[left.peek()] is the closest shorter bar.
width[i] += i - left.peek() - 1;
}
left.push(i);
}
for (int i = n-1; i >=0; i--) {
while(!right.isEmpty() && height[i] <= height[right.peek()]){
right.pop();
}
if(right.isEmpty()){
// all elements to the right are larger than height[i]
width[i] += n - 1 - i;
} else {
width[i] += right.peek() - i - 1;
}
right.push(i);
}
int max = Integer.MIN_VALUE;
for(int i = 0; i < n; i++){
// find the maximum value of all rectangle areas.
max = Math.max(max, width[i] * height[i]);
}
return max;
}
Implementation in Python of the #IVlad's answer O(n) solution:
from collections import namedtuple
Info = namedtuple('Info', 'start height')
def max_rectangle_area(histogram):
"""Find the area of the largest rectangle that fits entirely under
the histogram.
"""
stack = []
top = lambda: stack[-1]
max_area = 0
pos = 0 # current position in the histogram
for pos, height in enumerate(histogram):
start = pos # position where rectangle starts
while True:
if not stack or height > top().height:
stack.append(Info(start, height)) # push
elif stack and height < top().height:
max_area = max(max_area, top().height*(pos-top().start))
start, _ = stack.pop()
continue
break # height == top().height goes here
pos += 1
for start, height in stack:
max_area = max(max_area, height*(pos-start))
return max_area
Example:
>>> f = max_rectangle_area
>>> f([5,3,1])
6
>>> f([1,3,5])
6
>>> f([3,1,5])
5
>>> f([4,8,3,2,0])
9
>>> f([4,8,3,1,1,0])
9
Linear search using a stack of incomplete subproblems
Copy-paste algorithm's description (in case the page goes down):
We process the elements in
left-to-right order and maintain a
stack of information about started but
yet unfinished subhistograms. Whenever
a new element arrives it is subjected
to the following rules. If the stack
is empty we open a new subproblem by
pushing the element onto the stack.
Otherwise we compare it to the element
on top of the stack. If the new one is
greater we again push it. If the new
one is equal we skip it. In all these
cases, we continue with the next new
element. If the new one is less, we
finish the topmost subproblem by
updating the maximum area w.r.t. the
element at the top of the stack. Then,
we discard the element at the top, and
repeat the procedure keeping the
current new element. This way, all
subproblems are finished until the
stack becomes empty, or its top
element is less than or equal to the
new element, leading to the actions
described above. If all elements have
been processed, and the stack is not
yet empty, we finish the remaining
subproblems by updating the maximum
area w.r.t. to the elements at the
top.
For the update w.r.t. an element, we
find the largest rectangle that
includes that element. Observe that an
update of the maximum area is carried
out for all elements except for those
skipped. If an element is skipped,
however, it has the same largest
rectangle as the element on top of the
stack at that time that will be
updated later. The height of the
largest rectangle is, of course, the
value of the element. At the time of
the update, we know how far the
largest rectangle extends to the right
of the element, because then, for the
first time, a new element with smaller
height arrived. The information, how
far the largest rectangle extends to
the left of the element, is available
if we store it on the stack, too.
We therefore revise the procedure
described above. If a new element is
pushed immediately, either because the
stack is empty or it is greater than
the top element of the stack, the
largest rectangle containing it
extends to the left no farther than
the current element. If it is pushed
after several elements have been
popped off the stack, because it is
less than these elements, the largest
rectangle containing it extends to the
left as far as that of the most
recently popped element.
Every element is pushed and popped at
most once and in every step of the
procedure at least one element is
pushed or popped. Since the amount of
work for the decisions and the update
is constant, the complexity of the
algorithm is O(n) by amortized
analysis.
The other answers here have done a great job presenting the O(n)-time, O(n)-space solution using two stacks. There's another perspective on this problem that independently provides an O(n)-time, O(n)-space solution to the problem, and might provide a little bit more insight as to why the stack-based solution works.
The key idea is to use a data structure called a Cartesian tree. A Cartesian tree is a binary tree structure (though not a binary search tree) that's built around an input array. Specifically, the root of the Cartesian tree is built above the minimum element of the array, and the left and right subtrees are recursively constructed from the subarrays to the left and right of the minimum value.
For example, here's a sample array and its Cartesian tree:
+----------------------- 23 ------+
| |
+------------- 26 --+ +-- 79
| | |
31 --+ 53 --+ 84
| |
41 --+ 58 -------+
| |
59 +-- 93
|
97
+----+----+----+----+----+----+----+----+----+----+----+
| 31 | 41 | 59 | 26 | 53 | 58 | 97 | 93 | 23 | 84 | 79 |
+----+----+----+----+----+----+----+----+----+----+----+
The reason that Cartesian trees are useful in this problem is that the question at hand has a really nice recursive structure to it. Begin by looking at the lowest rectangle in the histogram. There are three options for where the maximum rectangle could end up being placed:
It could pass right under the minimum value in the histogram. In that case, to make it as large as possible, we'd want to make it as wide as the entire array.
It could be entirely to the left of the minimum value. In that case, we recursively want the answer formed from the subarray purely to the left of the minimum value.
It could be entirely to the right of the minimum value. In that case, we recursively want the answer formed from the subarray purely to the right of the minimum value.
Notice that this recursive structure - find the minimum value, do something with the subarrays to the left and the right of that value - perfectly matches the recursive structure of a Cartesian tree. In fact, if we can create a Cartesian tree for the overall array when we get started, we can then solve this problem by recursively walking the Cartesian tree from the root downward. At each point, we recursively compute the optimal rectangle in the left and right subarrays, along with the rectangle you'd get by fitting right under the minimum value, and then return the best option we find.
In pseudocode, this looks like this:
function largestRectangleUnder(int low, int high, Node root) {
/* Base case: If the range is empty, the biggest rectangle we
* can fit is the empty rectangle.
*/
if (low == high) return 0;
/* Assume the Cartesian tree nodes are annotated with their
* positions in the original array.
*/
return max {
(high - low) * root.value, // Widest rectangle under the minimum
largestRectangleUnder(low, root.index, root.left),
largestRectnagleUnder(root.index + 1, high, root.right)
}
}
Once we have the Cartesian tree, this algorithm takes time O(n), since we visit each node exactly once and do O(1) work per node.
It turns out that there's a simple, linear-time algorithm for building Cartesian trees. The "natural" way you'd probably think to build one would be to scan across the array, find the minimum value, then recursively build a Cartesian tree from the left and right subarrays. The problem is that the process of finding the minimum value is really expensive, and this can take time Θ(n2).
The "fast" way to build a Cartesian tree is by scanning the array from the left to the right, adding in one element at a time. This algorithm is based on the following observations about Cartesian trees:
First, Cartesian trees obey the heap property: every element is less than or equal to its children. The reason for this is that the Cartesian tree root is the smallest value in the overall array, and its children are the smallest elements in their subarrays, etc.
Second, if you do an inorder traversal of a Cartesian tree, you get back the elements of the array in the order in which they appear. To see why this is, notice that if you do an inorder traversal of a Cartesian tree, you first visit everything to the left of the minimum value, then the minimum value, then everything to the right of the minimum value. Those visitations are recursively done the same way, so everything ends up being visited in order.
These two rules give us a lot of information about what happens if we start with a Cartesian tree of the first k elements of the array and want to form a Cartesian tree for the first k+1 elements. That new element will have to end up on the right spine of the Cartesian tree - the part of the tree formed by starting at the root and only taking steps to the right - because otherwise something would come after it in an inorder traversal. And, within that right spine, it has to be placed in a way that makes it bigger than everything above it, since we need to obey the heap property.
The way that you actually add a new node to the Cartesian tree is to start at the rightmost node in the tree and walk upwards until you either hit the root of the tree or find a node that has a smaller value. You then make the new value have as its left child the last node it walked up on top of.
Here's a trace of that algorithm on a small array:
+---+---+---+---+
| 2 | 4 | 3 | 1 |
+---+---+---+---+
2 becomes the root.
2 --+
|
4
4 is bigger than 2, we can't move upwards. Append to right.
+---+---+---+---+
| 2 | 4 | 3 | 1 |
+---+---+---+---+
2 ------+
|
--- 3
|
4
3 is lesser than 4, climb over it. Can't climb further over 2, as it is smaller than 3. Climbed over subtree rooted at 4 goes to the left of new value 3 and 3 becomes rightmost node now.
+---+---+---+---+
| 2 | 4 | 3 | 1 |
+---+---+---+---+
+---------- 1
|
2 ------+
|
--- 3
|
4
1 climbs over the root 2, the entire tree rooted at 2 is moved to left of 1, and 1 is now the new root - and also the rightmost value.
+---+---+---+---+
| 2 | 4 | 3 | 1 |
+---+---+---+---+
Although this might not seem to run in linear time - wouldn't you potentially end up climbing all the way to the root of the tree over and over and over again? - you can show that this runs in linear time using a clever argument. If you climb up over a node in the right spine during an insertion, that node ends up getting moved off the right spine and therefore can't be rescanned in a future insertion. Therefore, every node is only ever scanned over at most once, so the total work done is linear.
And now the kicker - the standard way that you'd actually implement this approach is by maintaining a stack of the values that correspond to the nodes on the right spine. The act of "walking up" and over a node corresponds to popping a node off the stack. Therefore, the code for building a Cartesian tree looks something like this:
Stack s;
for (each array element x) {
pop s until it's empty or s.top > x
push x onto the stack.
do some sort of pointer rewiring based on what you just did.
}
The stack manipulations here might seem really familiar, and that's because these are the exact stack operations that you would do in the answers shown elsewhere here. In fact, you can think of what those approaches are doing as implicitly building the Cartesian tree and running the recursive algorithm shown above in the process of doing so.
The advantage, I think, of knowing about Cartesian trees is that it provides a really nice conceptual framework for seeing why this algorithm works correctly. If you know that what you're doing is running a recursive walk of a Cartesian tree, it's easier to see that you're guaranteed to find the largest rectangle. Plus, knowing that the Cartesian tree exists gives you a useful tool for solving other problems. Cartesian trees show up in the design of fast data structures for the range minimum query problem and are used to convert suffix arrays into suffix trees.
Here's some Java code that implements this idea, courtesy of #Azeem!
import java.util.Stack;
public class CartesianTreeMakerUtil {
private static class Node {
int val;
Node left;
Node right;
}
public static Node cartesianTreeFor(int[] nums) {
Node root = null;
Stack<Node> s = new Stack<>();
for(int curr : nums) {
Node lastJumpedOver = null;
while(!s.empty() && s.peek().val > curr) {
lastJumpedOver = s.pop();
}
Node currNode = this.new Node();
currNode.val = curr;
if(s.isEmpty()) {
root = currNode;
}
else {
s.peek().right = currNode;
}
currNode.left = lastJumpedOver;
s.push(currNode);
}
return root;
}
public static void printInOrder(Node root) {
if(root == null) return;
if(root.left != null ) {
printInOrder(root.left);
}
System.out.println(root.val);
if(root.right != null) {
printInOrder(root.right);
}
}
public static void main(String[] args) {
int[] nums = new int[args.length];
for (int i = 0; i < args.length; i++) {
nums[i] = Integer.parseInt(args[i]);
}
Node root = cartesianTreeFor(nums);
tester.printInOrder(root);
}
}
The easiest solution in O(N)
long long getMaxArea(long long hist[], long long n)
{
stack<long long> s;
long long max_area = 0;
long long tp;
long long area_with_top;
long long i = 0;
while (i < n)
{
if (s.empty() || hist[s.top()] <= hist[i])
s.push(i++);
else
{
tp = s.top(); // store the top index
s.pop(); // pop the top
area_with_top = hist[tp] * (s.empty() ? i : i - s.top() - 1);
if (max_area < area_with_top)
{
max_area = area_with_top;
}
}
}
while (!s.empty())
{
tp = s.top();
s.pop();
area_with_top = hist[tp] * (s.empty() ? i : i - s.top() - 1);
if (max_area < area_with_top)
max_area = area_with_top;
}
return max_area;
}
There is also another solution using Divide and Conquer. The algorithm for it is :
1) Divide the array into 2 parts with the smallest height as the breaking point
2) The maximum area is the maximum of :
a) Smallest height * size of the array
b) Maximum rectangle in left half array
c) Maximum rectangle in right half array
The time complexity comes to O(nlogn)
The stack solution is one of the most clever solutions I've seen till date. And it can be a little hard to understand why that works.
I've taken a jab at explaining the same in some detail here.
Summary points from the post:-
General way our brain thinks is :-
Create every situation and try to find the value of the contraint that is needed to solve the problem.
And we happily convert that to code as :- find the value of contraint(min) for each situation(pair(i,j))
The clever solutions tries to flip the problem.For each constraint/min value of tha area, what is the best possible left and right extremes ?
So if we traverse over each possible min in the array. What are the left and right extremes for each value ?
Little thought says, the first left most value less than the current min and similarly the first rightmost value that is lesser than the current min.
So now we need to see if we can find a clever way to find the first left and right values lesser than the current value.
To think: If we have traversed the array partially say till min_i, how can the solution to min_i+1 be built?
We need the first value less than min_i to its left.
Inverting the statement : we need to ignore all values to the left of min_i that are greater than min_i. We stop when we find the first value smaller than min_i (i) . The troughs in the curve hence become useless once we have crossed it. In histogram , (2 4 3) => if 3 is min_i, 4 being larger is not of interest.
Corrollary: in a range (i,j). j being the min value we are considering.. all values between j and its left value i are useless. Even for further calculations.
Any histogram on the right with a min value larger than j, will be binded at j. The values of interest on the left form a monotonically increasing sequence with j being the largest value. (Values of interest here being possible values that may be of interest for the later array)
Since, we are travelling from left to right, for each min value/ current value - we do not know whether the right side of the array will have an element smaller than it.
So we have to keep it in memory until we get to know this value is useless. (since a smaller value is found)
All this leads to a usage of our very own stack structure.
We keep on stack until we don't know its useless.
We remove from stack once we know the thing is crap.
So for each min value to find its left smaller value, we do the following:-
pop the elements larger to it (useless values)
The first element smaller than the value is the left extreme. The i to our min.
We can do the same thing from the right side of the array and we will get j to our min.
It's quite hard to explain this, but if this is making sense then I'd suggest read the complete article here since it has more insights and details.
I don't understand the other entries, but I think I know how to do it in O(n) as follows.
A) for each index find the largest rectangle inside the histogram ending at that index where the index column touches the top of the rectangle and remember where the rectangle starts. This can be done in O(n) using a stack based algorithm.
B) Similarly for each index find the largest rectangle starting at that index where the index column touches the top of the rectangle and remember where the rectangle ends. Also O(n) using the same method as (A) but scanning the histogram backwards.
C) For each index combine the results of (A) and (B) to determine the largest rectangle where the column at that index touches the top of the rectangle. O(n) like (A).
D) Since the largest rectangle must be touched by some column of the histogram the largest rectangle is the largest rectangle found in step (C).
The hard part is implementing (A) and (B), which I think is what JF Sebastian may have solved rather than the general problem stated.
I coded this one and felt little better in the sense:
import java.util.Stack;
class StackItem{
public int sup;
public int height;
public int sub;
public StackItem(int a, int b, int c){
sup = a;
height = b;
sub =c;
}
public int getArea(){
return (sup - sub)* height;
}
#Override
public String toString(){
return " from:"+sup+
" to:"+sub+
" height:"+height+
" Area ="+getArea();
}
}
public class MaxRectangleInHistogram {
Stack<StackItem> S;
StackItem curr;
StackItem maxRectangle;
public StackItem getMaxRectangleInHistogram(int A[], int n){
int i = 0;
S = new Stack();
S.push(new StackItem(0,0,-1));
maxRectangle = new StackItem(0,0,-1);
while(i<n){
curr = new StackItem(i,A[i],i);
if(curr.height > S.peek().height){
S.push(curr);
}else if(curr.height == S.peek().height){
S.peek().sup = i+1;
}else if(curr.height < S.peek().height){
while((S.size()>1) && (curr.height<=S.peek().height)){
curr.sub = S.peek().sub;
S.peek().sup = i;
decideMaxRectangle(S.peek());
S.pop();
}
S.push(curr);
}
i++;
}
while(S.size()>1){
S.peek().sup = i;
decideMaxRectangle(S.peek());
S.pop();
}
return maxRectangle;
}
private void decideMaxRectangle(StackItem s){
if(s.getArea() > maxRectangle.getArea() )
maxRectangle = s;
}
}
Just Note:
Time Complexity: T(n) < O(2n) ~ O(n)
Space Complexity S(n) < O(n)
I would like to thank #templatetypedef for his/her extremely detailed and intuitive answer. The Java code below is based on his suggestion to use Cartesian Trees and solves the problem in O(N) time and O(N) space. I suggest that you read #templatetypedef's answer above before reading the code below. The code is given in the format of the solution to the problem at leetcode: https://leetcode.com/problems/largest-rectangle-in-histogram/description/ and passes all 96 test cases.
class Solution {
private class Node {
int val;
Node left;
Node right;
int index;
}
public Node getCartesianTreeFromArray(int [] nums) {
Node root = null;
Stack<Node> s = new Stack<>();
for(int i = 0; i < nums.length; i++) {
int curr = nums[i];
Node lastJumpedOver = null;
while(!s.empty() && s.peek().val >= curr) {
lastJumpedOver = s.pop();
}
Node currNode = this.new Node();
currNode.val = curr;
currNode.index = i;
if(s.isEmpty()) {
root = currNode;
}
else {
s.peek().right = currNode;
}
currNode.left = lastJumpedOver;
s.push(currNode);
}
return root;
}
public int largestRectangleUnder(int low, int high, Node root, int [] nums) {
/* Base case: If the range is empty, the biggest rectangle we
* can fit is the empty rectangle.
*/
if(root == null) return 0;
if (low == high) {
if(0 <= low && low <= nums.length - 1) {
return nums[low];
}
return 0;
}
/* Assume the Cartesian tree nodes are annotated with their
* positions in the original array.
*/
int leftArea = -1 , rightArea= -1;
if(root.left != null) {
leftArea = largestRectangleUnder(low, root.index - 1 , root.left, nums);
}
if(root.right != null) {
rightArea = largestRectangleUnder(root.index + 1, high,root.right, nums);
}
return Math.max((high - low + 1) * root.val,
Math.max(leftArea, rightArea));
}
public int largestRectangleArea(int[] heights) {
if(heights == null || heights.length == 0 ) {
return 0;
}
if(heights.length == 1) {
return heights[0];
}
Node root = getCartesianTreeFromArray(heights);
return largestRectangleUnder(0, heights.length - 1, root, heights);
}
}
python-3
a=[3,4,7,4,6]
a.sort()
r=0
for i in range(len(a)):
if a[i]* (n-1) > r:
r = a[i]*(n-i)
print(r)
output:
16
I come across this question in one of interview. Was trying to solve this, resulting in observed following things -
Need to check consecutive left elements greater than current
element
Need to check consecutive right elements greater than
current element
Calculate area (number of left side max elements + number of right side max elements + 1) * current element
Check and replace existing maxArea if calculated area is greater than
maxArea
Following is the JS code implementing above pseudocode
function maxAreaCovered(arr) {
let maxArea = 0;
for (let index = 0; index < arr.length; index++) {
let l = index - 1;
let r = index + 1;
let maxEleCount = 0
while (l > -1) {
if (arr[l] >= arr[index]) {
maxEleCount++;
} else {
break;
}
l--;
}
while (r < arr.length) {
if (arr[r] >= arr[index]) {
maxEleCount++;
} else {
break;
}
r++;
}
let area = (maxEleCount + 1) * arr[index];
maxArea = Math.max(area, maxArea);
}
return maxArea
}
console.log(maxAreaCovered([6, 2, 5, 4, 5, 1, 6]));
You can use O(n) method which uses stack to calculate the maximum area under the histogram.
long long histogramArea(vector<int> &histo){
stack<int> s;
long long maxArea=0;
long long area= 0;
int i =0;
for (i = 0; i < histo.size();) {
if(s.empty() || histo[s.top()] <= histo[i]){
s.push(i++);
}
else{
int top = s.top(); s.pop();
area= histo[top]* (s.empty()?i:i-s.top()-1);
if(area >maxArea)
maxArea= area;
}
}
while(!s.empty()){
int top = s.top();s.pop();
area= histo[top]* (s.empty()?i:i-s.top()-1);
if(area >maxArea)
maxArea= area;
}
return maxArea;
}
For explanation you can read here http://www.geeksforgeeks.org/largest-rectangle-under-histogram/
Fenwick tree is a data-structure that gives an efficient way to answer to main queries:
add an element to a particular index of an array update(index, value)
find sum of elements from 1 to N find(n)
both operations are done in O(log(n)) time and I understand the logic and implementation. It is not hard to implement a bunch of other operations like find a sum from N to M.
I wanted to understand how to adapt Fenwick tree for RMQ. It is obvious to change Fenwick tree for first two operations. But I am failing to figure out how to find minimum on the range from N to M.
After searching for solutions majority of people think that this is not possible and a small minority claims that it actually can be done (approach1, approach2).
The first approach (written in Russian, based on my google translate has 0 explanation and only two functions) relies on three arrays (initial, left and right) upon my testing was not working correctly for all possible test cases.
The second approach requires only one array and based on the claims runs in O(log^2(n)) and also has close to no explanation of why and how should it work. I have not tried to test it.
In light of controversial claims, I wanted to find out whether it is possible to augment Fenwick tree to answer update(index, value) and findMin(from, to).
If it is possible, I would be happy to hear how it works.
Yes, you can adapt Fenwick Trees (Binary Indexed Trees) to
Update value at a given index in O(log n)
Query minimum value for a range in O(log n) (amortized)
We need 2 Fenwick trees and an additional array holding the real values for nodes.
Suppose we have the following array:
index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
value 1 0 2 1 1 3 0 4 2 5 2 2 3 1 0
We wave a magic wand and the following trees appear:
Note that in both trees each node represents the minimum value for all nodes within that subtree. For example, in BIT2 node 12 has value 0, which is the minimum value for nodes 12,13,14,15.
Queries
We can efficiently query the minimum value for any range by calculating the minimum of several subtree values and one additional real node value. For example, the minimum value for range [2,7] can be determined by taking the minimum value of BIT2_Node2 (representing nodes 2,3) and BIT1_Node7 (representing node 7), BIT1_Node6 (representing nodes 5,6) and REAL_4 - therefore covering all nodes in [2,7]. But how do we know which sub trees we want to look at?
Query(int a, int b) {
int val = infinity // always holds the known min value for our range
// Start traversing the first tree, BIT1, from the beginning of range, a
int i = a
while (parentOf(i, BIT1) <= b) {
val = min(val, BIT2[i]) // Note: traversing BIT1, yet looking up values in BIT2
i = parentOf(i, BIT1)
}
// Start traversing the second tree, BIT2, from the end of range, b
i = b
while (parentOf(i, BIT2) >= a) {
val = min(val, BIT1[i]) // Note: traversing BIT2, yet looking up values in BIT1
i = parentOf(i, BIT2)
}
val = min(val, REAL[i]) // Explained below
return val
}
It can be mathematically proven that both traversals will end in the same node. That node is a part of our range, yet it is not a part of any subtrees we have looked at. Imagine a case where the (unique) smallest value of our range is in that special node. If we didn't look it up our algorithm would give incorrect results. This is why we have to do that one lookup into the real values array.
To help understand the algorithm I suggest you simulate it with pen & paper, looking up data in the example trees above. For example, a query for range [4,14] would return the minimum of values BIT2_4 (rep. 4,5,6,7), BIT1_14 (rep. 13,14), BIT1_12 (rep. 9,10,11,12) and REAL_8, therefore covering all possible values [4,14].
Updates
Since a node represents the minimum value of itself and its children, changing a node will affect its parents, but not its children. Therefore, to update a tree we start from the node we are modifying and move up all the way to the fictional root node (0 or N+1 depending on which tree).
Suppose we are updating some node in some tree:
If new value < old value, we will always overwrite the value and move up
If new value == old value, we can stop since there will be no more changes cascading upwards
If new value > old value, things get interesting.
If the old value still exists somewhere within that subtree, we are done
If not, we have to find the new minimum value between real[node] and each tree[child_of_node], change tree[node] and move up
Pseudocode for updating node with value v in a tree:
while (node <= n+1) {
if (v > tree[node]) {
if (oldValue == tree[node]) {
v = min(v, real[node])
for-each child {
v = min(v, tree[child])
}
} else break
}
if (v == tree[node]) break
tree[node] = v
node = parentOf(node, tree)
}
Note that oldValue is the original value we replaced, whereas v may be reassigned multiple times as we move up the tree.
Binary Indexing
In my experiments Range Minimum Queries were about twice as fast as a Segment Tree implementation and updates were marginally faster. The main reason for this is using super efficient bitwise operations for moving between nodes. They are very well explained here. Segment Trees are really simple to code so think about is the performance advantage really worth it? The update method of my Fenwick RMQ is 40 lines and took a while to debug. If anyone wants my code I can put it on github. I also produced a brute and test generators to make sure everything works.
I had help understanding this subject & implementing it from the Finnish algorithm community. Source of the image is http://ioinformatics.org/oi/pdf/v9_2015_39_44.pdf, but they credit Fenwick's 1994 paper for it.
The Fenwick tree structure works for addition because addition is invertible. It doesn't work for minimum, because as soon as you have a cell that's supposed to be the minimum of two or more inputs, you've lost information potentially.
If you're willing to double your storage requirements, you can support RMQ with a segment tree that is constructed implicitly, like a binary heap. For an RMQ with n values, store the n values at locations [n, 2n) of an array. Locations [1, n) are aggregates, with the formula A(k) = min(A(2k), A(2k+1)). Location 2n is an infinite sentinel. The update routine should look something like this.
def update(n, a, i, x): # value[i] = x
i += n
a[i] = x
# update the aggregates
while i > 1:
i //= 2
a[i] = min(a[2*i], a[2*i+1])
The multiplies and divides here can be replaced by shifts for efficiency.
The RMQ pseudocode is more delicate. Here's another untested and unoptimized routine.
def rmq(n, a, i, j): # min(value[i:j])
i += n
j += n
x = inf
while i < j:
if i%2 == 0:
i //= 2
else:
x = min(x, a[i])
i = i//2 + 1
if j%2 == 0:
j //= 2
else:
x = min(x, a[j-1])
j //= 2
return x
I need to get all possible paths of a tree so I implemented a DFS like this:
void bisearch(std::vector<int> path, int steps,
int node, std::vector<std::vector<int>> *paths) {
int sum = 0;
if (path.size() == steps) {
for(std::vector<int>::iterator it=path.begin(); it != path.end(); ++it) {
sum += (*it);
}
if (sum == node)
paths->push_back(path);
}
else {
std::vector<int> uPath(path);
uPath.push_back(1);
bisearch(uPath, steps, node, paths);
std::vector<int> dPath(path);
dPath.push_back(0);
bisearch(dPath, steps, node, paths);
}
}
The above code gives me all paths to some ending node for a tree of length "steps". I then loop through all ending nodes and run this to get every path. Issue is it takes forever! I was thinking of maybe hardcoding all the possible combinations to speed it up, of course I couldn't do this by hand since for instance a tree with 25 steps would have 2^25 ~= 35 million possible combinations, but maybe I could print the output from the search and use that to hardcode? Or does anyone see any easy optimizations I could make that would make a big difference on the performance? Thanks.
EDIT: Let me clarify a little. I need the path, that is the sequence of movements along the tree where 1 represents a right hand move and 0 a left (or up/down whichever you prefer). So for instance a 2 step tree I need the four ordered pairs (1,0) (0,1) (1,1) (0,0).
Since "all the combinations" should mean just "the combinations of turning right / left at a certain level", you could just loop through 0 to 2 ^ n - 1, and the binary representation padded with 0 in the front might be just what you want.
If what you want is the count of paths with left turn count equals to a certain number k, then this just equals the numbers from 0 to 2 ^ n - 1 that has k bit equal to 1, and you could possibly use this to compute the result you want.
Given a binary tree which is huge and can not be placed in memory, how do you check if the tree is a mirror image.
I got this as an interview question
If a tree is a mirror image of another tree, the inorder traversal of one tree would be reverse of another.
So just do inorder traversal on the first tree and a reverse inorder traversal on another and check if all the elements are the same.
I can't take full credit for this reply of course; a handful of my colleagues helped with some assumptions and for poking holes in my original idea. Much thanks to them!
Assumptions
We can't have the entire tree in memory, so it's not ideal to use recursion. Let's assume, for simplicity's sake, that we can only hold a maximum of two nodes in memory.
We know n, the total number of levels in our tree.
We can perform seeks on the data with respect to the character or line position it's in.
The data that is on disk is ordered by depth. That is to say, the first entry on disk is the root, and the next two are its children, and the next four are its children's children, and so forth.
There are cases in which the data is perfectly mirrored, and cases in which it isn't. Blank data interlaced with non-blank data is considered "acceptable", unless otherwise specified.
We have freedom over using any data type we wish so long as the values can be compared for equivalence. Testing for object equivalence may not be ideal, so let's assume we're comparing primitives.
"Mirrored" means mirrored between the root's children. To use different terminologies, the grandparent's left child is mirrored with its right child, and the left child (parent)'s left child is mirrored with the grandparent's right child's right child. This is illustrated in the graph below; the matching symbols represent the mirroring we want to check for.
G
P* P*
C1& C2^ C3^ C4&
Approach
We know how many nodes on each level we should expect when we're reading from disk - some multiple of 2k. We can establish a double loop to iterate over the total depth of the tree, and the count of the nodes in each level. Inside of this, we can simply compare the outermost values for equivalence, and short-circuit if we find an unequal value.
We can determine the location of each outer location by using multiples of 2k. The leftmost child of any level will always be 2k, and the rightmost child of any level will always be 2k+1-1.
Small Proof: Outermost nodes on level 1 are 2 and 3; 21 = 2, 21+1-1 = 22-1 = 3. Outermost nodes on level 2 are 4 and 7; 22 = 4, 22+1-1 = 23-1 = 7. One could expand this all the way to the nth case.
Pseudocode
int k, i;
for(k = 1; k < n; k++) { // Skip root, trivially mirrored
for(i = 0; i < pow(2, k) / 2; i++) {
if(node_retrieve(i + pow(2, k)) != node_retrieve(pow(2, (k+1)-i)) {
return false;
}
}
}
return true;
Thoughts
This sort of question is a great interview question because, more than likely, they want to see how you would approach this problem. This approach may be horrible, it may be immaculate, but an employer would want you to take your time, draw things on a piece of paper or whiteboard, and ask them questions about how the data is stored, how it can be read, what limitations there are on seeks, etc etc.
It's not the coding aspect that interviewers are interested in, but the problem solving aspect.
Recursion is easy.
struct node {
struct node *left;
struct node *right;
int payload;
};
int is_not_mirror(struct node *one, struct node *two)
{
if (!one && !two) return 0;
if (!one) return 1;
if (!two) return 1;
if (compare(one->payload, two->payload)) return 1;
if (is_not_mirror(one->left, two->right)) return 1;
if (is_not_mirror(one->right, two->left)) return 1;
return 0;
}
What's the best way to create a balanced binary search tree from a sorted singly linked list?
How about creating nodes bottom-up?
This solution's time complexity is O(N). Detailed explanation in my blog post:
http://www.leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html
Two traversal of the linked list is all we need. First traversal to get the length of the list (which is then passed in as the parameter n into the function), then create nodes by the list's order.
BinaryTree* sortedListToBST(ListNode *& list, int start, int end) {
if (start > end) return NULL;
// same as (start+end)/2, avoids overflow
int mid = start + (end - start) / 2;
BinaryTree *leftChild = sortedListToBST(list, start, mid-1);
BinaryTree *parent = new BinaryTree(list->data);
parent->left = leftChild;
list = list->next;
parent->right = sortedListToBST(list, mid+1, end);
return parent;
}
BinaryTree* sortedListToBST(ListNode *head, int n) {
return sortedListToBST(head, 0, n-1);
}
You can't do better than linear time, since you have to at least read all the elements of the list, so you might as well copy the list into an array (linear time) and then construct the tree efficiently in the usual way, i.e. if you had the list [9,12,18,23,24,51,84], then you'd start by making 23 the root, with children 12 and 51, then 9 and 18 become children of 12, and 24 and 84 become children of 51. Overall, should be O(n) if you do it right.
The actual algorithm, for what it's worth, is "take the middle element of the list as the root, and recursively build BSTs for the sub-lists to the left and right of the middle element and attach them below the root".
Best isn't only about asynmptopic run time. The sorted linked list has all the information needed to create the binary tree directly, and I think this is probably what they are looking for
Note that the first and third entries become children of the second, then the fourth node has chidren of the second and sixth (which has children the fifth and seventh) and so on...
in psuedo code
read three elements, make a node from them, mark as level 1, push on stack
loop
read three elemeents and make a node of them
mark as level 1
push on stack
loop while top two enties on stack have same level (n)
make node of top two entries, mark as level n + 1, push on stack
while elements remain in list
(with a bit of adjustment for when there's less than three elements left or an unbalanced tree at any point)
EDIT:
At any point, there is a left node of height N on the stack. Next step is to read one element, then read and construct another node of height N on the stack. To construct a node of height N, make and push a node of height N -1 on the stack, then read an element, make another node of height N-1 on the stack -- which is a recursive call.
Actually, this means the algorithm (even as modified) won't produce a balanced tree. If there are 2N+1 nodes, it will produce a tree with 2N-1 values on the left, and 1 on the right.
So I think #sgolodetz's answer is better, unless I can think of a way of rebalancing the tree as it's built.
Trick question!
The best way is to use the STL, and advantage yourself of the fact that the sorted associative container ADT, of which set is an implementation, demands insertion of sorted ranges have amortized linear time. Any passable set of core data structures for any language should offer a similar guarantee. For a real answer, see the quite clever solutions others have provided.
What's that? I should offer something useful?
Hum...
How about this?
The smallest possible meaningful tree in a balanced binary tree is 3 nodes.
A parent, and two children. The very first instance of such a tree is the first three elements. Child-parent-Child. Let's now imagine this as a single node. Okay, well, we no longer have a tree. But we know that the shape we want is Child-parent-Child.
Done for a moment with our imaginings, we want to keep a pointer to the parent in that initial triumvirate. But it's singly linked!
We'll want to have four pointers, which I'll call A, B, C, and D. So, we move A to 1, set B equal to A and advance it one. Set C equal to B, and advance it two. The node under B already points to its right-child-to-be. We build our initial tree. We leave B at the parent of Tree one. C is sitting at the node that will have our two minimal trees as children. Set A equal to C, and advance it one. Set D equal to A, and advance it one. We can now build our next minimal tree. D points to the root of that tree, B points to the root of the other, and C points to the... the new root from which we will hang our two minimal trees.
How about some pictures?
[A][B][-][C]
With our image of a minimal tree as a node...
[B = Tree][C][A][D][-]
And then
[Tree A][C][Tree B]
Except we have a problem. The node two after D is our next root.
[B = Tree A][C][A][D][-][Roooooot?!]
It would be a lot easier on us if we could simply maintain a pointer to it instead of to it and C. Turns out, since we know it will point to C, we can go ahead and start constructing the node in the binary tree that will hold it, and as part of this we can enter C into it as a left-node. How can we do this elegantly?
Set the pointer of the Node under C to the node Under B.
It's cheating in every sense of the word, but by using this trick, we free up B.
Alternatively, you can be sane, and actually start building out the node structure. After all, you really can't reuse the nodes from the SLL, they're probably POD structs.
So now...
[TreeA]<-[C][A][D][-][B]
[TreeA]<-[C]->[TreeB][B]
And... Wait a sec. We can use this same trick to free up C, if we just let ourselves think of it as a single node instead of a tree. Because after all, it really is just a single node.
[TreeC]<-[B][A][D][-][C]
We can further generalize our tricks.
[TreeC]<-[B][TreeD]<-[C][-]<-[D][-][A]
[TreeC]<-[B][TreeD]<-[C]->[TreeE][A]
[TreeC]<-[B]->[TreeF][A]
[TreeG]<-[A][B][C][-][D]
[TreeG]<-[A][-]<-[C][-][D]
[TreeG]<-[A][TreeH]<-[D][B][C][-]
[TreeG]<-[A][TreeH]<-[D][-]<-[C][-][B]
[TreeG]<-[A][TreeJ]<-[B][-]<-[C][-][D]
[TreeG]<-[A][TreeJ]<-[B][TreeK]<-[D][-]<-[C][-]
[TreeG]<-[A][TreeJ]<-[B][TreeK]<-[D][-]<-[C][-]
We are missing a critical step!
[TreeG]<-[A]->([TreeJ]<-[B]->([TreeK]<-[D][-]<-[C][-]))
Becomes :
[TreeG]<-[A]->[TreeL->([TreeK]<-[D][-]<-[C][-])][B]
[TreeG]<-[A]->[TreeL->([TreeK]<-[D]->[TreeM])][B]
[TreeG]<-[A]->[TreeL->[TreeN]][B]
[TreeG]<-[A]->[TreeO][B]
[TreeP]<-[B]
Obviously, the algorithm can be cleaned up considerably, but I thought it would be interesting to demonstrate how one can optimize as you go by iteratively designing your algorithm. I think this kind of process is what a good employer should be looking for more than anything.
The trick, basically, is that each time we reach the next midpoint, which we know is a parent-to-be, we know that its left subtree is already finished. The other trick is that we are done with a node once it has two children and something pointing to it, even if all of the sub-trees aren't finished. Using this, we can get what I am pretty sure is a linear time solution, as each element is touched only 4 times at most. The problem is that this relies on being given a list that will form a truly balanced binary search tree. There are, in other words, some hidden constraints that may make this solution either much harder to apply, or impossible. For example, if you have an odd number of elements, or if there are a lot of non-unique values, this starts to produce a fairly silly tree.
Considerations:
Render the element unique.
Insert a dummy element at the end if the number of nodes is odd.
Sing longingly for a more naive implementation.
Use a deque to keep the roots of completed subtrees and the midpoints in, instead of mucking around with my second trick.
This is a python implementation:
def sll_to_bbst(sll, start, end):
"""Build a balanced binary search tree from sorted linked list.
This assumes that you have a class BinarySearchTree, with properties
'l_child' and 'r_child'.
Params:
sll: sorted linked list, any data structure with 'popleft()' method,
which removes and returns the leftmost element of the list. The
easiest thing to do is to use 'collections.deque' for the sorted
list.
start: int, start index, on initial call set to 0
end: int, on initial call should be set to len(sll)
Returns:
A balanced instance of BinarySearchTree
This is a python implementation of solution found here:
http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html
"""
if start >= end:
return None
middle = (start + end) // 2
l_child = sll_to_bbst(sll, start, middle)
root = BinarySearchTree(sll.popleft())
root.l_child = l_child
root.r_child = sll_to_bbst(sll, middle+1, end)
return root
Instead of the sorted linked list i was asked on a sorted array (doesn't matter though logically, but yes run-time varies) to create a BST of minimal height, following is the code i could get out:
typedef struct Node{
struct Node *left;
int info;
struct Node *right;
}Node_t;
Node_t* Bin(int low, int high) {
Node_t* node = NULL;
int mid = 0;
if(low <= high) {
mid = (low+high)/2;
node = CreateNode(a[mid]);
printf("DEBUG: creating node for %d\n", a[mid]);
if(node->left == NULL) {
node->left = Bin(low, mid-1);
}
if(node->right == NULL) {
node->right = Bin(mid+1, high);
}
return node;
}//if(low <=high)
else {
return NULL;
}
}//Bin(low,high)
Node_t* CreateNode(int info) {
Node_t* node = malloc(sizeof(Node_t));
memset(node, 0, sizeof(Node_t));
node->info = info;
node->left = NULL;
node->right = NULL;
return node;
}//CreateNode(info)
// call function for an array example: 6 7 8 9 10 11 12, it gets you desired
// result
Bin(0,6);
HTH Somebody..
This is the pseudo recursive algorithm that I will suggest.
createTree(treenode *root, linknode *start, linknode *end)
{
if(start == end or start = end->next)
{
return;
}
ptrsingle=start;
ptrdouble=start;
while(ptrdouble != end and ptrdouble->next !=end)
{
ptrsignle=ptrsingle->next;
ptrdouble=ptrdouble->next->next;
}
//ptrsignle will now be at the middle element.
treenode cur_node=Allocatememory;
cur_node->data = ptrsingle->data;
if(root = null)
{
root = cur_node;
}
else
{
if(cur_node->data (less than) root->data)
root->left=cur_node
else
root->right=cur_node
}
createTree(cur_node, start, ptrSingle);
createTree(cur_node, ptrSingle, End);
}
Root = null;
The inital call will be createtree(Root, list, null);
We are doing the recursive building of the tree, but without using the intermediate array.
To get to the middle element every time we are advancing two pointers, one by one element, other by two elements. By the time the second pointer is at the end, the first pointer will be at the middle.
The running time will be o(nlogn). The extra space will be o(logn). Not an efficient solution for a real situation where you can have R-B tree which guarantees nlogn insertion. But good enough for interview.
Similar to #Stuart Golodetz and #Jake Kurzer the important thing is that the list is already sorted.
In #Stuart's answer, the array he presented is the backing data structure for the BST. The find operation for example would just need to perform index array calculations to traverse the tree. Growing the array and removing elements would be the trickier part, so I'd prefer a vector or other constant time lookup data structure.
#Jake's answer also uses this fact but unfortunately requires you to traverse the list to find each time to do a get(index) operation. But requires no additional memory usage.
Unless it was specifically mentioned by the interviewer that they wanted an object structure representation of the tree, I would use #Stuart's answer.
In a question like this you'd be given extra points for discussing the tradeoffs and all the options that you have.
Hope the detailed explanation on this post helps:
http://preparefortechinterview.blogspot.com/2013/10/planting-trees_1.html
A slightly improved implementation from #1337c0d3r in my blog.
// create a balanced BST using #len elements starting from #head & move #head forward by #len
TreeNode *sortedListToBSTHelper(ListNode *&head, int len) {
if (0 == len) return NULL;
auto left = sortedListToBSTHelper(head, len / 2);
auto root = new TreeNode(head->val);
root->left = left;
head = head->next;
root->right = sortedListToBSTHelper(head, (len - 1) / 2);
return root;
}
TreeNode *sortedListToBST(ListNode *head) {
int n = length(head);
return sortedListToBSTHelper(head, n);
}
If you know how many nodes are in the linked list, you can do it like this:
// Gives path to subtree being built. If branch[N] is false, branch
// less from the node at depth N, if true branch greater.
bool branch[max depth];
// If rem[N] is true, then for the current subtree at depth N, it's
// greater subtree has one more node than it's less subtree.
bool rem[max depth];
// Depth of root node of current subtree.
unsigned depth = 0;
// Number of nodes in current subtree.
unsigned num_sub = Number of nodes in linked list;
// The algorithm relies on a stack of nodes whose less subtree has
// been built, but whose right subtree has not yet been built. The
// stack is implemented as linked list. The nodes are linked
// together by having the "greater" handle of a node set to the
// next node in the list. "less_parent" is the handle of the first
// node in the list.
Node *less_parent = nullptr;
// h is root of current subtree, child is one of its children.
Node *h, *child;
Node *p = head of the sorted linked list of nodes;
LOOP // loop unconditionally
LOOP WHILE (num_sub > 2)
// Subtract one for root of subtree.
num_sub = num_sub - 1;
rem[depth] = !!(num_sub & 1); // true if num_sub is an odd number
branch[depth] = false;
depth = depth + 1;
num_sub = num_sub / 2;
END LOOP
IF (num_sub == 2)
// Build a subtree with two nodes, slanting to greater.
// I arbitrarily chose to always have the extra node in the
// greater subtree when there is an odd number of nodes to
// split between the two subtrees.
h = p;
p = the node after p in the linked list;
child = p;
p = the node after p in the linked list;
make h and p into a two-element AVL tree;
ELSE // num_sub == 1
// Build a subtree with one node.
h = p;
p = the next node in the linked list;
make h into a leaf node;
END IF
LOOP WHILE (depth > 0)
depth = depth - 1;
IF (not branch[depth])
// We've completed a less subtree, exit while loop.
EXIT LOOP;
END IF
// We've completed a greater subtree, so attach it to
// its parent (that is less than it). We pop the parent
// off the stack of less parents.
child = h;
h = less_parent;
less_parent = h->greater_child;
h->greater_child = child;
num_sub = 2 * (num_sub - rem[depth]) + rem[depth] + 1;
IF (num_sub & (num_sub - 1))
// num_sub is not a power of 2
h->balance_factor = 0;
ELSE
// num_sub is a power of 2
h->balance_factor = 1;
END IF
END LOOP
IF (num_sub == number of node in original linked list)
// We've completed the full tree, exit outer unconditional loop
EXIT LOOP;
END IF
// The subtree we've completed is the less subtree of the
// next node in the sequence.
child = h;
h = p;
p = the next node in the linked list;
h->less_child = child;
// Put h onto the stack of less parents.
h->greater_child = less_parent;
less_parent = h;
// Proceed to creating greater than subtree of h.
branch[depth] = true;
num_sub = num_sub + rem[depth];
depth = depth + 1;
END LOOP
// h now points to the root of the completed AVL tree.
For an encoding of this in C++, see the build member function (currently at line 361) in https://github.com/wkaras/C-plus-plus-intrusive-container-templates/blob/master/avl_tree.h . It's actually more general, a template using any forward iterator rather than specifically a linked list.