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/
Given an n-ary tree of integers, the task is to find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should share a common edge in the tree.
Example:
1
/ \
2 5
/ \
3 4
Maximum non adjacent sum = 3 + 4 + 5 = 12
The following is the faulty extension of the algorithm outlined in http://www.geeksforgeeks.org/maximum-sum-such-that-no-two-elements-are-adjacent?
def max_sum(node, inc_sum, exc_sum):
for child in node.children:
exc_new = max(inc_sum, exc_sum)
inc_sum = exc_sum + child.val
exc_sum = exc_new
inc_sum, exc_sum = max(max_sum(child, inc_sum, exc_sum),
max_sum(child, inc_sum, inc_sum - node.val))
return exc_sum, inc_sum
But I wasn't sure if swapping exc_sum and inc_sum while returning is the right way to achieve the result and how do I keep track of the possible sums which can lead to a maximum sum, in this example, the maximum sum in the left subtree is (1+3+4) whereas the sum which leads to the final maximum is (3+4+5), so how should (3+4) be tracked? Should all the intermediary sums stored in a table?
Lets say dp[u][select] stores the answer: maximum sub sequence sum with no two nodes having edge such that we consider only the sub-tree rooted at node u ( such that u is selected or not ). Now you can write a recursive program where state of each recursion is (u,select) where u means root of the sub graph being considered and select means whether or not we select node u. So we get the following pseudo code
/* Initialize dp[][] to be -1 for all values (u,select) */
/* Select is 0 or 1 for false/true respectively */
int func(int node , int select )
{
if(dp[node][select] != -1)return dp[node][select];
int ans = 0,i;
// assuming value of node is same as node number
if(select)ans=node;
//edges[i] stores children of node i
for(i=0;i<edges[node].size();i++)
{
if(select)ans=ans+func(edges[node][i],1-select);
else ans=ans+max(func(edges[node][i],0),func(edges[node][i],1));
}
dp[node][select] = ans;
return ans;
}
// from main call, root is root of tree and answer is
// your final answer
answer = max(func(root,0),func(root,1));
We have used memoization in addition to recursion to reduce time complexity.Its O(V+E) in both space and time. You can see here a working version of
the code Code. Click on the fork on top right corner to run on test case
4 1
1 2
1 5
2 3
2 4
It gives output 12 as expected.
The input format is specified in comments in the code along with other clarifications. Its in C++ but there is not significant changes if you want it in python once you understand the code. Do post in comments if you have any doubts regarding the code.
The problem:
Given a particular sequence of up to 50 integers, which represent
nodes of a certain binary search tree (BST), how many permutations of
this sequence are there, which will also give rise to exactly the same
BST? Include the original sequence in your total count as 1 sequence.
For instance, answer = 6 for such a sequence [5,2,1,9,8]: [5,2,1,9,8]
(itself, the original sequence given), [5,9,8,2,1], [5,2,9,1,8],
[5,9,2,1,8], [5,2,9,8,1], [5,9,2,8,1]
Suppose you have your example sequence [5,2,1,9,8]. The first node will become the root of the binary tree, so we know the first node must be 5.
From then on, all the nodes smaller than 5 will go to the left child, and all the nodes greater than 5 will go to the right child.
So you can solve this recursively, count the number of ways of making the left subtree (consisting of [2,1]) multiply by the number of ways of making the right subtree (consisting of [9,8]) and multiply by the number of ways of ordering the arrival of integers on opposite sides.
Sample code
def nCr(n,r):
"""Return number of ways of choosing r objects from n"""
top, bottom = 1, 1
for i in xrange(r):
top *= n
n = n - 1
bottom *= r
r = r- 1
return top / bottom
def count_ways_to_make_bst(seq):
if len(seq)<=1:
return 1
num_remaining = len(seq) - 1
left = [r for r in seq[1:] if r>seq[0]]
right = [r for r in seq[1:] if r<seq[0]]
return count_ways_to_make_bst(left) * count_ways_to_make_bst(right) * nCr(num_remaining,len(left))
Assume that the given node sequence is the order in which we added those elements into the BST.
We can easily see that if two nodes a and b belong to two different branches, so the order of adding a and b will not affect the final structure.
We only need to ensure that the parent need to be added first, before its children is added.
For the example [5,2,1,9,8]: 5 should always added first.
There will be only two choices: add 2 or add 9
If we add 2: 1 can be added in any order
similarly, If we add 9: 8 can be added anytime after that
So, now we have our algo while travelling from root -> leaf:
int numberOfWays (Node node) {
if(node is leaf) return 1;
int a = numberOfWays(node.left);
int b = numberOfWays(node.right);
int x = number of node in the left;
int y = number of node in the right;
//nCr is combination, so we can choose x places in total of x + y positions
return nCr(x+y,x) * a * b;
}
Given a binary tree that might contain duplicate values, you need to find minimum distance between two given values. Note that the binary tree can be large.
For example:
5
/ \
1 7
/ \ / \
4 3 8 2
/ \
1 2
The function should return 2 for (1 and 2 as input).
(If duplicates are not present, we can find LCA and then calculate the distance.)
I've written the following code but I couldn't handle cases when the values are present in different subtrees and in the below cases:
root = 1, root.left = 4, root.left.left = 3, root.left.right = 2, root.left.left.left = 1
root = 1, root.left = 4, root.left.left = 3, root.left.left.left = 1, root.left.left.right = 2
void dist(struct node* root,int& min,int n1,int n2,int pos1,int pos2,int level) {
if(!root)
return;
if(root->data==n1){
pos1 = level;
if(pos2>=0)
if(pos1-pos2 < min)
min = pos1-pos2;
}
else if(root->data==n2){
pos2 = level;
if(pos1>=0)
if(pos2-pos1 < min)
min = pos2-pos1;
}
dist(root->left,min,n1,n2,pos1,pos2,level+1);
dist(root->right,min,n1,n2,pos1,pos2,level+1);
}
I think at each node we can find if that node is the LCA of the values or not. If that node is LCA then find the distance and update min accordingly, but this would take O(n2).
Following is an algorithm to solve the problem:-
traverse all of the tree and calculate paths for each node using binary strings representation and store into hash map
eg. For your tree the hashmap will be
1 => 0,000
2 => 001,11
3 => 01
...
When query for distance between (u,v) check for each pair and calculate distance between them. Remove common prefix from strings and then sum the remaining lengths
eg. u=1 and v=2
distance(0,001) = 2
distance(0,11) = 3
distance(000,001) = 2
distance(000,11) = 5
min = 2
Note: I think the second step can be made more efficient but need to do more research
You can compute the LCA of a set of nodes by computing LCA(x1, LCA(x2, LCA(x3... and all the nodes in the set will be somewhere below this LCA. If you compare the LCAs of two sets of nodes and one is not directly beneath the other then the minimum distance between any two nodes in different sets will be at least the distance between the LCAs. If one LCA is above the other then the minimum distance could be zero.
This allows a sort of branch and bound approach. At each point you have a best minimum distance so far (initialized as infinity). Given two sets of nodes, use their LCAs to work out a lower bound on their minimum distance and discard them if this is no better than the best answer so far. If not discarded, split each set into two plus a possible single depending on whether each node in the set is to the left of the LCA, to the right of the LCA, or is the LCA. Recursively check for the minimum distance in the (up to nine) pairs of split sets. If both splits in a pair are below some minimum size, just work out the LCAs and minimum distances of each pair of nodes across the two sets - at this point may find out that you have a new best answer and can update the best answer so far.
Looking at the example at the top of the question, the LCA of the 2s is the root of the tree, and the LCA of the 1s is the highest 1. So the minimum distance between these two sets could be close to zero. Now split each set in two. The left hand 2 is distance two from both of the two 1s. The LCA of the right hand 2 is itself, on the right hand branch of the tree, and the LCA of each of the two 1s is down on the left hand branch of the tree. So the distance between the two is at least two, and we could tell this even if we had a large number of 2s anywhere below the position of the existing right-hand two, and a large number of 1s anywhere on the left hand subtree.
Do a pre-order traversal of the tree (or any traversal should work).
During this process, simply keep track of the closest 1 and 2, and update the distance whenever you find a 2 and the closest 1 is closer than the closest distance so far, or vice versa.
Code (C++, untested first draft): (hardcoded 1 and 2 for simplicity)
int getLeastDistance(Node *n, int *distTo1, int *distTo2)
{
if (n == NULL)
return;
int dist = LARGE_VALUE;
// process current node
if (n->data == 1)
{
dist = *distTo2;
*distTo1 = 0;
}
else if (n->data == 2)
{
dist = *distTo1;
*distTo2 = 0;
}
// go left
int newDistTo1 = *distTo1 + 1,
newDistTo2 = *distTo2 + 1;
dist = min(dist, getLeastDistance(n->left, &newDistTo1, &newDistTo2));
// update distances
*distTo1 = min(*distTo1, newDistTo1 + 1);
*distTo2 = min(*distTo2, newDistTo2 + 1);
// go right
newDistTo1 = *distTo1 + 1;
newDistTo2 = *distTo2 + 1;
dist = min(dist, getLeastDistance(n->right, &newDistTo1, &newDistTo2));
}
Caller:
Node root = ...;
int distTo1 = LARGE_VALUE, distTo2 = LARGE_VALUE;
int dist = getLeastDistance(&root, &distTo1, &distTo2);
Just be sure to make LARGE_VALUE far enough from the maximum value for int such that it won't overflow if incremented (-1 is probably safer, but it requires more complex code).