Segment tree is basically taking the recursion pattern of merge sort and turning it into data structure. So, before we start with segment tree we have to know how merge sort works. If we see the diagram below;
So, what we are doing here is- we are recursing down the
array by diving it in half and then we are recursing back up and merging the
sorted array. On each level while recursing down it take linear amount of time
because we are loop over the array and diving in half which takes linear amount
of time, then while recursing up while combing the sorted list in liniear
amount of time. So we can use this information to analyze mergesort runtime. If
we see the width of the array we get at the end of splitting the array , we get
the width as n so, that’a O(n) and the height of the array is log n because we
are splitting the array in half each time and log of something basically means
how many time we can diving that in half. Now if we combine both width and
height we get the area a nlog n that means the complexity cof merge sort
is O(nlog n).
Now that we know how merge sort works the question is -how
can we turn this into data structure? What we’ll do is we’ll consider each
array of the merge sort as different nodes and use the nodes to form a tree.
So, I am taking the sequence of the recursion and turning it into data
structure. The tree will look like below:
:
If we try to solve this using linear search the if we have n
element and 1 query time complexity will be O(n*1) that means O(n).So, if we
have q queries the complexity will be O(nQ) . Now if n<=10^5 and q<=10^5
and the time limit is 1second, with linear search we’ll get a TLE. But if we
use segment tree, that we’ll help us reducing the time complexity. For this
problem we’ll draw a similar tree like before.
But in each node we’ll save the minimum element of the
entire range that node. So, the root node will contain minimum element of the
range [0-5] and likewise other nodes will contain minimum element of it’s
range. And we’ll start storing the minimum
value from the leaf nodes because the range size of the leaf is 1 from each
node, so the element of that index itself. First we’ll store the number 8 node
where range is 0-0 so we have to just put the element of 0th index
which is 1, then in range 1-1 we’ll store element of index 1 which is 3,
likewise we’ll fill all the leaf nodes with there values. Now we’ll move on to
the parent nodes of the leafs. First we’ll store the minim of range [0-1] which
is the minimum of range [0,0] and [1,1] that is 1, then minimum of range [3-4]
is minimum of range[3.3] and [3.4] which is -2. Then we’ll store the minimum of
range [3-5] which is -2 , minimum of range [0-2] is 1 and minimum of range
[0-5] is -2. Now that we’ve build our tree .At first we’ve filled the leaf
nodes and then in a bottom up way we’ve filled the rest of the nodes.
Now we’ll see how to build this tree using code. We will use
an array to build this tree. If you have noticed I have numbered the nodes in
the tree and where 10 and 11 is missing because node 5 can’t be divided
further. The numbering of the nodes is done so that we can make a relationship
between parent and child nodes. So, node 1’s children are 2 and 3, node 2’s
children are 4 and 5, node 3’s children are 6 and 7. Can you see a pattern
here? When we are getting node i it’s children are 2*I and 2*i+1 and if we have
children and want to find out it’s parent then if the child node is i, the
parent node will be i/2 because for children 2,3 we get parent 1 likewise
children 4,5 we get parent 2 . By using these notation we can store the
children node in the array. So, the array for our tree we’ll look like this:
The nodes are not in the tree I have kept those indexes empty.
So for n elements we’ll need total 2n-1
nodes in the tree but when my allocating
array that is larger than 2n-1 because we re also taking some extra indexes
which are empty, so the exact size of the array will be 2*(2^ceil(log n))+1 ≈ 4*n+1, this is the maximum size needed
to build a tree of size n.
Function for building
the tree is:
Now that we have made the segment tree, we have to do the query in the
tree. In this case we are finding the minimum element in a range. So, when the
query comes, there will be three cases .
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1)This is complete overlap because
the node range [0,5] is completely
inside the query range[0,5], so in complete overlap we just have to return
tree[index], that’s how we get -2 as output
To know the second case and third case let’s see an example-
If user does a query for range [0,2] then in the tree the root node is the
range [0,5] that means we’ve partial overlap Case is:
2)Partial overlap because the query range [0,2] is partially inside the root node range[0,5], So, in partial overlap we’ll go to
both sides left and right. First we’ll go to left in left we get the node range[0,2]and
our query range is also [0,2] this is complete overlap so we’ll return the
value which this node contains which is 1.
3) When the query range is outside the node range we get no
overlap, in this case we’ll return infinity
So, the minimum of 1 and infinity we’ll be 1. And if you see
our input array, the minimum element between [0,2] is also 1 that means we’re
getting the right output.
So, the three cases for segment tree is:
1) Complete overlap where output is element of that node
2) Partial overlap where we have to find output by going
both sides
3) No overlap, in this case we’ll return infinity
Now let’s see the code for the query:
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