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Applications, Advantages and Disadvantages of Segment Tree

First, let us understand why we need it prior to landing on the introduction so as to get why this concept was introduced. Suppose we are given an array and we need to find out the subarray 

Purpose of Segment Trees:



A segment tree is a data structure that deals with a range of queries over an array. It has a divide and conquers approach. Used to solve range minimum and maximum & Sum Queries and Range Update Queries in O (log n) time complexity.

Construction of Segment Tree:



array[]: 5, 3, 2, 4, 1, 8, 6 10

we’ll use divide and conquer

 

Now we’ll see how to segment the tree Construction is done.

Number of nodes will be = n + n/2 + n/4 + …… + 2+1

Geometric progression: common difference will be 2

let, number of terms be ‘x’, a=first term and r = common ratio.

(ar)^x-1 = n

a=1,r=2

(2)^x-1=n

log2(2)^x-1=log2n

x=1+log2n=number of levels

(2) – number of nodes = 1+2+4+….+n/2+n/4+n

1[(2)^1+logn – 1]/2-1 => (2n-1)

 

Let us take an example of returning and updating the sum of the subarray a[i…..j] of size n.

Example:

 

Query: output the sum from i=1 to i=5

Approach 1: Iterate from i = 1 to i = 5 and calculate the sum, update the element at i’th index, we’ll update a[i] = updated element. now, the time complexity of the query is O(n) and the update is O(1).

Approach 2: Prefix Sum approach

First, we’ll build the prefix sum array.

 

Query: Output the sum from i to j

sum[i….j] = {pref[j] – pref[i-1]} (if i!=0)

pref[j] (if i=0)

Time Complexity : O(1)

Updating value: Put a[i] = updated value.

To update the prefix array we need to change all prefix[i].

Array:

5 3 2 4 1 8 6 10

i=4,

Prefix sum array:

5 8 10  14 15 23 29 39

Let’s take an example to understand this.

Example: Update the 4th index element to 13.

Original array: arr[] 

5 3 2 4 13 8 6 10

Prefix sum: arr[] all elements get changed because the 4th index is updated so it will affect all elements after the 3rd index.

5 8 10 14 27 35 41 51

Time Complexity comparison table:

  Query Update
Approach-1 O(n) O(1)
Approach-2 O(1) O(n)
Segment Tree O(log(n)) O(log(n))

Requirement of log(n) time complexity: Many times, the number of queries and number of updates are of the order of 105-106, we will get TLE if we use Approach 1 or Approach 2.
 

Building a segment tree:

It is very simple to build a segment tree, we use the divide and conquer approach to build the segment tree.

pseudocode for implementation of Segment tree:

const int N = 1e5+2;
int a[N];
int tree[4*N];

void build(int node, int st, int en) {

if(st==en) {

tree[node] = a[st];
return;
}

int mid = (st+en)/2;
build(2*node, st, mid);
build(2*node+1, mid+1, en);

tree[node] = tree[2*node] + tree[2*node+1];
}

 

Applications of Segment trees:

Segment tree data structure can be used to solve various problems like:

Advantages and Disadvantages of segment trees:

S.No   Advantages Disadvantages
1 There is no need to know tree rotation because in the test cases a divide and conquer algorithm. Time Complexity of each and every query is O(log^2 max(n))
2 Fast execution of code in general test cases. Source code is longer than source using a balanced tree.
3 Allows processing interval or range queries in logarithmic time. Segment trees require a large amount of memory to store all the nodes of the tree.
4  It performs well for large datasets.  The implementation of segment trees can be complex and difficult to understand.
5 Segment trees can be used to perform both static and dynamic operations on an array. Slower for point update.

Related Topic: Segment Tree

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