# Segment Tree

Hello Everyone,

In this post, Range Minimum Query problem is discussed as another example where Segment Tree can be used. Following is problem statement.

We have an array arr[0 . . . n-1]. We should be able to efficiently find the minimum value from index qs (query start) to qe (query end) where 0 <= qs <= qe <= n-1 .

A simple solution is to run a loop from qs to qe and find minimum element in given range. This solution takes O(n) time in worst case.

Another solution is to create a 2D array where an entry [i, j] stores the minimum value in range arr[i…j]. Minimum of a given range can now be calculated in O(1) time, but preprocessing takes O(n^2) time. Also, this approach needs O(n^2) extra space which may become huge for large input arrays.

Segment tree can be used to do preprocessing and query in moderate time. With segment tree, preprocessing time is O(n) and time to for range minimum query is O(Logn). The extra space required is O(n) to store the segment tree.

Representation of Segment trees
1. Leaf Nodes are the elements of the input array.
2. Each internal node represents minimum of all leaves under it.

An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index 2i+1, right child at 2i+2 and the parent is at

Query for minimum value of given range
Once the tree is constructed, how to do range minimum query using the constructed segment tree. Following is algorithm to get the minimum.

// qs --> query start index, qe --> query end index int RMQ(node, qs, qe) { if range of node is within qs and qe return value in node else if range of node is completely outside qs and qe return INFINITE else return min( RMQ(node’s left child, qs, qe), RMQ(node’s right child, qs, qe) ) }

Implementation:

`// C++ program for range minimum`

`// query using segment tree `

`#include <bits/stdc++.h>`

`using` `namespace` `std;`

` `

`// A utility function to get minimum of two numbers `

`int` `minVal(` `int` `x, ` `int` `y) { ` `return` `(x < y)? x: y; } `

` `

`// A utility function to get the `

`// middle index from corner indexes. `

`int` `getMid(` `int` `s, ` `int` `e) { ` `return` `s + (e -s)/2; } `

` `

`/* A recursive function to get the`

`minimum value in a given range `

`of array indexes. The following `

`are parameters for this function. `

` `

` ` `st --> Pointer to segment tree `

` ` `index --> Index of current node in the `

` ` `segment tree. Initially 0 is `

` ` `passed as root is always at index 0 `

` ` `ss & se --> Starting and ending indexes `

` ` `of the segment represented `

` ` `by current node, i.e., st[index] `

` ` `qs & qe --> Starting and ending indexes of query range */`

`int` `RMQUtil(` `int` `*st, ` `int` `ss, ` `int` `se, ` `int` `qs, ` `int` `qe, ` `int` `index) `

`{ `

` ` `// If segment of this node is a part `

` ` `// of given range, then return `

` ` `// the min of the segment `

` ` `if` `(qs <= ss && qe >= se) `

` ` `return` `st[index]; `

` `

` ` `// If segment of this node`

` ` `// is outside the given range `

` ` `if` `(se < qs || ss > qe) `

` ` `return` `INT_MAX; `

` `

` ` `// If a part of this segment`

` ` `// overlaps with the given range `

` ` `int` `mid = getMid(ss, se); `

` ` `return` `minVal(RMQUtil(st, ss, mid, qs, qe, 2*index+1), `

` ` `RMQUtil(st, mid+1, se, qs, qe, 2*index+2)); `

`} `

` `

`// Return minimum of elements in range`

`// from index qs (query start) to `

`// qe (query end). It mainly uses RMQUtil() `

`int` `RMQ(` `int` `*st, ` `int` `n, ` `int` `qs, ` `int` `qe) `

`{ `

` ` `// Check for erroneous input values `

` ` `if` `(qs < 0 || qe > n-1 || qs > qe) `

` ` `{ `

` ` `cout<<` `"Invalid Input"` `; `

` ` `return` `-1; `

` ` `} `

` `

` ` `return` `RMQUtil(st, 0, n-1, qs, qe, 0); `

`} `

` `

`// A recursive function that constructs`

`// Segment Tree for array[ss..se]. `

`// si is index of current node in segment tree st `

`int` `constructSTUtil(` `int` `arr[], ` `int` `ss, ` `int` `se,`

` ` `int` `*st, ` `int` `si) `

`{ `

` ` `// If there is one element in array,`

` ` `// store it in current node of `

` ` `// segment tree and return `

` ` `if` `(ss == se) `

` ` `{ `

` ` `st[si] = arr[ss]; `

` ` `return` `arr[ss]; `

` ` `} `

` `

` ` `// If there are more than one elements, `

` ` `// then recur for left and right subtrees `

` ` `// and store the minimum of two values in this node `

` ` `int` `mid = getMid(ss, se); `

` ` `st[si] = minVal(constructSTUtil(arr, ss, mid, st, si*2+1), `

` ` `constructSTUtil(arr, mid+1, se, st, si*2+2)); `

` ` `return` `st[si]; `

`} `

` `

`/* Function to construct segment tree `

`from given array. This function allocates`

`memory for segment tree and calls constructSTUtil() to `

`fill the allocated memory */`

`int` `*constructST(` `int` `arr[], ` `int` `n) `

`{ `

` ` `// Allocate memory for segment tree `

` `

` ` `//Height of segment tree `

` ` `int` `x = (` `int` `)(` `ceil` `(log2(n))); `

` `

` ` `// Maximum size of segment tree `

` ` `int` `max_size = 2*(` `int` `)` `pow` `(2, x) - 1; `

` `

` ` `int` `*st = ` `new` `int` `[max_size]; `

` `

` ` `// Fill the allocated memory st `

` ` `constructSTUtil(arr, 0, n-1, st, 0); `

` `

` ` `// Return the constructed segment tree `

` ` `return` `st; `

`} `

` `

`// Driver program to test above functions `

`int` `main() `

`{ `

` ` `int` `arr[] = {1, 3, 2, 7, 9, 11}; `

` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr); `

` `

` ` `// Build segment tree from given array `

` ` `int` `*st = constructST(arr, n); `

` `

` ` `int` `qs = 1; ` `// Starting index of query range `

` ` `int` `qe = 5; ` `// Ending index of query range `

` `

` ` `// Print minimum value in arr[qs..qe] `

` ` `cout<<` `"Minimum of values in range ["` `<<qs<<` `", "` `<<qe<<` `"] "` `<<`

` ` `"is = "` `<<RMQ(st, n, qs, qe)<<endl; `

` `

` ` `return` `0; `

`} `

` `

Output:

Minimum of values in range [1, 5] is = 2

Time Complexity:
Time Complexity for tree construction is O(n). There are total 2n-1 nodes, and value of every node is calculated only once in tree construction.

Time complexity to query is O(Logn). To query a range minimum, we process at most two nodes at every level and number of levels is O(Logn).