# Segment Tree | Set 3 (XOR of given range)

We have an array arr[0 . . . n-1]. There are two type of queries

1. Find the XOR of elements from index l to r where 0 <= l <= r <= n-1
2. Change value of a specified element of the array to a new value x. We need to do arr[i] = x where 0 <= i <= n-1.

There will be total of q queries.

Input Constraint

``` n <= 10^5, q <= 10^5
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Solution 1
A simple solution is to run a loop from l to r and calculate xor of elements in given range. To update a value, simply do arr[i] = x. The first operation takes O(n) time and second operation takes O(1) time. Worst case time complexity is O(n*q) for q queries
which will take huge time for n ~ 10^5 and q ~ 10^5. Hence this solution will exceed time limit.

Solution 2
Another solution is to store xor in all possible ranges but there are O(n^2) possible ranges hence with n ~ 10^5 it wil exceed space complexity, hence without considering time complexity, we can state this solution will not work.

Solution 3 (Segment Tree)

Prerequesite : Segment Tree

We build a segment tree of given array such that array elements are at leaves and internal nodes store XOR of leaves covered under them.

 `// C program to show segment tree operations like construction, ` `// query and update ` `#include ` `#include ` `#include ` `  `  `// 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 xor of values in given range ` `    ``of the array. The following are parameters for this function. ` `  `  `    ``st    --> Pointer to segment tree ` `    ``si    --> 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[si] ` `    ``qs & qe  --> Starting and ending indexes of query range */` `int` `getXorUtil(``int` `*st, ``int` `ss, ``int` `se, ``int` `qs, ``int` `qe, ``int` `si) ` `{ ` `    ``// If segment of this node is a part of given range, then return ` `    ``// the xor of the segment ` `    ``if` `(qs <= ss && qe >= se) ` `        ``return` `st[si]; ` `  `  `    ``// If segment of this node is outside the given range ` `    ``if` `(se < qs || ss > qe) ` `        ``return` `0; ` `  `  `    ``// If a part of this segment overlaps with the given range ` `    ``int` `mid = getMid(ss, se); ` `    ``return` `getXorUtil(st, ss, mid, qs, qe, 2*si+1) ^ ` `           ``getXorUtil(st, mid+1, se, qs, qe, 2*si+2); ` `} ` `  `  `/* A recursive function to update the nodes which have the given  ` `   ``index in their range. The following are parameters ` `    ``st, si, ss and se are same as getXorUtil() ` `    ``i    --> index of the element to be updated. This index is  ` `             ``in input array. ` `   ``diff --> Value to be added to all nodes which have i in range */` `void` `updateValueUtil(``int` `*st, ``int` `ss, ``int` `se, ``int` `i, ``int` `diff, ``int` `si) ` `{ ` `    ``// Base Case: If the input index lies outside the range of  ` `    ``// this segment ` `    ``if` `(i < ss || i > se) ` `        ``return``; ` `  `  `    ``// If the input index is in range of this node, then update  ` `    ``// the value of the node and its children ` `    ``st[si] = st[si] + diff; ` `    ``if` `(se != ss) ` `    ``{ ` `        ``int` `mid = getMid(ss, se); ` `        ``updateValueUtil(st, ss, mid, i, diff, 2*si + 1); ` `        ``updateValueUtil(st, mid+1, se, i, diff, 2*si + 2); ` `    ``} ` `} ` `  `  `// The function to update a value in input array and segment tree. ` `// It uses updateValueUtil() to update the value in segment tree ` `void` `updateValue(``int` `arr[], ``int` `*st, ``int` `n, ``int` `i, ``int` `new_val) ` `{ ` `    ``// Check for erroneous input index ` `    ``if` `(i < 0 || i > n-1) ` `    ``{ ` `        ``printf``(``"Invalid Input"``); ` `        ``return``; ` `    ``} ` `  `  `    ``// Get the difference between new value and old value ` `    ``int` `diff = new_val - arr[i]; ` `  `  `    ``// Update the value in array ` `    ``arr[i] = new_val; ` `  `  `    ``// Update the values of nodes in segment tree ` `    ``updateValueUtil(st, 0, n-1, i, diff, 0); ` `} ` `  `  `// Return xor of elements in range from index qs (quey start) ` `// to qe (query end).  It mainly uses getXorUtil() ` `int` `getXor(``int` `*st, ``int` `n, ``int` `qs, ``int` `qe) ` `{ ` `    ``// Check for erroneous input values ` `    ``if` `(qs < 0 || qe > n-1 || qs > qe) ` `    ``{ ` `        ``printf``(``"Invalid Input"``); ` `        ``return` `-1; ` `    ``} ` `  `  `    ``return` `getXorUtil(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 xor of values in this node ` `    ``int` `mid = getMid(ss, se); ` `    ``st[si] =  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;  ` `  `  `    ``// Allocate memory ` `    ``int` `*st =  (``int` `*)``malloc``(``sizeof``(``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, 5, 7, 9, 11}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `  `  `    ``// Build segment tree from given array ` `    ``int` `*st = constructST(arr, n); ` `  `  `    ``// Print xor of values in array from index 1 to 3 ` `    ``printf``(``"Xor of values in given range = %d\n"``,  ` `            ``getXor(st, n, 1, 3)); ` `  `  `    ``// Update: set arr = 10 and update corresponding  ` `    ``// segment tree nodes ` `    ``updateValue(arr, st, n, 1, 10); ` `  `  `    ``// Find xor after the value is updated ` `    ``printf``(``"Updated xor of values in given range = %d\n"``, ` `             ``getXor(st, n, 1, 3)); ` `    ``return` `0; ` `} `

Output:

```Xor of values in given range = 1
Updated xor of values in given range = 8
```

Time and Space 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(log n).
The time complexity of update is also O(log n).

Total time Complexity is : O(n) for construction + O(log n) for each query = O(n) + O(n * log n) = O(n * log n)

```Time Complexity O(n * log n)
Auxiliary Space  O(n)
```

This article is contributed by Pratik Chhajer. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.