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

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 <stdio.h>
#include <stdlib.h>
#include <math.h>
// 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)
    // 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");
    // 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[].
// 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[0]);
    // 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[1] = 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;


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 or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Article Tags :
Practice Tags :