Find the XOR of the elements in the given range [L, R] with the value K for a given set of queries
Given an array arr[] and Q queries, the task is to find the resulting updated array after Q queries. There are two types of queries and the following operation is performed by them:
- Update(L, R, K): Perform XOR for each element within the range L to R with K.
- Display(): Display the current state of the given array.
Examples:
Input: arr[] = {1, 2, 3, 4, 5, 6}, Q = 2
Query 1: Update(1, 5, 3)
Query 2: Display()
Output: 2 1 0 7 6 6
Explanation:
The update query performs XOR of all elements in the range [1, 5].
After performing the update query, the above array is displayed for display query because:
1 ^ 3 = 2
2 ^ 3 = 1
3 ^ 3 = 0
4 ^ 3 = 7
5 ^ 3 = 6
Input: arr[] = {2, 4, 6, 8, 10}, Q = 3
Query 1: Update(1, 3, 2)
Query 2: Update(2, 4, 3)
Query 3: Display()
Output: 0 5 7 11 10
Explanation:
There are two update queries.
After performing the first update query, the given array is changed to {0, 6, 4, 8, 10}. This is because:
2 ^ 2 = 0
4 ^ 2 = 6
6 ^ 2 = 4
After obtaining this array, this array further gets changed for the second query as the {0, 5, 7, 11, 10}. This is because:
6 ^ 3 = 5
4 ^ 3 = 7
8 ^ 3 = 11
Naive Approach: The naive approach for this problem is for every update query, run a loop from L to R and perform the XOR operation with the given K with all the elements from arr[L] to arr[R]. In order to perform the display query, simply print the array arr[]. The time complexity for this approach is O(NQ) where N is the length of the array and Q is the number of queries.
Efficient Approach: The idea is to use a kadane’s algorithm for this problem.
- An empty array res[] is defined with the same length as the original array.
- For every update query {L, R, K}, the res[] array is modified as:
- XOR K to res[L]
- XOR K to res[R + 1]
- Whenever the user gives a display query:
- Initialize a counter variable ‘i’ to the index 1.
- Perform the operation:
res[i] = res[i] ^ res[i - 1]
- Initialize a counter variable ‘i’ to the index 0.
- For every index in the array’s the required answer is:
arr[i] ^ res[i]
Below is the implementation of the above approach:
C++
// C++ implementation to perform the // XOR range updates on an array #include <bits/stdc++.h> using namespace std; // Function to perform the update operation // on the given array void update( int res[], int L, int R, int K) { // Converting the indices to // 0 indexing. L -= 1; R -= 1; // Saving the XOR of K from the starting // index in the range [L, R]. res[L] ^= K; // Saving the XOR of K at the ending index // in the given [L, R]. res[R + 1] ^= K; } // Function to display the resulting array void display( int arr[], int res[], int n) { for ( int i = 1; i < n; i++) { // Finding the resultant value in the // result array res[i] = res[i] ^ res[i - 1]; } for ( int i = 0; i < n; i++) { // Combining the effects of the updates // with the original array without // changing the initial array. cout << (arr[i] ^ res[i]) << " " ; } cout << endl; } // Driver code int main() { int arr[] = { 2, 4, 6, 8, 10 }; int N = sizeof (arr) / sizeof (arr[0]); int res[N]; memset (res, 0, sizeof (res)); // Query 1 int L = 1, R = 3, K = 2; update(res, L, R, K); // Query 2 L = 2; R = 4; K = 3; update(res, L, R, K); // Query 3 display(arr, res, N); return 0; } |
Java
// Java implementation to perform the // XOR range updates on an array class GFG{ // Function to perform the update // operation on the given array static void update( int res[], int L, int R, int K) { // Converting the indices to // 0 indexing. L -= 1 ; R -= 1 ; // Saving the XOR of K from the // starting index in the range [L, R]. res[L] ^= K; // Saving the XOR of K at the ending // index in the given [L, R]. res[R + 1 ] ^= K; } // Function to display the resulting array static void display( int arr[], int res[], int n) { int i; for (i = 1 ; i < n; i++) { // Finding the resultant value // in the result array res[i] = res[i] ^ res[i - 1 ]; } for (i = 0 ; i < n; i++) { // Combining the effects of the // updates with the original array // without changing the initial array. System.out.print((arr[i] ^ res[i]) + " " ); } System.out.println(); } // Driver code public static void main(String []args) { int arr[] = { 2 , 4 , 6 , 8 , 10 }; int N = arr.length; int res[] = new int [N]; // Query 1 int L = 1 , R = 3 , K = 2 ; update(res, L, R, K); // Query 2 L = 2 ; R = 4 ; K = 3 ; update(res, L, R, K); // Query 3 display(arr, res, N); } } // This code is contributed by chitranayal |
Python3
# Python3 implementation to perform the # XOR range updates on an array # Function to perform the update operation # on the given array def update(res, L, R, K): # Converting the indices to # 0 indexing. L = L - 1 R = R - 1 # Saving the XOR of K from the starting # index in the range [L, R]. res[L] = res[L] ^ K # Saving the XOR of K at the ending index # in the given [L, R]. res[R + 1 ] = res[R + 1 ] ^ K # Function to display the resulting array def display(arr, res, n): for i in range ( 1 , n): # Finding the resultant value in the # result array res[i] = res[i] ^ res[i - 1 ] for i in range ( 0 , n): # Combining the effects of the updates # with the original array without # changing the initial array. print (arr[i] ^ res[i], end = " " ) # Driver code arr = [ 2 , 4 , 6 , 8 , 10 ] N = len (arr) res = [ 0 ] * N # Query 1 L = 1 R = 3 K = 2 update(res, L, R, K) # Query 2 L = 2 R = 4 K = 3 update(res, L, R, K) # Query 3 display(arr, res, N) # This code is contributed by Pratik Basu |
C#
// C# implementation to perform the // XOR range updates on an array using System; class GFG{ // Function to perform the update // operation on the given array static void update( int []res, int L, int R, int K) { // Converting the indices to // 0 indexing. L -= 1; R -= 1; // Saving the XOR of K from the // starting index in the range [L, R]. res[L] ^= K; // Saving the XOR of K at the ending // index in the given [L, R]. res[R + 1] ^= K; } // Function to display the resulting array static void display( int []arr, int []res, int n) { int i; for (i = 1; i < n; i++) { // Finding the resultant value // in the result array res[i] = res[i] ^ res[i - 1]; } for (i = 0; i < n; i++) { // Combining the effects of the // updates with the original array // without changing the initial array. Console.Write((arr[i] ^ res[i]) + " " ); } Console.WriteLine(); } // Driver code public static void Main() { int []arr = { 2, 4, 6, 8, 10 }; int N = arr.Length; int []res = new int [N]; // Query 1 int L = 1, R = 3, K = 2; update(res, L, R, K); // Query 2 L = 2; R = 4; K = 3; update(res, L, R, K); // Query 3 display(arr, res, N); } } // This code is contributed by Code_Mech |
Javascript
<script> // Javascript implementation to perform the // XOR range updates on an array // Function to perform the update // operation on the given array function update(res, L, R, K) { // Converting the indices to // 0 indexing. L -= 1; R -= 1; // Saving the XOR of K from the // starting index in the range [L, R]. res[L] ^= K; // Saving the XOR of K at the ending // index in the given [L, R]. res[R + 1] ^= K; } // Function to display the resulting array function display(arr, res, n) { let i; for (i = 1; i < n; i++) { // Finding the resultant value // in the result array res[i] = res[i] ^ res[i - 1]; } for (i = 0; i < n; i++) { // Combining the effects of the // updates with the original array // without changing the initial array. document.write((arr[i] ^ res[i]) + " " ); } document.write( "<br/>" ); } // Driver Code let arr = [ 2, 4, 6, 8, 10 ]; let N = arr.length; let res = Array.from({length: N}, (_, i) => 0); // Query 1 let L = 1, R = 3, K = 2; update(res, L, R, K); // Query 2 L = 2; R = 4; K = 3; update(res, L, R, K); // Query 3 display(arr, res, N); </script> |
0 5 7 11 10
Time Complexity:
- The time complexity for the update is O(1).
- The time complexity for displaying the array is O(N).
Auxiliary Space: O(N)
Note: This approach works very well when the update queries are very high compared to the display queries.
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