XOR in a range of a binary array

Given a binary array arr[] of size N and some queries. Each query represents an index range [l, r]. The task is to find the xor of the elements in the given index range for each query i.e. arr[l] ^ arr[l + 1] ^ … ^ arr[r].

Examples:

Input: arr[] = {1, 0, 1, 1, 0, 1, 1}, q[][] = {{0, 3}, {0, 2}}
Output:
1
0
Query 1: arr ^ arr ^ arr ^ arr = 1 ^ 0 ^ 1 ^ 1 = 1
Query 1: arr ^ arr ^ arr = 1 ^ 0 ^ 1 = 0

Input: arr[] = {1, 0, 1, 1, 0, 1, 1}, q[][] = {{1, 1}}
Output: 0

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The main observation is that the required answer will always be either 0 or 1. If the number of 1’s in the given range are odd then the answer will be 1. Otherwise 0. To answer multiple queries in constant time, use a prefix sum array array pre[] where pre[i] stores the number of 1’s in the original array in the index range [0, i] which can be used to find the number of 1’s in any index range of the given array.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function to return Xor in a range // of a binary array int xorRange(int pre[], int l, int r) {        // To store the count of 1s     int cntOnes = pre[r];     if (l - 1 >= 0)         cntOnes -= pre[l - 1];        // If number of ones are even     if (cntOnes % 2 == 0)         return 0;        // If number of ones are odd     else         return 1; }    // Function to perform the queries void performQueries(int queries[], int q,                     int a[], int n) {     // To store prefix sum     int pre[n];        // pre[i] stores the number of     // 1s from pre to pre[i]     pre = a;     for (int i = 1; i < n; i++)         pre[i] = pre[i - 1] + a[i];        // Perform queries     for (int i = 0; i < q; i++)         cout << xorRange(pre, queries[i],                          queries[i])              << endl; }    // Driver code int main() {     int a[] = { 1, 0, 1, 1, 0, 1, 1 };     int n = sizeof(a) / sizeof(a);        // Given queries     int queries[] = { { 0, 3 }, { 0, 2 } };     int q = sizeof(queries) / sizeof(queries);        performQueries(queries, q, a, n);        return 0; }

Java

 // Java implementation of the approach  import java.util.*;    class GFG {    // Function to return Xor in a range // of a binary array static int xorRange(int pre[], int l, int r) {        // To store the count of 1s     int cntOnes = pre[r];     if (l - 1 >= 0)         cntOnes -= pre[l - 1];        // If number of ones are even     if (cntOnes % 2 == 0)         return 0;        // If number of ones are odd     else         return 1; }    // Function to perform the queries static void performQueries(int queries[][], int q,                            int a[], int n) {     // To store prefix sum     int []pre = new int[n];        // pre[i] stores the number of     // 1s from pre to pre[i]     pre = a;     for (int i = 1; i < n; i++)         pre[i] = pre[i - 1] + a[i];        // Perform queries     for (int i = 0; i < q; i++)         System.out.println(xorRange(pre, queries[i],                                           queries[i])); }    // Driver code public static void main(String[] args)  {     int a[] = { 1, 0, 1, 1, 0, 1, 1 };     int n = a.length;        // Given queries     int queries[][] = { { 0, 3 }, { 0, 2 } };     int q = queries.length;        performQueries(queries, q, a, n); } }    // This code is contributed by Princi Singh

Python3

 # Python3 implementation of the approach    # Function to return Xor in a range # of a binary array def xorRange(pre, l, r):        # To store the count of 1s     cntOnes = pre[r]     if (l - 1 >= 0):         cntOnes -= pre[l - 1]        # If number of ones are even     if (cntOnes % 2 == 0):         return 0        # If number of ones are odd     else:         return 1    # Function to perform the queries def performQueries(queries, q, a, n):            # To store prefix sum     pre = [0 for i in range(n)]        # pre[i] stores the number of     # 1s from pre to pre[i]     pre = a     for i in range(1, n):         pre[i] = pre[i - 1] + a[i]        # Perform queries     for i in range(q):         print(xorRange(pre, queries[i],                              queries[i]))    # Driver code a = [ 1, 0, 1, 1, 0, 1, 1 ] n = len(a)    # Given queries queries = [[ 0, 3 ], [ 0, 2 ]] q = len(queries)    performQueries(queries, q, a, n)    # This code is contributed by Mohit Kumar

C#

 // C# implementation of the approach  using System;    class GFG {    // Function to return Xor in a range // of a binary array static int xorRange(int []pre, int l, int r) {        // To store the count of 1s     int cntOnes = pre[r];     if (l - 1 >= 0)         cntOnes -= pre[l - 1];        // If number of ones are even     if (cntOnes % 2 == 0)         return 0;        // If number of ones are odd     else         return 1; }    // Function to perform the queries static void performQueries(int [,]queries, int q,                            int []a, int n) {     // To store prefix sum     int []pre = new int[n];        // pre[i] stores the number of     // 1s from pre to pre[i]     pre = a;     for (int i = 1; i < n; i++)         pre[i] = pre[i - 1] + a[i];        // Perform queries     for (int i = 0; i < q; i++)         Console.WriteLine(xorRange(pre, queries[i, 0],                                          queries[i, 1])); }    // Driver code public static void Main()  {     int []a = { 1, 0, 1, 1, 0, 1, 1 };     int n = a.Length;        // Given queries     int [,]queries = { { 0, 3 }, { 0, 2 } };     int q = queries.Length;        performQueries(queries, q, a, n); } }    // This code is contributed // by Akanksha Rai

Output:

1
0

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