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Author: BHAVYAghub

x.cpp

@@ -0,0 +1,64 @@
+// C++ Program to count all subarrays having
+// XOR of elements as given value m with
+// O(n) time complexity.
+#include <bits/stdc++.h>
+using namespace std;
+
+// Returns count of subarrays of arr with XOR
+// value equals to m
+long long subarrayXor(int arr[], int n, int m)
+{
+	long long ans = 0; // Initialize answer to be returned
+
+	// Create a prefix xor-sum array such that
+	// xorArr[i] has value equal to XOR
+	// of all elements in arr[0 ..... i]
+	int* xorArr = new int[n];
+
+	// Create map that stores number of prefix array
+	// elements corresponding to a XOR value
+	unordered_map<int, int> mp;
+
+	// Initialize first element of prefix array
+	xorArr[0] = arr[0];
+
+	// Computing the prefix array.
+	for (int i = 1; i < n; i++)
+		xorArr[i] = xorArr[i - 1] ^ arr[i];
+
+	// Calculate the answer
+	for (int i = 0; i < n; i++) {
+		// Find XOR of current prefix with m.
+		int tmp = m ^ xorArr[i];
+
+		// If above XOR exists in map, then there
+		// is another previous prefix with same
+		// XOR, i.e., there is a subarray ending
+		// at i with XOR equal to m.
+		ans = ans + ((long long)mp[tmp]);
+
+		// If this subarray has XOR equal to m itself.
+		if (xorArr[i] == m)
+			ans++;
+
+		// Add the XOR of this subarray to the map
+		mp[xorArr[i]]++;
+	}
+
+	// Return total count of subarrays having XOR of
+	// elements as given value m
+	return ans;
+}
+
+// Driver program to test above function
+int main()
+{
+	int arr[] = { 4, 2, 2, 6, 4 };
+	int n = sizeof(arr) / sizeof(arr[0]);
+	int m = 6;
+
+	cout << "Number of subarrays having given XOR is "
+		<< subarrayXor(arr, n, m);
+	return 0;
+}
+