Add sliding window XOR implementation with tests in C++

Author: DataWorshipper

bit_manipulation/sliding_window_xor.cpp

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+/**
+ * @file
+ * @brief Implementation to [calculate XOR of sliding window of size k in an array of n integers]
+ * (https://cses.fi/problemset/task/3426)
+ *
+ * @details
+ * We are given an array of n integers. Our task is to calculate the bitwise XOR of each
+ * window of k elements, from left to right, and cumulatively XOR the results into a single value.
+ * 
+ * Worst Case Time Complexity: O(n)
+ * Space Complexity: O(n)
+ * 
+ * @author [Abhiraj Mandal](https://github.com/DataWorshipper)
+ */
+
+#include <cassert>   /// for assert
+#include <cstdint>   
+#include <vector>   
+#include <iostream>  /// for IO operations
+
+/**
+ * @namespace bit_manipulation
+ * @brief Bit manipulation algorithms
+ */
+namespace bit_manipulation {
+/**
+ * @namespace sliding_window_xor
+ * @brief Functions for cumulative XOR of sliding windows in arrays
+ */
+namespace sliding_window_xor {
+
+/**
+ * @brief Computes cumulative XOR of all windows of size k
+ * 
+ * @param n Size of the array
+ * @param k Window size
+ * @param x Initial value to generate the array
+ * @param a Multiplier in array generation
+ * @param b Increment in array generation
+ * @param c Modulo in array generation
+ * @returns std::uint64_t The cumulative XOR of all windows of size k
+ * 
+ * @details
+ * This function generates the array using the recurrence:
+ *   arr[0] = x
+ *   arr[i] = (a * arr[i-1] + b) % c
+ * 
+ * It maintains a sliding window of size k using two pointers l and r:
+ * - x1 stores the XOR of the current window
+ * - x2 stores the cumulative XOR of all valid windows
+ * 
+ * This approach ensures that the algorithm runs in O(n) time.
+ */
+std::uint64_t compute(
+    std::uint64_t n,
+    std::uint64_t k,
+    std::uint64_t x,
+    std::uint64_t a,
+    std::uint64_t b,
+    std::uint64_t c) {
+
+    // Generate the array of n elements
+    std::vector<std::uint64_t> arr(n);
+    arr[0] = x;  // First element of the array
+
+    for (std::uint64_t i = 1; i < n; ++i) {
+        arr[i] = (a * arr[i - 1] + b) % c;  // recurrence relation
+    }
+
+    std::uint64_t x1 = 0;  // XOR of the current window
+    std::uint64_t x2 = 0;  // Cumulative XOR of all windows of size k
+    std::uint64_t l = 0;   // Left pointer of sliding window
+    std::uint64_t r = 0;   // Right pointer of sliding window
+
+    // Slide the window over the array
+    while (r < n) {
+        x1 ^= arr[r];  // include current element in window XOR
+
+        // Shrink window from left if size exceeds k
+        while (r - l + 1 > k) {
+            x1 ^= arr[l];  // remove leftmost element from window XOR
+            ++l;
+        }
+
+        // If window size equals k, add it to cumulative XOR
+        if (r - l + 1 == k) {
+            x2 ^= x1;
+        }
+
+        ++r;  // Move right pointer
+    }
+
+    return x2;  // Return cumulative XOR of all windows
+}
+
+}  // namespace sliding_window_xor
+}  // namespace bit_manipulation
+
+/**
+ * @brief Self-test implementation
+ */
+static void test() {
+    using bit_manipulation::sliding_window_xor::compute;
+
+    // Testcase 1: n = 100, k = 20, expected = 1019
+    assert(compute(100, 20, 3, 7, 1, 997) == 1019);
+
+    // Testcase 2: n = 2, k = 1, expected = 2
+    assert(compute(2, 1, 2, 3, 4, 5) == 2);
+
+    // Testcase 3: n = 5, k = 2
+    assert(compute(5, 2, 1, 1, 1, 100) == 0 ^ 3 ^ 1 ^ 7); 
+
+    // Testcase 4: n = 3, k = 5, expected = 0
+    assert(compute(3, 5, 5, 2, 1, 100) == 0);
+
+    // Testcase 5: n = 4, k = 4, expected = 0
+    assert(compute(4, 4, 3, 1, 0, 10) == 0);
+
+    std::cout << "All test cases successfully passed!" << std::endl;
+}
+
+/**
+ * @brief Main function
+ * @returns 0 on exit
+ */
+int main() {
+    test();  // run self-test implementations
+    return 0;
+}