Create AlternateXOR.cpp

Cpp/AlternateXOR.cpp

@@ -0,0 +1,126 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+#define MAX 100000
+
+// Function to find the maximum and the
+// minimum elements from the array after
+// performing the given operation k times
+void xorOnSortedArray(int arr[], int n, int k, int x)
+{
+
+    // To store the current sequence of elements
+    int arr1[MAX + 1] = { 0 };
+
+    // To store the next sequence of elements
+    // after xoring with current elements
+    int arr2[MAX + 1] = { 0 };
+    int xor_val[MAX + 1];
+
+    // Store the frequency of elements of arr[] in arr1[]
+    for (int i = 0; i < n; i++)
+        arr1[arr[i]]++;
+
+    // Storing all precomputed XOR values so that
+    // we don't have to do it again and again
+    // as XOR is a costly operation
+    for (int i = 0; i <= MAX; i++)
+        xor_val[i] = i ^ x;
+
+    // Perform the operations k times
+    while (k--) {
+
+        // The value of count decide on how many elements
+        // we have to apply XOR operation
+        int count = 0;
+        for (int i = 0; i <= MAX; i++) {
+            int store = arr1[i];
+
+            // If current element is present in
+            // the array to be modified
+            if (arr1[i] > 0) {
+
+                // Suppose i = m and arr1[i] = num, it means
+                // 'm' appears 'num' times
+                // If the count is even we have to perform
+                // XOR operation on alternate 'm' starting
+                // from the 0th index because count is even
+                // and we have to perform XOR operations
+                // starting with initial 'm'
+                // Hence there will be ceil(num/2) operations on
+                // 'm' that will change 'm' to xor_val[m] i.e. m^x
+                if (count % 2 == 0) {
+                    int div = ceil((float)arr1[i] / 2);
+
+                    // Decrease the frequency of 'm' from arr1[]
+                    arr1[i] = arr1[i] - div;
+
+                    // Increase the frequency of 'm^x' in arr2[]
+                    arr2[xor_val[i]] += div;
+                }
+
+                // If the count is odd we have to perform
+                // XOR operation on alternate 'm' starting
+                // from the 1st index because count is odd
+                // and we have to leave the 0th 'm'
+                // Hence there will be (num/2) XOR operations on
+                // 'm' that will change 'm' to xor_val[m] i.e. m^x
+                else if (count % 2 != 0) {
+                    int div = arr1[i] / 2;
+                    arr1[i] = arr1[i] - div;
+                    arr2[xor_val[i]] += div;
+                }
+            }
+
+            // Updating the count by frequency of
+            // the current elements as we have
+            // processed that many elements
+            count = count + store;
+        }
+
+        // Updating arr1[] which will now store the
+        // next sequence of elements
+        // At this time, arr1[] stores the remaining
+        // 'm' on which XOR was not performed and
+        // arr2[] stores the frequency of 'm^x' i.e.
+        // those 'm' on which operation was performed
+        // Updating arr1[] with frequency of remaining
+        // 'm' & frequency of 'm^x' from arr2[]
+        // With help of arr2[], we prevent sorting of
+        // the array again and again
+        for (int i = 0; i <= MAX; i++) {
+            arr1[i] = arr1[i] + arr2[i];
+
+            // Resetting arr2[] for next iteration
+            arr2[i] = 0;
+        }
+    }
+
+    // Finding the maximum and the minimum element
+    // from the modified array after the operations
+    int min = INT_MAX;
+    int max = INT_MIN;
+    for (int i = 0; i <= MAX; i++) {
+        if (arr1[i] > 0) {
+            if (min > i)
+                min = i;
+            if (max < i)
+                max = i;
+        }
+    }
+
+    // Printing the max and the min element
+    cout << min << " " << max << endl;
+}
+
+// Driver code
+int main()
+{
+    int arr[] = { 605, 986 };
+    int n = sizeof(arr) / sizeof(arr[0]);
+    int k = 548, x = 569;
+
+    xorOnSortedArray(arr, n, k, x);
+
+    return 0;
+}