Added task 2317.

Author: javadev

README.md

@@ -1485,6 +1485,7 @@ implementation 'com.github.javadev:leetcode-in-java:1.11'
 
 | #    |      Title     | Difficulty  | Tag         | Time, ms | Time, %
 |------|----------------|-------------|-------------|----------|---------
+| 2317 |[Maximum XOR After Operations](src/main/java/g2301_2400/s2317_maximum_xor_after_operations/Solution.java)| Medium || 1 | 100.00
 | 2316 |[Count Unreachable Pairs of Nodes in an Undirected Graph](src/main/java/g2301_2400/s2316_count_unreachable_pairs_of_nodes_in_an_undirected_graph/Solution.java)| Medium || 32 | 100.00
 | 2315 |[Count Asterisks](src/main/java/g2301_2400/s2315_count_asterisks/Solution.java)| Easy || 1 | 100.00
 | 2312 |[Selling Pieces of Wood](src/main/java/g2301_2400/s2312_selling_pieces_of_wood/Solution.java)| Hard | Backtracking | 78 | 63.64

src/main/java/g2301_2400/s2317_maximum_xor_after_operations/Solution.java

@@ -0,0 +1,13 @@
+package g2301_2400.s2317_maximum_xor_after_operations;
+
+// #Medium #2022_06_26_Time_1_ms_(100.00%)_Space_53.3_MB_(100.00%)
+
+public class Solution {
+    public int maximumXOR(int[] nums) {
+        int max = 0;
+        for (int n : nums) {
+            max |= n;
+        }
+        return max;
+    }
+}

src/main/java/g2301_2400/s2317_maximum_xor_after_operations/readme.md

@@ -0,0 +1,40 @@
+2317\. Maximum XOR After Operations
+
+Medium
+
+You are given a **0-indexed** integer array `nums`. In one operation, select **any** non-negative integer `x` and an index `i`, then **update** `nums[i]` to be equal to `nums[i] AND (nums[i] XOR x)`.
+
+Note that `AND` is the bitwise AND operation and `XOR` is the bitwise XOR operation.
+
+Return _the **maximum** possible bitwise XOR of all elements of_ `nums` _after applying the operation **any number** of times_.
+
+**Example 1:**
+
+**Input:** nums = [3,2,4,6]
+
+**Output:** 7
+
+**Explanation:** Apply the operation with x = 4 and i = 3, num[3] = 6 AND (6 XOR 4) = 6 AND 2 = 2.
+
+Now, nums = [3, 2, 4, 2] and the bitwise XOR of all the elements = 3 XOR 2 XOR 4 XOR 2 = 7.
+
+It can be shown that 7 is the maximum possible bitwise XOR.
+
+Note that other operations may be used to achieve a bitwise XOR of 7.
+
+**Example 2:**
+
+**Input:** nums = [1,2,3,9,2]
+
+**Output:** 11
+
+**Explanation:** Apply the operation zero times.
+
+The bitwise XOR of all the elements = 1 XOR 2 XOR 3 XOR 9 XOR 2 = 11.
+
+It can be shown that 11 is the maximum possible bitwise XOR.
+
+**Constraints:**
+
+*   <code>1 <= nums.length <= 10<sup>5</sup></code>
+*   <code>0 <= nums[i] <= 10<sup>8</sup></code>

src/test/java/g2301_2400/s2317_maximum_xor_after_operations/SolutionTest.java

@@ -0,0 +1,18 @@
+package g2301_2400.s2317_maximum_xor_after_operations;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void maximumXOR() {
+        assertThat(new Solution().maximumXOR(new int[] {3, 2, 4, 6}), equalTo(7));
+    }
+
+    @Test
+    void maximumXOR2() {
+        assertThat(new Solution().maximumXOR(new int[] {1, 2, 3, 9, 2}), equalTo(11));
+    }
+}