1829. Maximum XOR for Each Query: AC

src/solution/mod.rs

@@ -1378,3 +1378,4 @@ mod s1824_minimum_sideway_jumps;
 mod s1825_finding_mk_average;
 mod s1827_minimum_operations_to_make_the_array_increasing;
 mod s1828_queries_on_number_of_points_inside_a_circle;
+mod s1829_maximum_xor_for_each_query;

src/solution/s1829_maximum_xor_for_each_query.rs

@@ -0,0 +1,103 @@
+/**
+ * [1829] Maximum XOR for Each Query
+ *
+ * You are given a sorted array nums of n non-negative integers and an integer maximumBit. You want to perform the following query n times:
+ * <ol>
+ * 	Find a non-negative integer k < 2^maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized. k is the answer to the i^th query.
+ * 	Remove the last element from the current array nums.
+ * </ol>
+ * Return an array answer, where answer[i] is the answer to the i^th query.
+ *  
+ * Example 1:
+ *
+ * Input: nums = [0,1,1,3], maximumBit = 2
+ * Output: [0,3,2,3]
+ * Explanation: The queries are answered as follows:
+ * 1^st query: nums = [0,1,1,3], k = 0 since 0 XOR 1 XOR 1 XOR 3 XOR 0 = 3.
+ * 2^nd query: nums = [0,1,1], k = 3 since 0 XOR 1 XOR 1 XOR 3 = 3.
+ * 3^rd query: nums = [0,1], k = 2 since 0 XOR 1 XOR 2 = 3.
+ * 4^th query: nums = [0], k = 3 since 0 XOR 3 = 3.
+ *
+ * Example 2:
+ *
+ * Input: nums = [2,3,4,7], maximumBit = 3
+ * Output: [5,2,6,5]
+ * Explanation: The queries are answered as follows:
+ * 1^st query: nums = [2,3,4,7], k = 5 since 2 XOR 3 XOR 4 XOR 7 XOR 5 = 7.
+ * 2^nd query: nums = [2,3,4], k = 2 since 2 XOR 3 XOR 4 XOR 2 = 7.
+ * 3^rd query: nums = [2,3], k = 6 since 2 XOR 3 XOR 6 = 7.
+ * 4^th query: nums = [2], k = 5 since 2 XOR 5 = 7.
+ *
+ * Example 3:
+ *
+ * Input: nums = [0,1,2,2,5,7], maximumBit = 3
+ * Output: [4,3,6,4,6,7]
+ *
+ *  
+ * Constraints:
+ *
+ * 	nums.length == n
+ * 	1 <= n <= 10^5
+ * 	1 <= maximumBit <= 20
+ * 	0 <= nums[i] < 2^maximumBit
+ * 	nums​​​ is sorted in ascending order.
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/maximum-xor-for-each-query/
+// discuss: https://leetcode.com/problems/maximum-xor-for-each-query/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn get_maximum_xor(nums: Vec<i32>, maximum_bit: i32) -> Vec<i32> {
+        let mut s = nums.iter().fold(0, |s, n| s ^ *n);
+        let m = (1 << maximum_bit) - 1; // maximum
+        let mut result = Vec::with_capacity(nums.len());
+
+        for &n in nums.iter().rev() {
+            result.push(s ^ m);
+            s ^= n;
+        }
+
+        result
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1829_example_1() {
+        let nums = vec![0, 1, 1, 3];
+        let maximum_bit = 2;
+
+        let result = vec![0, 3, 2, 3];
+
+        assert_eq!(Solution::get_maximum_xor(nums, maximum_bit), result);
+    }
+
+    #[test]
+    fn test_1829_example_2() {
+        let nums = vec![2, 3, 4, 7];
+        let maximum_bit = 3;
+
+        let result = vec![5, 2, 6, 5];
+
+        assert_eq!(Solution::get_maximum_xor(nums, maximum_bit), result);
+    }
+
+    #[test]
+    fn test_1829_example_3() {
+        let nums = vec![0, 1, 2, 2, 5, 7];
+        let maximum_bit = 3;
+
+        let result = vec![4, 3, 6, 4, 6, 7];
+
+        assert_eq!(Solution::get_maximum_xor(nums, maximum_bit), result);
+    }
+}