Add solution #1707

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,491 LeetCode solutions in JavaScript
+# 1,492 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1316,6 +1316,7 @@
 1704|[Determine if String Halves Are Alike](./solutions/1704-determine-if-string-halves-are-alike.js)|Easy|
 1705|[Maximum Number of Eaten Apples](./solutions/1705-maximum-number-of-eaten-apples.js)|Medium|
 1706|[Where Will the Ball Fall](./solutions/1706-where-will-the-ball-fall.js)|Medium|
+1707|[Maximum XOR With an Element From Array](./solutions/1707-maximum-xor-with-an-element-from-array.js)|Hard|
 1716|[Calculate Money in Leetcode Bank](./solutions/1716-calculate-money-in-leetcode-bank.js)|Easy|
 1718|[Construct the Lexicographically Largest Valid Sequence](./solutions/1718-construct-the-lexicographically-largest-valid-sequence.js)|Medium|
 1726|[Tuple with Same Product](./solutions/1726-tuple-with-same-product.js)|Medium|

solutions/1707-maximum-xor-with-an-element-from-array.js

@@ -0,0 +1,59 @@
+/**
+ * 1707. Maximum XOR With an Element From Array
+ * https://leetcode.com/problems/maximum-xor-with-an-element-from-array/
+ * Difficulty: Hard
+ *
+ * You are given an array nums consisting of non-negative integers. You are also given a queries
+ * array, where queries[i] = [xi, mi].
+ *
+ * The answer to the ith query is the maximum bitwise XOR value of xi and any element of nums that
+ * does not exceed mi. In other words, the answer is max(nums[j] XOR xi) for all j such that
+ * nums[j] <= mi. If all elements in nums are larger than mi, then the answer is -1.
+ *
+ * Return an integer array answer where answer.length == queries.length and answer[i] is the
+ * answer to the ith query.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number[][]} queries
+ * @return {number[]}
+ */
+var maximizeXor = function(nums, queries) {
+  nums.sort((a, b) => a - b);
+  const sortedQueries = queries.map((q, i) => [q[0], q[1], i]).sort((a, b) => a[1] - b[1]);
+  const result = new Array(queries.length).fill(-1);
+
+  const trie = {};
+  let numIndex = 0;
+
+  for (const [x, m, queryIndex] of sortedQueries) {
+    while (numIndex < nums.length && nums[numIndex] <= m) {
+      let node = trie;
+      for (let bit = 30; bit >= 0; bit--) {
+        const bitValue = (nums[numIndex] >> bit) & 1;
+        if (!node[bitValue]) node[bitValue] = {};
+        node = node[bitValue];
+      }
+      numIndex++;
+    }
+
+    if (numIndex === 0) continue;
+
+    let maxXor = 0;
+    let node = trie;
+    for (let bit = 30; bit >= 0; bit--) {
+      const bitValue = (x >> bit) & 1;
+      const oppositeBit = bitValue ^ 1;
+      if (node[oppositeBit]) {
+        maxXor |= (1 << bit);
+        node = node[oppositeBit];
+      } else {
+        node = node[bitValue];
+      }
+    }
+    result[queryIndex] = maxXor;
+  }
+
+  return result;
+};