Add solution #2997

Author: JoshCrozier

README.md

@@ -2115,6 +2115,7 @@
 2976|[Minimum Cost to Convert String I](./solutions/2976-minimum-cost-to-convert-string-i.js)|Medium|
 2980|[Check if Bitwise OR Has Trailing Zeros](./solutions/2980-check-if-bitwise-or-has-trailing-zeros.js)|Easy|
 2981|[Find Longest Special Substring That Occurs Thrice I](./solutions/2981-find-longest-special-substring-that-occurs-thrice-i.js)|Medium|
+2997|[Minimum Number of Operations to Make Array XOR Equal to K](./solutions/2997-minimum-number-of-operations-to-make-array-xor-equal-to-k.js)|Medium|
 2999|[Count the Number of Powerful Integers](./solutions/2999-count-the-number-of-powerful-integers.js)|Hard|
 3024|[Type of Triangle](./solutions/3024-type-of-triangle.js)|Easy|
 3042|[Count Prefix and Suffix Pairs I](./solutions/3042-count-prefix-and-suffix-pairs-i.js)|Easy|

solutions/2997-minimum-number-of-operations-to-make-array-xor-equal-to-k.js

@@ -0,0 +1,38 @@
+/**
+ * 2997. Minimum Number of Operations to Make Array XOR Equal to K
+ * https://leetcode.com/problems/minimum-number-of-operations-to-make-array-xor-equal-to-k/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array nums and a positive integer k.
+ *
+ * You can apply the following operation on the array any number of times:
+ * - Choose any element of the array and flip a bit in its binary representation. Flipping a bit
+ *   means changing a 0 to 1 or vice versa.
+ *
+ * Return the minimum number of operations required to make the bitwise XOR of all elements of the
+ * final array equal to k.
+ *
+ * Note that you can flip leading zero bits in the binary representation of elements. For example,
+ * for the number (101)2 you can flip the fourth bit and obtain (1101)2.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var minOperations = function(nums, k) {
+  let xorResult = 0;
+  for (const num of nums) {
+    xorResult ^= num;
+  }
+
+  let result = 0;
+  let diff = xorResult ^ k;
+  while (diff > 0) {
+    result += diff & 1;
+    diff >>= 1;
+  }
+
+  return result;
+};