Add solution #2044

JoshCrozier · JoshCrozier · commit b927d90c859c · 2025-05-08T12:50:55.000-04:00
diff --git a/README.md b/README.md
@@ -1,4 +1,4 @@
-# 1,706 LeetCode solutions in JavaScript
+# 1,707 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1567,6 +1567,7 @@
 2040|[Kth Smallest Product of Two Sorted Arrays](./solutions/2040-kth-smallest-product-of-two-sorted-arrays.js)|Hard|
 2042|[Check if Numbers Are Ascending in a Sentence](./solutions/2042-check-if-numbers-are-ascending-in-a-sentence.js)|Easy|
 2043|[Simple Bank System](./solutions/2043-simple-bank-system.js)|Medium|
+2044|[Count Number of Maximum Bitwise-OR Subsets](./solutions/2044-count-number-of-maximum-bitwise-or-subsets.js)|Medium|
 2047|[Number of Valid Words in a Sentence](./solutions/2047-number-of-valid-words-in-a-sentence.js)|Easy|
 2053|[Kth Distinct String in an Array](./solutions/2053-kth-distinct-string-in-an-array.js)|Medium|
 2071|[Maximum Number of Tasks You Can Assign](./solutions/2071-maximum-number-of-tasks-you-can-assign.js)|Hard|
diff --git a/solutions/2044-count-number-of-maximum-bitwise-or-subsets.js b/solutions/2044-count-number-of-maximum-bitwise-or-subsets.js
@@ -0,0 +1,34 @@
+/**
+ * 2044. Count Number of Maximum Bitwise-OR Subsets
+ * https://leetcode.com/problems/count-number-of-maximum-bitwise-or-subsets/
+ * Difficulty: Medium
+ *
+ * Given an integer array nums, find the maximum possible bitwise OR of a subset of nums and return
+ * the number of different non-empty subsets with the maximum bitwise OR.
+ *
+ * An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero)
+ * elements of b. Two subsets are considered different if the indices of the elements chosen are
+ * different.
+ *
+ * The bitwise OR of an array a is equal to a[0] OR a[1] OR ... OR a[a.length - 1] (0-indexed).
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var countMaxOrSubsets = function(nums) {
+  const maxOr = nums.reduce((or, num) => or | num, 0);
+  let count = 0;
+
+  backtrack(0, 0);
+
+  return count;
+
+  function backtrack(index, currentOr) {
+    if (currentOr === maxOr) count++;
+    for (let i = index; i < nums.length; i++) {
+      backtrack(i + 1, currentOr | nums[i]);
+    }
+  }
+};
