Sync LeetCode submission Runtime - 235 ms (68.11%), Memory - 17.7 MB (74.49%)

Author: jimit105

2170-count-number-of-maximum-bitwise-or-subsets/README.md

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+<p>Given an integer array <code>nums</code>, find the <strong>maximum</strong> possible <strong>bitwise OR</strong> of a subset of <code>nums</code> and return <em>the <strong>number of different non-empty subsets</strong> with the maximum bitwise OR</em>.</p>
+
+<p>An array <code>a</code> is a <strong>subset</strong> of an array <code>b</code> if <code>a</code> can be obtained from <code>b</code> by deleting some (possibly zero) elements of <code>b</code>. Two subsets are considered <strong>different</strong> if the indices of the elements chosen are different.</p>
+
+<p>The bitwise OR of an array <code>a</code> is equal to <code>a[0] <strong>OR</strong> a[1] <strong>OR</strong> ... <strong>OR</strong> a[a.length - 1]</code> (<strong>0-indexed</strong>).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,1]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> The maximum possible bitwise OR of a subset is 3. There are 2 subsets with a bitwise OR of 3:
+- [3]
+- [3,1]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,2,2]
+<strong>Output:</strong> 7
+<strong>Explanation:</strong> All non-empty subsets of [2,2,2] have a bitwise OR of 2. There are 2<sup>3</sup> - 1 = 7 total subsets.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,2,1,5]
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> The maximum possible bitwise OR of a subset is 7. There are 6 subsets with a bitwise OR of 7:
+- [3,5]
+- [3,1,5]
+- [3,2,5]
+- [3,2,1,5]
+- [2,5]
+- [2,1,5]</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 16</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+</ul>

2170-count-number-of-maximum-bitwise-or-subsets/solution.py

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+# Approach 1: Recursion
+
+# Time: O(2^n)
+# Space: O(n)
+
+class Solution:
+    def countMaxOrSubsets(self, nums: List[int]) -> int:
+        max_or_value = 0
+        for num in nums:
+            max_or_value |= num
+        return self._count_subsets(nums, 0, 0, max_or_value)
+
+    def _count_subsets(self, nums, index, current_or, target_or):
+        # Base case: reached end of array
+        if index == len(nums):
+            return 1 if current_or == target_or else 0
+
+        # Don't include current number
+        count_without = self._count_subsets(nums, index + 1, current_or, target_or)
+
+        # Include current number
+        count_with = self._count_subsets(nums, index + 1, current_or | nums[index], target_or)
+
+        # Return sum of both cases
+        return count_without + count_with
+