Sync LeetCode submission Runtime - 92 ms (55.49%), Memory - 33.6 MB (26.22%)

Author: jimit105

2478-longest-nice-subarray/README.md

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+<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers.</p>
+
+<p>We call a subarray of <code>nums</code> <strong>nice</strong> if the bitwise <strong>AND</strong> of every pair of elements that are in <strong>different</strong> positions in the subarray is equal to <code>0</code>.</p>
+
+<p>Return <em>the length of the <strong>longest</strong> nice subarray</em>.</p>
+
+<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>
+
+<p><strong>Note</strong> that subarrays of length <code>1</code> are always considered nice.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,8,48,10]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The longest nice subarray is [3,8,48]. This subarray satisfies the conditions:
+- 3 AND 8 = 0.
+- 3 AND 48 = 0.
+- 8 AND 48 = 0.
+It can be proven that no longer nice subarray can be obtained, so we return 3.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,1,5,11,13]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The length of the longest nice subarray is 1. Any subarray of length 1 can be chosen.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+</ul>

2478-longest-nice-subarray/solution.py

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+# Approach 2: Sliding Window
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def longestNiceSubarray(self, nums: List[int]) -> int:
+        used_bits = 0 # Track bits used in current window
+        window_start = 0
+        max_length = 0
+
+        for window_end in range(len(nums)):
+            # If current number shares bits with window, shrink window from left until there's no bit conflict
+            while used_bits & nums[window_end] != 0:
+                used_bits ^= nums[window_start] # Remove leftmost element's bits
+                window_start += 1
+
+            # Add current number to window
+            used_bits |= nums[window_end]
+
+            max_length = max(max_length, window_end - window_start + 1)
+
+        return max_length
+
+