Sync LeetCode submission Runtime - 47 ms (28.71%), Memory - 18.2 MB (28.52%)

Author: jimit105

0487-max-consecutive-ones-ii/README.md

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+<p>Given a binary array <code>nums</code>, return <em>the maximum number of consecutive </em><code>1</code><em>&#39;s in the array if you can flip at most one</em> <code>0</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,0,1,1,0]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 
+- If we flip the first zero, nums becomes [1,1,1,1,0] and we have 4 consecutive ones.
+- If we flip the second zero, nums becomes [1,0,1,1,1] and we have 3 consecutive ones.
+The max number of consecutive ones is 4.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,0,1,1,0,1]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 
+- If we flip the first zero, nums becomes [1,1,1,1,0,1] and we have 4 consecutive ones.
+- If we flip the second zero, nums becomes [1,0,1,1,1,1] and we have 4 consecutive ones.
+The max number of consecutive ones is 4.
+</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>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong> What if the input numbers come in one by one as an infinite stream? In other words, you can&#39;t store all numbers coming from the stream as it&#39;s too large to hold in memory. Could you solve it efficiently?</p>

0487-max-consecutive-ones-ii/solution.py

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+# Approach 2: Sliding Window
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
+        longest_sequence = 0
+        left, right = 0, 0
+        num_zeros = 0
+
+        while right < len(nums):
+            if nums[right] == 0:
+                num_zeros += 1
+
+            while num_zeros == 2: # if the window is invalid, contract the window
+                if nums[left] == 0:
+                    num_zeros -= 1
+                left += 1
+
+            longest_sequence = max(longest_sequence, right - left + 1)
+            right += 1 # expand the sequence
+
+        return longest_sequence
+