Sync LeetCode submission Runtime - 203 ms (65.63%), Memory - 28.9 MB (50.60%)

Author: jimit105

1813-maximum-erasure-value/README.md

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+<p>You are given an array of positive integers <code>nums</code> and want to erase a subarray containing&nbsp;<strong>unique elements</strong>. The <strong>score</strong> you get by erasing the subarray is equal to the <strong>sum</strong> of its elements.</p>
+
+<p>Return <em>the <strong>maximum score</strong> you can get by erasing <strong>exactly one</strong> subarray.</em></p>
+
+<p>An array <code>b</code> is called to be a <span class="tex-font-style-it">subarray</span> of <code>a</code> if it forms a contiguous subsequence of <code>a</code>, that is, if it is equal to <code>a[l],a[l+1],...,a[r]</code> for some <code>(l,r)</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [4,2,4,5,6]
+<strong>Output:</strong> 17
+<strong>Explanation:</strong> The optimal subarray here is [2,4,5,6].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [5,2,1,2,5,2,1,2,5]
+<strong>Output:</strong> 8
+<strong>Explanation:</strong> The optimal subarray here is [5,2,1] or [1,2,5].
+</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>4</sup></code></li>
+</ul>

1813-maximum-erasure-value/solution.py

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+# Approach: Two Pointer
+
+# Time: O(n)
+# Space: O(k), k = size of the unique elements in largest valid subarray
+
+class Solution:
+    def maximumUniqueSubarray(self, nums: List[int]) -> int:
+        seen = set()
+        curr_sum = 0
+        max_sum = 0
+        left = 0
+
+        for right in range(len(nums)):
+            while nums[right] in seen:
+                curr_sum -= nums[left]
+                seen.remove(nums[left])
+                left += 1
+
+            seen.add(nums[right])
+            curr_sum += nums[right]
+
+            max_sum = max(max_sum, curr_sum)
+
+        return max_sum