Sync LeetCode submission Runtime - 230 ms (12.07%), Memory - 30.3 MB (17.24%)

Author: jimit105

3307-find-the-maximum-sum-of-node-values/README.md

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+<p>There exists an <strong>undirected</strong> tree with <code>n</code> nodes numbered <code>0</code> to <code>n - 1</code>. You are given a <strong>0-indexed</strong> 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> in the tree. You are also given a <strong>positive</strong> integer <code>k</code>, and a <strong>0-indexed</strong> array of <strong>non-negative</strong> integers <code>nums</code> of length <code>n</code>, where <code>nums[i]</code> represents the <strong>value</strong> of the node numbered <code>i</code>.</p>
+
+<p>Alice wants the sum of values of tree nodes to be <strong>maximum</strong>, for which Alice can perform the following operation <strong>any</strong> number of times (<strong>including zero</strong>) on the tree:</p>
+
+<ul>
+	<li>Choose any edge <code>[u, v]</code> connecting the nodes <code>u</code> and <code>v</code>, and update their values as follows:
+
+	<ul>
+		<li><code>nums[u] = nums[u] XOR k</code></li>
+		<li><code>nums[v] = nums[v] XOR k</code></li>
+	</ul>
+	</li>
+</ul>
+
+<p>Return <em>the <strong>maximum</strong> possible <strong>sum</strong> of the <strong>values</strong> Alice can achieve by performing the operation <strong>any</strong> number of times</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2023/11/09/screenshot-2023-11-10-012513.png" style="width: 300px; height: 277px;padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>Input:</strong> nums = [1,2,1], k = 3, edges = [[0,1],[0,2]]
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> Alice can achieve the maximum sum of 6 using a single operation:
+- Choose the edge [0,2]. nums[0] and nums[2] become: 1 XOR 3 = 2, and the array nums becomes: [1,2,1] -&gt; [2,2,2].
+The total sum of values is 2 + 2 + 2 = 6.
+It can be shown that 6 is the maximum achievable sum of values.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2024/01/09/screenshot-2024-01-09-220017.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 300px; height: 239px;" />
+<pre>
+<strong>Input:</strong> nums = [2,3], k = 7, edges = [[0,1]]
+<strong>Output:</strong> 9
+<strong>Explanation:</strong> Alice can achieve the maximum sum of 9 using a single operation:
+- Choose the edge [0,1]. nums[0] becomes: 2 XOR 7 = 5 and nums[1] become: 3 XOR 7 = 4, and the array nums becomes: [2,3] -&gt; [5,4].
+The total sum of values is 5 + 4 = 9.
+It can be shown that 9 is the maximum achievable sum of values.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2023/11/09/screenshot-2023-11-10-012641.png" style="width: 600px; height: 233px;padding: 10px; background: #fff; border-radius: .5rem;" />
+<pre>
+<strong>Input:</strong> nums = [7,7,7,7,7,7], k = 3, edges = [[0,1],[0,2],[0,3],[0,4],[0,5]]
+<strong>Output:</strong> 42
+<strong>Explanation:</strong> The maximum achievable sum is 42 which can be achieved by Alice performing no operations.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n == nums.length &lt;= 2 * 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>edges.length == n - 1</code></li>
+	<li><code>edges[i].length == 2</code></li>
+	<li><code>0 &lt;= edges[i][0], edges[i][1] &lt;= n - 1</code></li>
+	<li>The input is generated such that <code>edges</code> represent&nbsp;a valid tree.</li>
+</ul>

3307-find-the-maximum-sum-of-node-values/solution.py

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+class Solution:
+    def maxSumOfNodes(self, index, isEven, nums, k, memo):
+        if index == len(nums):
+            # If the operation is performed on an odd number of elements return INT_MIN
+            return 0 if isEven == 1 else -float("inf")
+        if memo[index][isEven] != -1:
+            return memo[index][isEven]
+
+        # No operation performed on the element
+        noXorDone = nums[index] + self.maxSumOfNodes(index + 1, isEven, nums, k, memo)
+        # XOR operation is performed on the element
+        xorDone = (nums[index] ^ k) + self.maxSumOfNodes(
+            index + 1, isEven ^ 1, nums, k, memo
+        )
+
+        # Memoize and return the result
+        memo[index][isEven] = max(xorDone, noXorDone)
+        return memo[index][isEven]
+
+    def maximumValueSum(self, nums: List[int], k: int, edges: List[List[int]]) -> int:
+        memo = [[-1] * 2 for _ in range(len(nums))]
+        return self.maxSumOfNodes(0, 1, nums, k, memo)