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This approach involves checking all possible operations on each edge and calculating the resulting sum by applying the XOR operation. We must continuously update and evaluate for every operation on each pair of nodes linked by an edge to see if the resulting sum after the operation is beneficial.
This naive strategy will consider 2^n operations (since each edge can have 2 XOR states per pair), making it inefficient for larger datasets. Ideal for understanding XOR effects but not recommended for large n.
Time Complexity: O(n), where n is the number of nodes. Each iteration is constant time.
Space Complexity: O(1), since additional space used is constant.
1function getMaxSum(nums, k, edges) {
2 let sum = nums.reduce((a, b) => a + b, 0);
3 let maxSum = sum;
4 edges.forEach(([u, v]) => {
5 let newSum = sum - (nums[u] + nums[v]) + ((nums[u] ^ k) + (nums[v] ^ k));
6 if (newSum > maxSum) {
7 maxSum = newSum;
8 }
9 });
10 return maxSum;
11}
12
13const nums = [1, 2, 1];
14const k = 3;
15const edges = [[0, 1], [0, 2]];
16console.log(getMaxSum(nums, k, edges));This JavaScript version uses the same concept as previous solutions, seeking the nodal value results post-XOR and choosing the better among them.
Instead of brute checking each edge operation individually, observe that XOR operation symmetry allows flipping of node values to be optimal only when advantageous. Analyze and preprocess which nodes when XOR'd with k will increase their value and prioritize these changes globally without redundant individual edge checks.
This idea is crucially about observing which nodes independently contribute more value when toggled and using the tree structure to apply XOR operations optimally.
Time Complexity: O(n), as it processes each node individually once.
Space Complexity: O(1), using a fixed amount of extra space.
#include <vector>
int getMaxSumOptimized(std::vector<int>& nums, int k) {
int sum = 0, xor_sum = 0;
for (int num : nums) {
sum += num;
xor_sum += (num ^ k);
}
return std::max(sum, xor_sum);
}
int main() {
std::vector<int> nums = {1, 2, 1};
int k = 3;
std::cout << getMaxSumOptimized(nums, k) << std::endl;
return 0;
}This C++ solution tackles each node value independently and simply contrasts the potential gain from converting any node value globally with XOR. Thus, yields either the natural sum or total XOR-oriented benefit.