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Diffstat (limited to 'other/find_max_sum_node_val.cpp')
-rw-r--r-- | other/find_max_sum_node_val.cpp | 43 |
1 files changed, 43 insertions, 0 deletions
diff --git a/other/find_max_sum_node_val.cpp b/other/find_max_sum_node_val.cpp new file mode 100644 index 0000000..60c43b8 --- /dev/null +++ b/other/find_max_sum_node_val.cpp @@ -0,0 +1,43 @@ +// 3068. Find the Maximum Sum of Node Values +#include "leetcode.h" + +/* + * there exists an undirected tree with 'n' nodes numbered 0 to 'n - 1'. you are + * given a 0-indexed 2d integer array 'edges' of length 'n - 1', where 'edges[i] + * = [ui,vi]' indicates that there is an edge between nodes 'ui' and 'vi' in the + * tree. you are also given a positive integer 'k' and a 0 indexed array of non + * negative integers 'nums' of length 'n' where 'nums[i]' represents the value + * of the node numbered 'i'. return the maximum possible sum of the values alice + * can achieve by performing the operation any number of times. + */ + +class Solution { +public: + long long maximumValueSum(vector<int> &nums, int k, + vector<vector<int>> &edges) { + int change = 0; + long long sum = 0LL, ans = 0LL; + for (auto &n : nums) { + if ((n ^ k) > n) { + change++; + sum += (n ^ k); + } else + sum += n; + } + if (!(change & 1)) + return sum; + for (int i = 0; i < nums.size(); i++) + ans = max(ans, sum - abs(nums[i] - (nums[i] ^ k))); + return ans; + } +}; + +int main() { + Solution obj; + vector<int> n1 = {1, 2, 1}, n2 = {2, 3}, n3 = {7, 7, 7, 7, 7, 7}; + vvd e1 = {{0, 1}, {0, 2}}, e2 = {{0, 1}}, + e3 = {{0, 1}, {0, 2}, {0, 3}, {0, 4}, {0, 5}}; + printf("%lld\n", obj.maximumValueSum(n1, 3, e1)); // expect: 6 + printf("%lld\n", obj.maximumValueSum(n2, 7, e2)); // expect: 9 + printf("%lld\n", obj.maximumValueSum(n3, 3, e3)); // expect: 42 +} |