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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, 0 insertions, 43 deletions
diff --git a/other/find_max_sum_node_val.cpp b/other/find_max_sum_node_val.cpp deleted file mode 100644 index 60c43b8..0000000 --- a/other/find_max_sum_node_val.cpp +++ /dev/null @@ -1,43 +0,0 @@ -// 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 -} |