23_Find_the_Maximum_Sum_of_Node_Values.cpp

// 3068. Find the Maximum Sum of Node Values

// 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.

// Alice wants the sum of values of tree nodes to be maximum, for which Alice can perform the following operation any number of times (including zero) on the tree:

// Choose any edge [u, v] connecting the nodes u and v, and update their values as follows:
// nums[u] = nums[u] XOR k
// nums[v] = nums[v] XOR k
// Return the maximum possible sum of the values Alice can achieve by performing the operation any number of times.

 

// Example 1:


// Input: nums = [1,2,1], k = 3, edges = [[0,1],[0,2]]
// Output: 6
// Explanation: 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] -> [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.
// Example 2:


// Input: nums = [2,3], k = 7, edges = [[0,1]]
// Output: 9
// Explanation: 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] -> [5,4].
// The total sum of values is 5 + 4 = 9.
// It can be shown that 9 is the maximum achievable sum of values.
// Example 3:


// Input: nums = [7,7,7,7,7,7], k = 3, edges = [[0,1],[0,2],[0,3],[0,4],[0,5]]
// Output: 42
// Explanation: The maximum achievable sum is 42 which can be achieved by Alice performing no operations.
 

// Constraints:

// 2 <= n == nums.length <= 2 * 104
// 1 <= k <= 109
// 0 <= nums[i] <= 109
// edges.length == n - 1
// edges[i].length == 2
// 0 <= edges[i][0], edges[i][1] <= n - 1
// The input is generated such that edges represent a valid tree.

class Solution
{
public:
    static long long maximumValueSum(vector<int> &nums, const int k, vector<vector<int>> &edges)
    {
        const int n = nums.size();
        long long dp0 = 0, dp1 = INT_MIN;
        for (int i = 1; i <= n; i++)
        {
            const long long x = nums[i - 1], xk = x ^ k;
            const long long dp_0 = max(x + dp0, xk + dp1);
            dp1 = max(x + dp1, xk + dp0);
            dp0 = dp_0;
        }
        return dp0;
    }
};

auto init = []()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 'c';
}();

/*
This solution uses dynamic programming to find the maximum possible sum after performing XOR operations.
dp0 represents the maximum sum when even number of XOR operations are performed
dp1 represents the maximum sum when odd number of XOR operations are performed
For each number, we try both possibilities:
1. Keep the number as is (x)
2. XOR it with k (xk)
And update dp0 and dp1 accordingly based on previous states.
Finally return dp0 as we want even number of XOR operations for valid result.

Time Complexity: O(n) where n is length of nums array
Space Complexity: O(1) as we only use two variables dp0 and dp1
*/