LeetCode 2440: Create Components With Same Value Solution

Master LeetCode problem 2440 (Create Components With Same Value), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2440. Create Components With Same Value

Hard

Array Math Tree Depth-First Search Enumeration

Problem Explanation

Explanation:

To solve this problem, we can use a depth-first search (DFS) algorithm to traverse the tree and compute the sum of values in each subtree. We can keep track of the count of edges that need to be deleted to make the subtree have the same value. If all subtrees under a node have the same value, we can delete all edges connecting those subtrees except for one, which will be the edge connecting the current node to its parent.

Perform a DFS traversal starting from the root node.
At each node, recursively calculate the sum of values in all its child subtrees.
If the sum of values in all child subtrees is the same, we can delete all edges connecting those subtrees except one.
Keep track of the count of edges that need to be deleted to make the subtree have the same value.

Time Complexity: O(n), where n is the number of nodes in the tree. Space Complexity: O(n).

Solution Code

class Solution {
    private int result = 0;

    public int deleteTreeEdges(int[] nums, int[][] edges) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (int[] edge : edges) {
            graph.computeIfAbsent(edge[0], k -> new ArrayList<>()).add(edge[1]);
            graph.computeIfAbsent(edge[1], k -> new ArrayList<>()).add(edge[0]);
        }
        
        dfs(0, -1, nums, graph);
        
        return result;
    }
    
    private int dfs(int node, int parent, int[] nums, Map<Integer, List<Integer>> graph) {
        int sum = nums[node];
        List<Integer> childSums = new ArrayList<>();
        
        for (int child : graph.getOrDefault(node, new ArrayList<>())) {
            if (child == parent) continue;
            int childSum = dfs(child, node, nums, graph);
            sum += childSum;
            childSums.add(childSum);
        }
        
        for (int childSum : childSums) {
            if (childSum * 2 == sum) {
                result++;
            }
        }
        
        return sum;
    }
}

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