LeetCode 3331: Find Subtree Sizes After Changes Solution

Master LeetCode problem 3331 (Find Subtree Sizes After Changes), a medium 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.

3331. Find Subtree Sizes After Changes

Medium

Array Hash Table String Tree Depth-First Search

Problem Explanation

Explanation

To solve this problem, we can perform a post-order traversal of the tree. For each node x, we recursively find the subtree size rooted at x and update the parent of x according to the rules mentioned in the problem. We can maintain a map to store the subtree sizes of each node. When updating the parent of a node x, we merge the subtree sizes of x with the subtree sizes of its new parent y.

Algorithm:

Perform a post-order traversal of the tree.
For each node x:
- Recursively find the subtree size rooted at x.
- Identify the closest ancestor y such that s[x] == s[y].
- If y exists, update the parent of x to y and merge the subtree sizes of x with y.
Return the array of subtree sizes.

Time Complexity:

The time complexity of this algorithm is O(n) where n is the number of nodes in the tree.

Space Complexity:

The space complexity is O(n) for the array and map used to store the subtree sizes.

Solution Code

class Solution {
    public int[] countSubTrees(int n, int[] parent, String s) {
        Map<Integer, List<Integer>> tree = new HashMap<>();
        for (int i = 0; i < n; i++) {
            tree.put(i, new ArrayList<>());
        }
        
        for (int i = 1; i < n; i++) {
            tree.get(parent[i]).add(i);
        }
        
        Map<Integer, Integer> subtreeSizes = new HashMap<>();
        postOrder(0, tree, parent, s, subtreeSizes);
        
        int[] result = new int[n];
        for (int i = 0; i < n; i++) {
            result[i] = subtreeSizes.get(i);
        }
        
        return result;
    }
    
    private int postOrder(int node, Map<Integer, List<Integer>> tree, int[] parent, String s, Map<Integer, Integer> subtreeSizes) {
        int size = 1;
        for (int child : tree.get(node)) {
            size += postOrder(child, tree, parent, s, subtreeSizes);
        }
        
        int val = 1;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == s.charAt(node)) {
                val += subtreeSizes.getOrDefault(i, 0);
            }
        }
        
        subtreeSizes.put(node, val);
        
        return size;
    }
}

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