Formatted question description: https://leetcode.ca/all/2246.html

2246. Longest Path With Different Adjacent Characters

Difficulty: Hard.
Related Topics: Array, String, Tree, Depth-First Search, Graph, Topological Sort.
Similar Questions: Diameter of Binary Tree, Longest Univalue Path, Choose Edges to Maximize Score in a Tree.

Problem

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.

You are also given a string s of length n, where s[i] is the character assigned to node i.

Return the length of the **longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them.**

Example 1:

Input: parent = [-1,0,0,1,1,2], s = "abacbe"
Output: 3
Explanation: The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned.
It can be proven that there is no longer path that satisfies the conditions. 

Example 2:

Input: parent = [-1,0,0,0], s = "aabc"
Output: 3
Explanation: The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.

Constraints:

n == parent.length == s.length
1 <= n <= 105
0 <= parent[i] <= n - 1 for all i >= 1
parent[0] == -1
parent represents a valid tree.
s consists of only lowercase English letters.

Solution

class Solution {
    public int longestPath(int[] parent, String s) {
        // for first max length
        int[] first = new int[s.length()];
        Arrays.fill(first, 0);
        // for second max length
        int[] second = new int[s.length()];
        Arrays.fill(second, 0);
        // for number of children for this node
        int[] children = new int[s.length()];
        Arrays.fill(children, 0);
        for (int i = 1; i != s.length(); i++) {
            // calculate all children for each node
            children[parent[i]]++;
        }
        // for traversal from leafs to root
        LinkedList<Integer> st = new LinkedList<>();
        // put all leafs
        for (int i = 1; i != s.length(); i++) {
            if (children[i] == 0) {
                st.add(i);
            }
        }
        // traversal from leafs to root
        while (!st.isEmpty()) {
            // fetch current node
            int i = st.pollLast();
            // if we in root - ignore it
            if (i == 0) {
                continue;
            }
            if (--children[parent[i]] == 0) {
                // decrease number of children by parent node and if number of children
                st.add(parent[i]);
            }
            // is equal 0 - our parent became a leaf
            // if letters isn't equal
            if (s.charAt(parent[i]) != s.charAt(i)) {
                // fetch maximal path from node
                int maxi = 1 + Math.max(first[i], second[i]);
                // and update maximal first and second path from parent
                if (maxi >= first[parent[i]]) {
                    second[parent[i]] = first[parent[i]];
                    first[parent[i]] = maxi;
                } else if (second[parent[i]] < maxi) {
                    second[parent[i]] = maxi;
                }
            }
        }
        // fetch answer
        int ans = 0;
        for (int i = 0; i != s.length(); i++) {
            ans = Math.max(ans, first[i] + second[i]);
        }
        return ans + 1;
    }
}

# 2246. Longest Path With Different Adjacent Characters
# https://leetcode.com/problems/longest-path-with-different-adjacent-characters/

class Solution:
    def longestPath(self, parent: List[int], s: str) -> int:
        n = len(parent)
        graph = defaultdict(list)
        res = 0
        
        for x, y in enumerate(parent):
            if y == -1 or s[x] == s[y]: continue
                
            graph[y].append(x)
            graph[x].append(y)
            
        @cache
        def go(node, prev):
            nonlocal res
            
            path = [1]
            
            for nei in graph[node]:
                if nei != prev:
                    path.append(go(nei, node) + 1)
            
            path.sort(reverse = 1)
            
            if len(path) == 1:
                res = max(res, path[0])
            else:
                res = max(res, path[0] + path[1] - 1)
            
            return path[0]

        for i in range(n):
            go(i, -1)
        
        return res

function longestPath(parent: number[], s: string): number {
    const n = parent.length;
    let graph = Array.from({ length: n }, v => []);
    for (let i = 1; i < n; i++) {
        graph[parent[i]].push(i);
    }
    let ans = 0;
    function dfs(x: number): number {
        let maxLen = 0;
        for (let y of graph[x]) {
            let len = dfs(y) + 1;
            if (s.charAt(x) !== s.charAt(y)) {
                ans = Math.max(maxLen + len, ans);
                maxLen = Math.max(len, maxLen);
            }
        }
        return maxLen;
    }
    dfs(0);
    return ans + 1;
}

class Solution {
public:
    int longestPath(vector<int>& parent, string s) {
        int n = parent.size();
        vector<int> g[n];
        for (int i = 1; i < n; ++i) {
            g[parent[i]].push_back(i);
        }
        int ans = 0;
        function<int(int)> dfs = [&](int i) -> int {
            int mx = 0;
            for (int j : g[i]) {
                int x = dfs(j) + 1;
                if (s[i] != s[j]) {
                    ans = max(ans, mx + x);
                    mx = max(mx, x);
                }
            }
            return mx;
        };
        dfs(0);
        return ans + 1;
    }
};

func longestPath(parent []int, s string) int {
	n := len(parent)
	g := make([][]int, n)
	for i := 1; i < n; i++ {
		g[parent[i]] = append(g[parent[i]], i)
	}
	ans := 0
	var dfs func(int) int
	dfs = func(i int) int {
		mx := 0
		for _, j := range g[i] {
			x := dfs(j) + 1
			if s[i] != s[j] {
				ans = max(ans, x+mx)
				mx = max(mx, x)
			}
		}
		return mx
	}
	dfs(0)
	return ans + 1
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).

2246 - Longest Path With Different Adjacent Characters

2246. Longest Path With Different Adjacent Characters

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2246. Longest Path With Different Adjacent Characters

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