Longest Special Path - Problem

Array Hash Table Tree Depth-First Search Prefix Sum Hard

You are given an undirected tree rooted at node 0 with n nodes numbered from 0 to n - 1, represented by a 2D array edges of length n - 1, where edges[i] = [u_i, v_i, length_i] indicates an edge between nodes u_i and v_i with length length_i.

You are also given an integer array nums, where nums[i] represents the value at node i.

A special path is defined as a downward path from an ancestor node to a descendant node such that all the values of the nodes in that path are unique.

Note that a path may start and end at the same node.

Return an array result of size 2, where result[0] is the length of the longest special path, and result[1] is the minimum number of nodes in all possible longest special paths.

Input & Output

Example 1 — Basic Tree

$ Input: edges = [[0,1,5],[0,2,3],[1,3,2]], nums = [1,2,3,2]

› Output: [8,3]

💡 Note: The longest special path is 2→0→1 with unique values [3,1,2] and total length 3+5=8 using 3 nodes. This gives the result [8,3].

Example 2 — All Unique Values

$ Input: edges = [[0,1,2],[1,2,3]], nums = [1,2,3]

› Output: [5,3]

💡 Note: Path 0→1→2 has unique values [1,2,3] and length 2+3=5 with 3 nodes total.

Example 3 — Single Node

$ Input: edges = [], nums = [1]

› Output: [0,1]

💡 Note: Only one node, so path length is 0 and minimum nodes is 1.

Constraints

1 ≤ n ≤ 10⁵
0 ≤ edges.length ≤ n - 1
edges[i].length = 3
0 ≤ u_i, v_i ≤ n - 1
1 ≤ length_i ≤ 10⁵
1 ≤ nums[i] ≤ 10⁵

Visualization

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Asked in

G Google 12 a Amazon 8 M Microsoft 6

The key insight is to use DFS with backtracking to explore all downward paths while maintaining a set of visited values to ensure uniqueness. Best approach uses DFS with visited set tracking for efficient path exploration. Time: O(n²), Space: O(n)

Common Approaches

✓ Frequency Count

⏱️ Time: N/A Space: N/A

Sort First

⏱️ Time: N/A Space: N/A

Brute Force DFS

⏱️ Time: O(n²) Space: O(n)

For each node, explore all possible downward paths and check if values are unique. Track the maximum path length and count paths with that length.

Optimized DFS with Backtracking

⏱️ Time: O(n²) Space: O(n)

Build the tree structure and perform DFS from root, using backtracking to efficiently explore all unique paths without redundant work.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

typedef struct Edge {
    int to;
    int length;
    struct Edge* next;
} Edge;

Edge* adj[100005];
int nums[100005];
int maxLength = 0;
int minNodes = 1000000;

void addEdge(int from, int to, int length) {
    Edge* edge = (Edge*)malloc(sizeof(Edge));
    edge->to = to;
    edge->length = length;
    edge->next = adj[from];
    adj[from] = edge;
}

void dfs(int node, int parent, bool used[], int currentLength, int nodeCount) {
    if (used[nums[node]]) {
        return;
    }
    
    // Include current node
    used[nums[node]] = true;
    nodeCount++;
    
    // Update result
    if (currentLength > maxLength || (currentLength == maxLength && nodeCount < minNodes)) {
        maxLength = currentLength;
        minNodes = nodeCount;
    }
    
    // Try extending to children
    for (Edge* edge = adj[node]; edge != NULL; edge = edge->next) {
        if (edge->to != parent) {
            dfs(edge->to, node, used, currentLength + edge->length, nodeCount);
        }
    }
    
    // Backtrack
    used[nums[node]] = false;
}

void solve(int n) {
    bool used[100001] = {false};
    
    // Try starting from each node
    for (int i = 0; i < n; i++) {
        dfs(i, -1, used, 0, 0);
    }
}

int main() {
    char line[10000];
    
    // Initialize
    for (int i = 0; i < 100005; i++) {
        adj[i] = NULL;
    }
    maxLength = 0;
    minNodes = 1000000;
    
    // Read edges
    fgets(line, sizeof(line), stdin);
    
    // Parse edges
    int edgeCount = 0;
    char* p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == '[') {
            p++;
            int u = (int)strtol(p, &p, 10);
            while (*p == ',' || *p == ' ') p++;
            int v = (int)strtol(p, &p, 10);
            while (*p == ',' || *p == ' ') p++;
            int len = (int)strtol(p, &p, 10);
            while (*p == ']') p++;
            
            addEdge(u, v, len);
            addEdge(v, u, len);
            edgeCount++;
        } else {
            break;
        }
    }
    
    // Read nums
    fgets(line, sizeof(line), stdin);
    int n = 0;
    p = line;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        nums[n++] = (int)strtol(p, &p, 10);
    }
    
    solve(n);
    
    // If no path found or only single nodes, return [0,1]
    if (minNodes == 1000000) {
        minNodes = 1;
    }
    
    printf("[%d,%d]\n", maxLength, minNodes);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

2.2K Views

Medium Frequency

~35 min Avg. Time

89 Likes

Ln 1, Col 1

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