Longest Special Path

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

✓ Sort First

⏱️ Time: N/A Space: N/A

Frequency Count

⏱️ 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>

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

bool solution(int* target, int* arr, int n) {
    // Create copies to avoid modifying original arrays
    int* targetCopy = malloc(n * sizeof(int));
    int* arrCopy = malloc(n * sizeof(int));
    
    memcpy(targetCopy, target, n * sizeof(int));
    memcpy(arrCopy, arr, n * sizeof(int));
    
    // Sort both arrays
    qsort(targetCopy, n, sizeof(int), compare);
    qsort(arrCopy, n, sizeof(int), compare);
    
    // Compare sorted arrays
    bool result = true;
    for (int i = 0; i < n; i++) {
        if (targetCopy[i] != arrCopy[i]) {
            result = false;
            break;
        }
    }
    
    free(targetCopy);
    free(arrCopy);
    return result;
}

int main() {
    // Read input arrays dynamically
    char line[10000];
    
    // Read target array
    fgets(line, sizeof(line), stdin);
    int target[1000];
    int n = 0;
    char* token = strtok(line, "[],\n ");
    while (token != NULL) {
        target[n++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    // Read arr array
    fgets(line, sizeof(line), stdin);
    int arr[1000];
    int m = 0;
    token = strtok(line, "[],\n ");
    while (token != NULL) {
        arr[m++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    bool result = solution(target, arr, n);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚡ Linearithmic Space

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Medium Frequency

~35 min Avg. Time

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Ln 1, Col 1

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