LeetCode #3534 — HARD

Path Existence Queries in a Graph II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an integer n representing the number of nodes in a graph, labeled from 0 to n - 1.

You are also given an integer array nums of length n and an integer maxDiff.

An undirected edge exists between nodes i and j if the absolute difference between nums[i] and nums[j] is at most maxDiff (i.e., |nums[i] - nums[j]| <= maxDiff).

You are also given a 2D integer array queries. For each queries[i] = [ui, vi], find the minimum distance between nodes ui and vi. If no path exists between the two nodes, return -1 for that query.

Return an array answer, where answer[i] is the result of the ith query.

Note: The edges between the nodes are unweighted.

Example 1:

Input: n = 5, nums = [1,8,3,4,2], maxDiff = 3, queries = [[0,3],[2,4]]

Output: [1,1]

Explanation:

The resulting graph is:

Query Shortest Path Minimum Distance
[0, 3] 0 → 3 1
[2, 4] 2 → 4 1

Thus, the output is [1, 1].

Example 2:

Input: n = 5, nums = [5,3,1,9,10], maxDiff = 2, queries = [[0,1],[0,2],[2,3],[4,3]]

Output: [1,2,-1,1]

Explanation:

The resulting graph is:

Query Shortest Path Minimum Distance
[0, 1] 0 → 1 1
[0, 2] 0 → 1 → 2 2
[2, 3] None -1
[4, 3] 3 → 4 1

Thus, the output is [1, 2, -1, 1].

Example 3:

Input: n = 3, nums = [3,6,1], maxDiff = 1, queries = [[0,0],[0,1],[1,2]]

Output: [0,-1,-1]

Explanation:

There are no edges between any two nodes because:

  • Nodes 0 and 1: |nums[0] - nums[1]| = |3 - 6| = 3 > 1
  • Nodes 0 and 2: |nums[0] - nums[2]| = |3 - 1| = 2 > 1
  • Nodes 1 and 2: |nums[1] - nums[2]| = |6 - 1| = 5 > 1

Thus, no node can reach any other node, and the output is [0, -1, -1].

Constraints:

  • 1 <= n == nums.length <= 105
  • 0 <= nums[i] <= 105
  • 0 <= maxDiff <= 105
  • 1 <= queries.length <= 105
  • queries[i] == [ui, vi]
  • 0 <= ui, vi < n
Patterns Used
👉Two Pointers→🔍Modified Binary Search→📊Dynamic Programming→🎯Greedy→⚡Bit Manipulation→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer n representing the number of nodes in a graph, labeled from 0 to n - 1. You are also given an integer array nums of length n and an integer maxDiff. An undirected edge exists between nodes i and j if the absolute difference between nums[i] and nums[j] is at most maxDiff (i.e., |nums[i] - nums[j]| <= maxDiff). You are also given a 2D integer array queries. For each queries[i] = [ui, vi], find the minimum distance between nodes ui and vi. If no path exists between the two nodes, return -1 for that query. Return an array answer, where answer[i] is the result of the ith query. Note: The edges between the nodes are unweighted.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search · Dynamic Programming · Greedy · Bit Manipulation

Example 1

5
[1,8,3,4,2]
3
[[0,3],[2,4]]

Example 2

5
[5,3,1,9,10]
2
[[0,1],[0,2],[2,3],[4,3]]

Example 3

3
[3,6,1]
1
[[0,0],[0,1],[1,2]]
Step 02

Core Insight

What unlocks the optimal approach

  • Sort the nodes according to <code>nums[i]</code>.
  • Can we use binary jumping?
  • Use binary jumping with a sparse table data structure.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3534: Path Existence Queries in a Graph II
class Solution {
  public int[] pathExistenceQueries(int n, int[] nums, int maxDiff, int[][] queries) {
    int[] ans = new int[queries.length];
    int[] indexMap = new int[n];
    int[] sortedNums = new int[n];
    Pair<Integer, Integer>[] sortedNumAndIndexes = new Pair[n];

    for (int i = 0; i < n; ++i)
      sortedNumAndIndexes[i] = new Pair<>(nums[i], i);

    Arrays.sort(sortedNumAndIndexes, Comparator.comparingInt(Pair::getKey));

    for (int i = 0; i < n; ++i) {
      final int num = sortedNumAndIndexes[i].getKey();
      final int sortedIndex = sortedNumAndIndexes[i].getValue();
      sortedNums[i] = num;
      indexMap[sortedIndex] = i;
    }

    final int maxLevel = Integer.SIZE - Integer.numberOfLeadingZeros(n) + 1;
    // jump[i][j] := the index of the j-th ancestor of i
    int[][] jump = new int[n][maxLevel];

    int right = 0;
    for (int i = 0; i < n; ++i) {
      while (right + 1 < n && sortedNums[right + 1] - sortedNums[i] <= maxDiff)
        ++right;
      jump[i][0] = right;
    }

    for (int level = 1; level < maxLevel; ++level)
      for (int i = 0; i < n; ++i) {
        final int prevJump = jump[i][level - 1];
        jump[i][level] = jump[prevJump][level - 1];
      }

    for (int i = 0; i < queries.length; ++i) {
      final int u = queries[i][0];
      final int v = queries[i][1];
      final int uIndex = indexMap[u];
      final int vIndex = indexMap[v];
      final int start = Math.min(uIndex, vIndex);
      final int end = Math.max(uIndex, vIndex);
      final int res = minJumps(jump, start, end, maxLevel - 1);
      ans[i] = res == Integer.MAX_VALUE ? -1 : res;
    }

    return ans;
  }

  // Returns the minimum number of jumps from `start` to `end` using binary
  // lifting.
  private int minJumps(int[][] jump, int start, int end, int level) {
    if (start == end)
      return 0;
    if (jump[start][0] >= end)
      return 1;
    if (jump[start][level] < end)
      return Integer.MAX_VALUE;
    int j = level;
    for (; j >= 0; --j)
      if (jump[start][j] < end)
        break;
    return (1 << j) + minJumps(jump, jump[start][j], end, j);
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS
O(n) time
O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Related Problems
#2035Partition Array Into Two Arrays to Minimize Sum Differencehard→#3801Minimum Cost to Merge Sorted Listshard→#287Find the Duplicate Numbermedium→#410Split Array Largest Sumhard→#611Valid Triangle Numbermedium→