LeetCode #3532 — MEDIUM

Path Existence Queries in a Graph I

Move from brute-force thinking to an efficient approach using array strategy.

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 sorted in non-decreasing order, 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] = [u_i, v_i], determine whether there exists a path between nodes u_i and v_i.

Return a boolean array answer, where answer[i] is true if there exists a path between u_i and v_i in the i^th query and false otherwise.

Example 1:

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

Output: [true,false]

Explanation:

Query [0,0]: Node 0 has a trivial path to itself.
Query [0,1]: There is no edge between Node 0 and Node 1 because |nums[0] - nums[1]| = |1 - 3| = 2, which is greater than maxDiff.
Thus, the final answer after processing all the queries is [true, false].

Example 2:

Input: n = 4, nums = [2,5,6,8], maxDiff = 2, queries = [[0,1],[0,2],[1,3],[2,3]]

Output: [false,false,true,true]

Explanation:

The resulting graph is:

Query [0,1]: There is no edge between Node 0 and Node 1 because |nums[0] - nums[1]| = |2 - 5| = 3, which is greater than maxDiff.
Query [0,2]: There is no edge between Node 0 and Node 2 because |nums[0] - nums[2]| = |2 - 6| = 4, which is greater than maxDiff.
Query [1,3]: There is a path between Node 1 and Node 3 through Node 2 since |nums[1] - nums[2]| = |5 - 6| = 1 and |nums[2] - nums[3]| = |6 - 8| = 2, both of which are within maxDiff.
Query [2,3]: There is an edge between Node 2 and Node 3 because |nums[2] - nums[3]| = |6 - 8| = 2, which is equal to maxDiff.
Thus, the final answer after processing all the queries is [false, false, true, true].

Constraints:

1 <= n == nums.length <= 10⁵
0 <= nums[i] <= 10⁵
nums is sorted in non-decreasing order.
0 <= maxDiff <= 10⁵
1 <= queries.length <= 10⁵
queries[i] == [u_i, v_i]
0 <= u_i, v_i < n

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 sorted in non-decreasing order, 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], determine whether there exists a path between nodes ui and vi. Return a boolean array answer, where answer[i] is true if there exists a path between ui and vi in the ith query and false otherwise.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search · Union-Find

Example 1

2
[1,3]
1
[[0,0],[0,1]]

Example 2

4
[2,5,6,8]
2
[[0,1],[0,2],[1,3],[2,3]]

Step 02

Core Insight

What unlocks the optimal approach

How do the connected components look? Do they appear in segments (i.e., are they continuous)?
Preprocess the connected components.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Upper-end input sizes

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 #3532: Path Existence Queries in a Graph I
class Solution {
    public boolean[] pathExistenceQueries(int n, int[] nums, int maxDiff, int[][] queries) {
        int[] g = new int[n];
        int cnt = 0;
        for (int i = 1; i < n; ++i) {
            if (nums[i] - nums[i - 1] > maxDiff) {
                cnt++;
            }
            g[i] = cnt;
        }

        int m = queries.length;
        boolean[] ans = new boolean[m];
        for (int i = 0; i < m; ++i) {
            int u = queries[i][0];
            int v = queries[i][1];
            ans[i] = g[u] == g[v];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3532: Path Existence Queries in a Graph I
func pathExistenceQueries(n int, nums []int, maxDiff int, queries [][]int) (ans []bool) {
	g := make([]int, n)
	cnt := 0
	for i := 1; i < n; i++ {
		if nums[i]-nums[i-1] > maxDiff {
			cnt++
		}
		g[i] = cnt
	}

	for _, q := range queries {
		u, v := q[0], q[1]
		ans = append(ans, g[u] == g[v])
	}
	return
}

# Accepted solution for LeetCode #3532: Path Existence Queries in a Graph I
class Solution:
    def pathExistenceQueries(
        self, n: int, nums: List[int], maxDiff: int, queries: List[List[int]]
    ) -> List[bool]:
        g = [0] * n
        cnt = 0
        for i in range(1, n):
            if nums[i] - nums[i - 1] > maxDiff:
                cnt += 1
            g[i] = cnt
        return [g[u] == g[v] for u, v in queries]

// Accepted solution for LeetCode #3532: Path Existence Queries in a Graph I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3532: Path Existence Queries in a Graph I
// class Solution {
//     public boolean[] pathExistenceQueries(int n, int[] nums, int maxDiff, int[][] queries) {
//         int[] g = new int[n];
//         int cnt = 0;
//         for (int i = 1; i < n; ++i) {
//             if (nums[i] - nums[i - 1] > maxDiff) {
//                 cnt++;
//             }
//             g[i] = cnt;
//         }
// 
//         int m = queries.length;
//         boolean[] ans = new boolean[m];
//         for (int i = 0; i < m; ++i) {
//             int u = queries[i][0];
//             int v = queries[i][1];
//             ans[i] = g[u] == g[v];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3532: Path Existence Queries in a Graph I
function pathExistenceQueries(
    n: number,
    nums: number[],
    maxDiff: number,
    queries: number[][],
): boolean[] {
    const g: number[] = Array(n).fill(0);
    let cnt = 0;

    for (let i = 1; i < n; ++i) {
        if (nums[i] - nums[i - 1] > maxDiff) {
            ++cnt;
        }
        g[i] = cnt;
    }

    return queries.map(([u, v]) => g[u] === g[v]);
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.