LeetCode #3288 — HARD

Length of the Longest Increasing Path

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 a 2D array of integers coordinates of length n and an integer k, where 0 <= k < n.

coordinates[i] = [xi, yi] indicates the point (xi, yi) in a 2D plane.

An increasing path of length m is defined as a list of points (x1, y1), (x2, y2), (x3, y3), ..., (xm, ym) such that:

  • xi < xi + 1 and yi < yi + 1 for all i where 1 <= i < m.
  • (xi, yi) is in the given coordinates for all i where 1 <= i <= m.

Return the maximum length of an increasing path that contains coordinates[k].

Example 1:

Input: coordinates = [[3,1],[2,2],[4,1],[0,0],[5,3]], k = 1

Output: 3

Explanation:

(0, 0), (2, 2), (5, 3) is the longest increasing path that contains (2, 2).

Example 2:

Input: coordinates = [[2,1],[7,0],[5,6]], k = 2

Output: 2

Explanation:

(2, 1), (5, 6) is the longest increasing path that contains (5, 6).

Constraints:

  • 1 <= n == coordinates.length <= 105
  • coordinates[i].length == 2
  • 0 <= coordinates[i][0], coordinates[i][1] <= 109
  • All elements in coordinates are distinct.
  • 0 <= k <= n - 1
Patterns Used
🔍Modified Binary Search→

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 a 2D array of integers coordinates of length n and an integer k, where 0 <= k < n. coordinates[i] = [xi, yi] indicates the point (xi, yi) in a 2D plane. An increasing path of length m is defined as a list of points (x1, y1), (x2, y2), (x3, y3), ..., (xm, ym) such that: xi < xi + 1 and yi < yi + 1 for all i where 1 <= i < m. (xi, yi) is in the given coordinates for all i where 1 <= i <= m. Return the maximum length of an increasing path that contains coordinates[k].

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

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

Example 2

[[2,1],[7,0],[5,6]]
2
Step 02

Core Insight

What unlocks the optimal approach

  • Only keep coordinates with both <code>x</code> and <code>y</code> being strictly less than <code>coordinates[k]</code>.
  • Sort them by <code>x</code>’s, in the case of equal, the <code>y</code> values should be decreasing.
  • Calculate LIS only using <code>y</code> values.
  • Do the same for coordinates with both <code>x</code> and <code>y</code> being strictly larger than <code>coordinates[k]</code>.
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 #3288: Length of the Longest Increasing Path
class Solution {
  public int maxPathLength(int[][] coordinates, int k) {
    final int xk = coordinates[k][0];
    final int yk = coordinates[k][1];
    List<int[]> leftCoordinates = new ArrayList<>();
    List<int[]> rightCoordinates = new ArrayList<>();

    for (int[] coordinate : coordinates) {
      final int x = coordinate[0];
      final int y = coordinate[1];
      if (x < xk && y < yk)
        leftCoordinates.add(new int[] {x, y});
      else if (x > xk && y > yk)
        rightCoordinates.add(new int[] {x, y});
    }

    return 1 + lengthOfLIS(leftCoordinates) + lengthOfLIS(rightCoordinates);
  }

  // Similar to 300. Longest Increasing Subsequence
  private int lengthOfLIS(List<int[]> coordinates) {
    coordinates.sort(Comparator.comparingInt((int[] coordinate) -> coordinate[0])
                         .thenComparingInt((int[] coordinate) -> - coordinate[1]));
    // tails[i] := the minimum tail of all the increasing subsequences having
    // length i + 1
    List<Integer> tails = new ArrayList<>();
    for (int[] coordinate : coordinates) {
      final int y = coordinate[1];
      if (tails.isEmpty() || y > tails.get(tails.size() - 1))
        tails.add(y);
      else
        tails.set(firstGreaterEqual(tails, y), y);
    }
    return tails.size();
  }

  private int firstGreaterEqual(List<Integer> arr, int target) {
    final int i = Collections.binarySearch(arr, target);
    return i < 0 ? -i - 1 : i;
  }
}
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(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.

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.

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