What is the time complexity of Maximum Number of Points From Grid Queries?

The optimal solution has a time complexity of O(mnlog(mn) + klog(k)).

What is the space complexity of Maximum Number of Points From Grid Queries?

The space complexity is O(mn + k).

What data structures are used to solve Maximum Number of Points From Grid Queries?

This problem uses Array, Two Pointers, Breadth-First Search, Union Find, Sorting, Heap (Priority Queue), Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Number of Points From Grid Queries?

Insight 1: Sorting the queries allows processing them in increasing order, gradually expanding the visited area of the grid and utilizing the union-find data structure efficiently. Insight 2: Using Union-Find (Disjoint Set Union) is crucial to efficiently track connected components of visited cells, thus avoiding redundant counting of cells connected to the starting point (0, 0).

Which companies ask LeetCode 2503. Maximum Number of Points From Grid Queries in interviews?

This problem is frequently asked at J.P. Morgan.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximum Number of Points From Grid Queries - LeetCode 2503 Solution

Maximum Number of Points From Grid Queries - Complete Solution Guide

Maximum Number of Points From Grid Queries is LeetCode problem 2503, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an m x n integer matrix grid and an array queries of size k . Find an array answer of size k such that for each integer queries[i] you start in the top left cell of the matrix and repeat the following process: If queries[i] is strictly greater than the value of the current cell that you are in, then you get one point if it is your first time visiting this cell, and you can move to any adjacent cell in all 4 directions: up, down, left, and right. Otherwise, you do not get any points

Detailed Explanation

The problem asks us to find the maximum number of points achievable for each query in a given set of queries, based on traversing an m x n grid. We start from the top-left cell (0, 0) and can move to adjacent cells (up, down, left, right) if the query value is strictly greater than the value of the current cell. Each cell visited for the first time adds a point. The goal is to determine the maximum points obtainable for each query, allowing revisits to cells.

Solution Approach

The provided solution uses a combination of sorting, Union-Find, and incremental processing. First, the queries are sorted based on their values, and the grid cells are sorted based on their values. Then, we iterate through the sorted queries. For each query, we process grid cells whose values are less than the current query value and add them to the visited area. The Union-Find data structure maintains connected components of visited cells. Finally, the size of the connected component containing the starting cell (0,0) represents the maximum points achievable for that query.

Step-by-Step Algorithm

Step 1: Create pairs of (query value, original index) and sort them based on the query value. This allows tracking the original order for the final answer.
Step 2: Create a list of (cell value, row index, column index) from the grid and sort it based on the cell value. This allows processing cells in increasing order of their values.
Step 3: Initialize a Union-Find data structure to track connected components in the grid. Each cell is initially its own parent, and the size of each component is 1.
Step 4: Iterate through the sorted queries. For each query:
Step 5: Process cells from the sorted cell list whose values are less than the current query. Mark each such cell as visited and use Union-Find to connect it to its adjacent visited neighbors.
Step 6: If the starting cell (0, 0) has been visited, find the root of its connected component. The size of this component is the number of points achievable for the current query.
Step 7: Store the number of points (component size) in the correct position of the answer array, using the original index of the query.
Step 8: Return the answer array.

Key Insights

Insight 1: Sorting the queries allows processing them in increasing order, gradually expanding the visited area of the grid and utilizing the union-find data structure efficiently.
Insight 2: Using Union-Find (Disjoint Set Union) is crucial to efficiently track connected components of visited cells, thus avoiding redundant counting of cells connected to the starting point (0, 0).
Insight 3: Pre-sorting the cells of the grid based on their values enables us to process them only once, when the corresponding query value is greater than the cell value. This avoids re-evaluating cells for each query and is key to optimization.

Complexity Analysis

Time Complexity: O(mnlog(mn) + klog(k))

Space Complexity: O(mn + k)

Topics

This problem involves: Array, Two Pointers, Breadth-First Search, Union Find, Sorting, Heap (Priority Queue), Matrix.

Companies

Asked at: J.P. Morgan.

Maximum Number of Points From Grid Queries | LeetCode 2503

Hard

Array Two Pointers Breadth-First Search Union Find Sorting Heap (Priority Queue)Matrix

Time: O(mnlog(mn) + klog(k))Space: O(mn + k)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;
import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[] maxPoints(int[][] grid, int[] queries) {
        int m = grid.length;
        int n = grid[0].length;
        int k = queries.length;

        int[][] indexedQueries = new int[k][2];
        for (int i = 0; i < k; i++) {
            indexedQueries[i][0] = queries[i];
            indexedQueries[i][1] = i;
        }
        Arrays.sort(indexedQueries, Comparator.comparingInt(a -> a[0]));

        List<int[]> sortedCells = new ArrayList<>();
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                sortedCells.add(new int[]{grid[r][c], r, c});
            }
        }
        sortedCells.sort(Comparator.comparingInt(a -> a[0]));

        int[] parent = new int[m * n];
        for (int i = 0; i < m * n; i++) {
            parent[i] = i;
        }
        int[] sz = new int[m * n];
        Arrays.fill(sz, 1);

        class UnionFind {
            public int find(int i) {
                if (parent[i] == i) {
                    return i;
                }
                parent[i] = find(parent[i]);
                return parent[i];
            }

            public void union(int i, int j) {
                int root_i = find(i);
                int root_j = find(j);
                if (root_i != root_j) {
                    if (sz[root_i] < sz[root_j]) {
                        int temp = root_i;
                        root_i = root_j;
                        root_j = temp;
                    }
                    parent[root_j] = root_i;
                    sz[root_i] += sz[root_j];
                }
            }
        }
        
        UnionFind uf = new UnionFind();


        int[] answer = new int[k];

        int cellPtr = 0;
        boolean[][] processed = new boolean[m][n];

        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        for (int[] indexedQuery : indexedQueries) {
            int q = indexedQuery[0];
            int originalIdx = indexedQuery[1];

            while (cellPtr < m * n && sortedCells.get(cellPtr)[0] < q) {
                int r = sortedCells.get(cellPtr)[1];
                int c = sortedCells.get(cellPtr)[2];
                processed[r][c] = true;
                int cell1D = r * n + c;

                for (int[] direction : directions) {
                    int dr = direction[0];
                    int dc = direction[1];
                    int nr = r + dr;
                    int nc = c + dc;
                    if (0 <= nr && nr < m && 0 <= nc && nc < n && processed[nr][nc]) {
                        int neighbor1D = nr * n + nc;
                        uf.union(cell1D, neighbor1D);
                    }
                }

                cellPtr++;
            }

            if (processed[0][0]) {
                int rootOfStart = uf.find(0);
                answer[originalIdx] = sz[rootOfStart];
            }
        }

        return answer;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2503: Maximum Number of Points From Grid Queries - Detailed explanation and solution approaches.

Topics

Array Two Pointers Breadth-First Search Union Find Sorting Heap (Priority Queue)Matrix

Companies

J.P. Morgan

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2503

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Create pairs of (query value, original index) and sort them based on the query value. This allows tracking the original order for the final answer.
2
Step 2: Create a list of (cell value, row index, column index) from the grid and sort it based on the cell value. This allows processing cells in increasing order of their values.
3
Step 3: Initialize a Union-Find data structure to track connected components in the grid. Each cell is initially its own parent, and the size of each component is 1.
4
Step 4: Iterate through the sorted queries. For each query:
5
Step 5: Process cells from the sorted cell list whose values are less than the current query. Mark each such cell as visited and use Union-Find to connect it to its adjacent visited neighbors.
6
Step 6: If the starting cell (0, 0) has been visited, find the root of its connected component. The size of this component is the number of points achievable for the current query.
7
Step 7: Store the number of points (component size) in the correct position of the answer array, using the original index of the query.
8
Step 8: Return the answer array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(mnlog(mn) + klog(k))

💾

Space Complexity

O(mn + k)

📝

Why?

Sorting the grid cells takes O(mnlog(mn)) time, and sorting the queries takes O(klog(k)) time. The Union-Find operations take close to constant time per operation in practice. The loop iterating through queries and processing cells takes O(mn) time in the worst case, but this cost is amortized due to the pre-sorting of cells. The space complexity is O(mn) for the parent and size arrays in Union-Find and processed array, plus O(k) for storing the answer and indexed queries. In total the time complexity is therefore O(mnlog(mn) + klog(k)). The space complexity is dominated by the size of the grid i.e. O(mn+k).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS) for each query

For each query, perform a BFS starting from the top-left cell, visiting adjacent cells if their value is less than the query. Count the visited cells. Reset the visited array for each query.

Time: O(k * mn)Space: O(mn)

Trade-offs: This approach is simpler to implement but less efficient than the Union-Find approach, especially when k is large. The Union-Find approach amortizes the cost of traversing the grid across multiple queries.

Depth-First Search (DFS) for each query

Similar to BFS, but using DFS. This approach would be even less efficient for larger grids.

Time: O(k * mn)Space: O(mn) in worst case (entire grid is a long path)

Trade-offs: DFS has the same time complexity as BFS but can lead to stack overflow issues with large grids in some languages.

✨

Key Insights

1Insight 1: Sorting the queries allows processing them in increasing order, gradually expanding the visited area of the grid and utilizing the union-find data structure efficiently.
2Insight 2: Using Union-Find (Disjoint Set Union) is crucial to efficiently track connected components of visited cells, thus avoiding redundant counting of cells connected to the starting point (0, 0).
3Insight 3: Pre-sorting the cells of the grid based on their values enables us to process them only once, when the corresponding query value is greater than the cell value. This avoids re-evaluating cells for each query and is key to optimization.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: The query value is less than or equal to the value of the top-left cell. In this case, no points can be obtained, and the answer should be 0. The code correctly handles this because it only processes cells with values strictly less than the query.
Edge case 2: The grid is a single cell. The solution still works because sorting and the other data structures operate correctly even with small inputs.
Edge case 3: All query values are the same. The code processes all cells up to that value once and correctly calculates the component size for each query.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the queries by value. Processing queries in an arbitrary order does not allow incremental building of the connected components and results in inefficient recomputation.

✗Common mistake 2: Not using Union-Find to efficiently track connected components. Using separate BFS/DFS for each query leads to significantly higher time complexity.

✗Common mistake 3: Incorrectly implementing Union-Find, especially the `union` and `find` operations, which can lead to incorrect results.

✗Common mistake 4: Not considering edge cases like the starting cell value being greater than the query value, leading to incorrect point calculation.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network connectivity analysis. The grid can represent a network, and the query values can represent connection thresholds. Determining the size of the connected component from a starting node given a threshold is a common network analysis task.

🏦Real-world scenario 2: Image segmentation. The grid can represent an image, and the query values can represent intensity thresholds. Identifying regions of connected pixels with intensities below a threshold is a core image segmentation task.

🎮Real-world scenario 3: Flood fill in games or image editing tools. Simulating the spread of color or other effects based on a threshold value, similar to how the points are calculated in this problem.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 11, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Number of Points From Grid Queries"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?