What is the time complexity of Most Stones Removed with Same Row or Column?

The optimal solution has a time complexity of O(nα(n)).

What data structures are used to solve Most Stones Removed with Same Row or Column?

This problem uses Hash Table, Depth-First Search, Union Find, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Most Stones Removed with Same Row or Column?

Insight 1: The problem can be modeled as finding the number of connected components in a graph where stones are nodes and edges connect stones sharing a row or column. Insight 2: Union Find (Disjoint Set Union) is an efficient data structure for determining connected components in a graph.

Which companies ask LeetCode 947. Most Stones Removed with Same Row or Column in interviews?

This problem is frequently asked at PhonePe, Tekion, thoughtspot.

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Most Stones Removed with Same Row or Column - LeetCode 947 Solution

Most Stones Removed with Same Row or Column - Complete Solution Guide

Most Stones Removed with Same Row or Column is LeetCode problem 947, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

On a 2D plane, we place n stones at some integer coordinate points. Each coordinate point may have at most one stone. A stone can be removed if it shares either the same row or the same column as another stone that has not been removed. Given an array stones of length n where stones[i] = [x i , y i ] represents the location of the i th stone, return the largest possible number of stones that can be removed . Example 1: Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]] Output: 5 Explanation:

Detailed Explanation

The problem asks us to find the maximum number of stones that can be removed from a 2D plane, given a list of stone coordinates. A stone can be removed if it shares the same row or column with another stone that hasn't been removed yet. The goal is to strategically remove stones until no more stones can be removed according to the given rule.

Solution Approach

The solution uses the Union Find algorithm to determine the number of connected components. It iterates through the stones and for each stone, it checks if there is any other stone in the same row or column. If there is, it unions the two stones together, indicating they belong to the same connected component. After processing all stones, the number of connected components is calculated. The answer is then the total number of stones minus the number of connected components.

Step-by-Step Algorithm

Step 1: Initialize a Union Find data structure with each stone representing its own set (connected component).
Step 2: Iterate through the list of stones.
Step 3: For each stone, check if there's another stone sharing the same row. If yes, perform a union operation to merge their respective sets.
Step 4: Repeat the same check for stones sharing the same column, performing a union operation if necessary.
Step 5: After processing all stones, determine the number of connected components by counting the number of distinct roots in the Union Find data structure. However, to optimize for Time complexity, we incremented `merged_count` during the `union` operation. Therefore, the number of merged component will be the `merged_count`.
Step 6: Return the number of stones that can be removed, which is `total number of stones - number of components that stay` which is equivalent to merged_count = `number of stones - number of components`

Key Insights

Insight 1: The problem can be modeled as finding the number of connected components in a graph where stones are nodes and edges connect stones sharing a row or column.
Insight 2: Union Find (Disjoint Set Union) is an efficient data structure for determining connected components in a graph.
Insight 3: The maximum number of removable stones is equal to the total number of stones minus the number of connected components. Each connected component needs one stone to stay at the end.

Complexity Analysis

Time Complexity: O(nα(n))

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Depth-First Search, Union Find, Graph.

Companies

Asked at: PhonePe, Tekion, thoughtspot.

Most Stones Removed with Same Row or Column | LeetCode 947

Medium

Hash Table Depth-First Search Union Find Graph

Time: O(nα(n))Space: O(n)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int removeStones(int[][] stones) {
        int n = stones.length;
        int[] parent = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }

        function find = new function(parent);

        function2 union = new function2(parent, find);

        Map<Integer, Integer> row_map = new HashMap<>();
        Map<Integer, Integer> col_map = new HashMap<>();

        int merged_count = 0;
        for (int i = 0; i < n; i++) {
            int r = stones[i][0];
            int c = stones[i][1];
            if (row_map.containsKey(r)) {
                merged_count += union.apply(i, row_map.get(r));
            } else {
                row_map.put(r, i);
            }

            if (col_map.containsKey(c)) {
                merged_count += union.apply(i, col_map.get(c));
            } else {
                col_map.put(c, i);
            }
        }

        return merged_count;
    }

    static class function {
        private final int[] parent;

        function(int[] parent) {
            this.parent = parent;
        }

        public int apply(int i) {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = apply(parent[i]);
            return parent[i];
        }
    }

    static class function2 {
        private final int[] parent;
        private final function find;

        function2(int[] parent, function find) {
            this.parent = parent;
            this.find = find;
        }

        public int apply(int i, int j) {
            int root_i = find.apply(i);
            int root_j = find.apply(j);
            if (root_i != root_j) {
                parent[root_j] = root_i;
                return 1;
            }
            return 0;
        }
    }
}

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Problem Info

Description

Problem 947: Most Stones Removed with Same Row or Column - Detailed explanation and solution approaches.

Topics

Hash Table Depth-First Search Union Find Graph

Companies

PhonePeTekionthoughtspot

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:947

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

6 steps

1
Step 1: Initialize a Union Find data structure with each stone representing its own set (connected component).
2
Step 2: Iterate through the list of stones.
3
Step 3: For each stone, check if there's another stone sharing the same row. If yes, perform a union operation to merge their respective sets.
4
Step 4: Repeat the same check for stones sharing the same column, performing a union operation if necessary.
5
Step 5: After processing all stones, determine the number of connected components by counting the number of distinct roots in the Union Find data structure. However, to optimize for Time complexity, we incremented `merged_count` during the `union` operation. Therefore, the number of merged component will be the `merged_count`.
6
Step 6: Return the number of stones that can be removed, which is `total number of stones - number of components that stay` which is equivalent to merged_count = `number of stones - number of components`

💡

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(nα(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The Union Find algorithm takes O(α(n)) time for each find and union operation. We iterate through all the stones in the nested loop, so the overall time complexity is O(nα(n)). The space complexity is O(n) because the parent array stores the parent of each node, and the row/col maps store up to n entries.

📖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

Depth-First Search (DFS)

DFS can be used to find the connected components. Iterate through the stones. For each unvisited stone, start a DFS to visit all stones in the same row or column. Each DFS traversal represents a connected component.

Time: O(n^2)Space: O(n)

Trade-offs: DFS has a higher time complexity than Union Find (O(n^2) vs O(nα(n))). Union Find is generally preferred for its efficiency. DFS would involve recursion, while UnionFind is iterative. UnionFind also is easily parallelizable.

✨

Key Insights

1Insight 1: The problem can be modeled as finding the number of connected components in a graph where stones are nodes and edges connect stones sharing a row or column.
2Insight 2: Union Find (Disjoint Set Union) is an efficient data structure for determining connected components in a graph.
3Insight 3: The maximum number of removable stones is equal to the total number of stones minus the number of connected components. Each connected component needs one stone to stay at the end.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: `stones = [[0,0]]`. This is handled correctly because no unions will occur, and merged_count will be 0, which is the correct output.
Edge case 2: `stones = [[0,0],[0,0]]`. The problem statement specifies that 'No two stones are at the same coordinate point', but if they were present, the code would handle this correctly since they would be unioned at first step.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not recognizing that the problem can be solved using connected components.

✗Common mistake 2: Not using the Union Find data structure effectively, leading to higher time complexity. Forgetting path compression optimization.

✅

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: Determining how many devices on a network are connected, where devices sharing the same network segment play the role of stones.

🏦Real-world scenario 2: Image segmentation: Grouping pixels in an image that have similar properties (e.g., color) and are spatially connected, the pixels are stones sharing color attribute.

🎮Real-world scenario 3: Social network analysis: Finding the number of distinct communities in a social network where users with common interests and connections represent the stones.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: July 18, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Most Stones Removed with Same Row or Column"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?