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

The optimal solution for Most Stones Removed with Same Row or Column runs in O(n²) time and uses O(1) space. This complexity arises because we are comparing each stone with every other stone, leading to a quadratic number of comparisons.

What algorithm or data structure is used to solve Most Stones Removed with Same Row or Column?

Most Stones Removed with Same Row or Column uses the following concepts: Hash Table, Depth-First Search, Union-Find, Graph Theory. The recommended patterns to study are: Union-Find, Graph Theory.

What are the interview tips for Most Stones Removed with Same Row or Column?

Understand the problem context and visualize the connections Practice union-find problems to get comfortable with the concept Think about edge cases, like when there's only one stone.

What are common mistakes when solving Most Stones Removed with Same Row or Column?

Confusing rows and columns Not considering all stones in the same component

Most Stones Removed with Same Row or Column — LeetCode #947 (Medium)

Tags: Hash Table, Depth-First Search, Union-Find, Graph Theory

Related patterns: Union-Find, Graph Theory

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking every stone and attempting to remove it by checking if it shares a row or column with any other stone. This method is straightforward but inefficient as it requires comparing each stone with every other stone.

This complexity arises because we are comparing each stone with every other stone, leading to a quadratic number of comparisons.

Step 1: Initialize a counter for removed stones.
Step 2: For each stone, check if there is another stone in the same row or column.
Step 3: If a stone can be removed, increment the counter and mark it as removed.
Step 4: Repeat until no more stones can be removed.

For stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]: 1. Check [0,0]: can remove [0,1] -> removed = 1 2. Check [0,1]: can remove [0,0] -> already counted 3. Check [1,0]: can remove [1,2] -> removed = 2 4. Check [1,2]: can remove [1,0] -> already counted 5. Check [2,1]: can remove [2,2] -> removed = 3 6. Check [2,2]: can remove [2,1] -> already counted Final count of removed stones = 5.

Common Mistakes

Confusing rows and columns
Not considering all stones in the same component

Interview Tips

Understand the problem context and visualize the connections
Practice union-find problems to get comfortable with the concept
Think about edge cases, like when there's only one stone.

#947

Most Stones Removed with Same Row or Column

Medium

Hash Table↗Depth-First Search↗Union-Find↗Graph Theory↗Union-Find↗Graph Theory↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter for removed stones.
2Step 2: For each stone, check if there is another stone in the same row or column.
3Step 3: If a stone can be removed, increment the counter and mark it as removed.
4Step 4: Repeat until no more stones can be removed.

solution.py11 lines

1# Full working Python code
2class Solution:
3    def removeStones(self, stones):
4        removed = 0
5        n = len(stones)
6        for i in range(n):
7            for j in range(n):
8                if i != j and (stones[i][0] == stones[j][0] or stones[i][1] == stones[j][1]):
9                    removed += 1
10                    break
11        return removed

ℹ

Complexity note: This complexity arises because we are comparing each stone with every other stone, leading to a quadratic number of comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.