Most Stones Removed with Same Row or Column

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GraphsDynamic ProgrammingGreedy Algorithms

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: One way to remove 5 stones is as follows:
1. Remove stone [2,2] because it shares the same row as [2,1].
2. Remove stone [2,1] because it shares the same column as [0,1].
3. Remove stone [1,2] because it shares the same row as [1,0].
4. Remove stone [1,0] because it shares the same column as [0,0].
5. Remove stone [0,1] because it shares the same row as [0,0].
Stone [0,0] cannot be removed since it does not share a row/column with another stone still on the plane.

Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3
Explanation: One way to make 3 moves is as follows:
1. Remove stone [2,2] because it shares the same row as [2,0].
2. Remove stone [2,0] because it shares the same column as [0,0].
3. Remove stone [0,2] because it shares the same row as [0,0].
Stones [0,0] and [1,1] cannot be removed since they do not share a row/column with another stone still on the plane.

Example 3:

Input: stones = [[0,0]]
Output: 0
Explanation: [0,0] is the only stone on the plane, so you cannot remove it.

Constraints:

1 <= stones.length <= 1000
0 <= x_i, y_i <= 10⁴
No two stones are at the same coordinate point.

Try coding it in LeetCode

Most Stones Removed with Same Row or Column

Medium

Asked by:

53 views

Topics:

GraphsDynamic ProgrammingGreedy Algorithms

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: One way to remove 5 stones is as follows:
1. Remove stone [2,2] because it shares the same row as [2,1].
2. Remove stone [2,1] because it shares the same column as [0,1].
3. Remove stone [1,2] because it shares the same row as [1,0].
4. Remove stone [1,0] because it shares the same column as [0,0].
5. Remove stone [0,1] because it shares the same row as [0,0].
Stone [0,0] cannot be removed since it does not share a row/column with another stone still on the plane.

Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3
Explanation: One way to make 3 moves is as follows:
1. Remove stone [2,2] because it shares the same row as [2,0].
2. Remove stone [2,0] because it shares the same column as [0,0].
3. Remove stone [0,2] because it shares the same row as [0,0].
Stones [0,0] and [1,1] cannot be removed since they do not share a row/column with another stone still on the plane.

Example 3:

Input: stones = [[0,0]]
Output: 0
Explanation: [0,0] is the only stone on the plane, so you cannot remove it.

Constraints:

1 <= stones.length <= 1000
0 <= x_i, y_i <= 10⁴
No two stones are at the same coordinate point.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the number of stones (the size of the input array)?
Are the coordinates of the stones guaranteed to be non-negative integers?
If there are no stones that can be removed, should I return 0?
If multiple sets of stone removals result in the same maximum number of stones removed, is any such set acceptable?
Are the stone coordinates guaranteed to be unique? Can two stones have the exact same coordinates?

Brute Force Solution

Approach

The brute force approach is to explore every possible combination of removing stones. We will start by considering each stone and deciding whether to remove it or keep it, then check if the remaining stones satisfy the problem's condition.

Here's how the algorithm would work step-by-step:

Consider each stone one at a time.
For each stone, imagine two possibilities: either we remove it, or we keep it.
For every combination of removals (some stones removed, some kept), check if the remaining stones are connected as the problem requires.
To check if the remaining stones are connected, for each stone, see if there is another stone in the same row or column. If there isn't for any stone, then the remaining stones are not connected.
If the remaining stones are connected, then this is a valid solution. Count how many stones we removed to achieve this configuration.
Repeat this process for all possible combinations of stone removals.
Compare the number of removed stones for all the valid solutions we found. Pick the configuration with the most stones removed.

Code Implementation

def most_stones_removed_brute_force(stones):
    number_of_stones = len(stones)
    maximum_stones_removed = 0

    for i in range(2**number_of_stones):
        removed_stones_indices = []
        remaining_stones = []
        stones_removed_count = 0

        for stone_index in range(number_of_stones):
            # Check if the stone should be removed based on the current bitmask
            if (i >> stone_index) & 1:
                removed_stones_indices.append(stone_index)
                stones_removed_count += 1
            else:
                remaining_stones.append(stones[stone_index])

        # Check if the remaining stones are connected
        if not remaining_stones:
            maximum_stones_removed = max(maximum_stones_removed, stones_removed_count)
            continue

        is_connected = True
        for current_stone_index in range(len(remaining_stones)):
            stone = remaining_stones[current_stone_index]
            found_neighbor = False

            for neighbor_stone_index in range(len(remaining_stones)):
                if current_stone_index == neighbor_stone_index:
                    continue

                neighbor = remaining_stones[neighbor_stone_index]

                if stone[0] == neighbor[0] or stone[1] == neighbor[1]:
                    found_neighbor = True
                    break

            # If no neighbor is found, the stones are not connected
            if not found_neighbor:
                is_connected = False
                break

        # Update max stones removed if remaining stones are connected
        if is_connected:
            maximum_stones_removed = max(maximum_stones_removed, stones_removed_count)

    return maximum_stones_removed

Big(O) Analysis

Time Complexity

O(2^n * n^2) – The brute force approach explores all possible subsets of stones, leading to 2^n combinations (where n is the number of stones). For each subset, we check if the remaining stones are connected. Checking connectivity involves, for each remaining stone (at most n), searching for another stone in the same row or column, which could take O(n) time in the worst case. Therefore, each subset takes O(n * n) = O(n^2) time to validate. Consequently, the overall time complexity is O(2^n * n^2).

Space Complexity

O(N) – The brute force approach explores all possible combinations of removing stones using recursion. Each recursive call considers whether to remove a stone or keep it, leading to a binary decision tree. The maximum depth of this recursion tree is N, where N is the number of stones. Therefore, the recursion stack can grow up to a size proportional to N. In addition, the algorithm needs to store a boolean array of size N representing stones which are kept or removed, hence adding O(N) space. Overall, the auxiliary space is O(N).

Optimal Solution

Approach

The goal is to find the maximum number of stones we can remove. The key idea is to view stones connected by the same row or column as belonging to a group and then use the principle that in each group, we can remove all stones except for one. By identifying these connected components, we determine the number of removable stones.

Here's how the algorithm would work step-by-step:

Imagine each stone as a node, and create connections between stones that share the same row or column.
These connections form groups of stones. Find all of these groups.
For each group of stones, you can remove all but one stone.
Count the total number of stones in all the groups.
Subtract the number of groups from the total count of stones. The result is the maximum number of stones that can be removed.

Code Implementation

def mostStonesRemoved(stones):
    number_of_stones = len(stones)
    visited = [False] * number_of_stones
    number_of_connected_components = 0

    def depthFirstSearch(stone_index):
        visited[stone_index] = True
        for next_stone_index in range(number_of_stones):
            if not visited[next_stone_index] and \
               (stones[stone_index][0] == stones[next_stone_index][0] or \
                stones[stone_index][1] == stones[next_stone_index][1]):

                depthFirstSearch(next_stone_index)

    # Find all connected components.
    for stone_index in range(number_of_stones):
        if not visited[stone_index]:
            depthFirstSearch(stone_index)
            number_of_connected_components += 1

    # The maximum number of stones that can be removed.
    # This is stones count minus connected components.
    return number_of_stones - number_of_connected_components

def removeStones(stones):
    # Need to use disjoint set data structure to solve this problem.
    number_of_stones = len(stones)
    parent = list(range(number_of_stones))

    def find(stone_index):
        # Path compression to optimize find operation
        if parent[stone_index] != stone_index:
            parent[stone_index] = find(parent[stone_index])
        return parent[stone_index]

    def union(stone_index_one, stone_index_two):
        root_one = find(stone_index_one)
        root_two = find(stone_index_two)
        if root_one != root_two:
            parent[root_one] = root_two
            return 1
        return 0

    number_of_unions = 0

    # Iterate over all pairs of stones
    # The crux of the disjoint set strategy.
    for stone_index_one in range(number_of_stones):
        for stone_index_two in range(stone_index_one + 1, number_of_stones):
            # Check if the stones are in the same row or column
            if stones[stone_index_one][0] == stones[stone_index_two][0] or stones[stone_index_one][1] == stones[stone_index_two][1]:
                # If so, union them
                number_of_unions += union(stone_index_one, stone_index_two)

    # The number of removable stones is the number of unions that were performed
    return number_of_unions

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each stone (n total stones) and compares it with every other stone to determine if they share a row or column. This comparison process for each stone requires examining up to n-1 other stones. Therefore, the total number of comparisons approximates n * (n-1), which simplifies to n² - n. Dropping the lower-order term (n), the time complexity is O(n²).

Space Complexity

O(N) – The space complexity is primarily determined by the data structures used to represent the connections between stones and track which stones belong to the same group. In the worst-case scenario, each stone could potentially be in its own separate group, leading to the need for data structures like a Disjoint Set Union (DSU) or adjacency list representation that scales linearly with the number of stones, N. Therefore, the auxiliary space used grows proportionally with N, where N is the number of stones. The space required to store the group information would be O(N).

Edge Cases

Case	How to Handle
Null or empty input list of stones	Return 0 as no stones can be removed from an empty set.
Single stone in the input list	Return 0 because a single stone cannot be removed since there are no other stones in the same row or column.
Maximum number of stones allowed by constraints	The union-find algorithm should handle this efficiently without stack overflow, ensuring reasonable execution time.
All stones are in the same row or column	The union-find will connect all stones together, and the result will be n-1.
Duplicate stone coordinates (same row and column)	The algorithm should treat these as a single stone, effectively ignoring the duplicates due to the nature of the problem.
Stones form disconnected components; no two stones share a row or column	The union-find algorithm will not connect any stones, and the return value will be 0.
Integer overflow when calculating hash keys for coordinates	Use long or appropriate data type to prevent overflow when combining row and column values into a unique ID.
Stones form a cycle (e.g., stone A connects to B, B to C, and C to A)	The union-find algorithm correctly handles cycles by merging the sets of the connected components.

Null or empty input list of stones

How to Handle:

Return 0 as no stones can be removed from an empty set.

Single stone in the input list

How to Handle:

Return 0 because a single stone cannot be removed since there are no other stones in the same row or column.

Maximum number of stones allowed by constraints

How to Handle:

The union-find algorithm should handle this efficiently without stack overflow, ensuring reasonable execution time.

All stones are in the same row or column

How to Handle:

The union-find will connect all stones together, and the result will be n-1.

Duplicate stone coordinates (same row and column)

How to Handle:

The algorithm should treat these as a single stone, effectively ignoring the duplicates due to the nature of the problem.

Stones form disconnected components; no two stones share a row or column

How to Handle:

The union-find algorithm will not connect any stones, and the return value will be 0.

Integer overflow when calculating hash keys for coordinates

How to Handle:

Use long or appropriate data type to prevent overflow when combining row and column values into a unique ID.

Stones form a cycle (e.g., stone A connects to B, B to C, and C to A)

How to Handle:

The union-find algorithm correctly handles cycles by merging the sets of the connected components.