Approach Summary
Union stones sharing the same row or column. Stones in the same component can all be removed except one. Answer = n - (number of components).
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Most Stones Removed with Same Row or Column optimally, we apply the union find algorithmic strategy. Union stones sharing the same row or column. Stones in the same component can all be removed except one. Answer = n - (number of components). A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n α(n)) time, utilizing O(n) auxiliary memory.
Time Complexity: O(n α(n)) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(n) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
import java.util.*;
class Solution {
// Time: O(n α(n)), Space: O(n)
public int[] mostStonesRemoved(int numNodes, int[][] edges) {
List<List<Integer>> adj = new ArrayList<>();
for (int i = 0; i < numNodes; i++) adj.add(new ArrayList<>());
int[] indegree = new int[numNodes];
for (int[] edge : edges) {
adj.get(edge[0]).add(edge[1]);
indegree[edge[1]]++;
}
Queue<Integer> queue = new ArrayDeque<>();
for (int i = 0; i < numNodes; i++) {
if (indegree[i] == 0) queue.offer(i);
}
int[] order = new int[numNodes];
int idx = 0;
while (!queue.isEmpty()) {
int node = queue.poll();
order[idx++] = node;
for (int next : adj.get(node)) {
if (--indegree[next] == 0) queue.offer(next);
}
}
return idx == numNodes ? order : new int[0];
}
}#include <vector>
#include <queue>
using namespace std;
class Solution {
public:
// Time: O(n α(n)), Space: O(n)
vector<int> mostStonesRemoved(int numNodes, vector<vector<int>>& edges) {
vector<vector<int>> adj(numNodes);
vector<int> indegree(numNodes, 0);
for (const auto& edge : edges) {
adj[edge[0]].push_back(edge[1]);
indegree[edge[1]]++;
}
queue<int> q;
for (int i = 0; i < numNodes; ++i) {
if (indegree[i] == 0) q.push(i);
}
vector<int> order;
while (!q.empty()) {
int node = q.front(); q.pop();
order.push_back(node);
for (int next : adj[node]) {
if (--indegree[next] == 0) q.push(next);
}
}
return order.size() == numNodes ? order : vector<int>();
}
};from collections import deque, defaultdict
from typing import List
class Solution:
def most_stones_removed(self, numNodes: int, edges: List[List[int]], *args) -> object:
# Time: O(n α(n)), Space: O(n)
adj = defaultdict(list)
indegree = [0] * numNodes
for u, v in edges:
adj[u].append(v)
indegree[v] += 1
queue = deque([i for i in range(numNodes) if indegree[i] == 0])
order = []
while queue:
node = queue.popleft()
order.append(node)
for neighbor in adj[node]:
indegree[neighbor] -= 1
if indegree[neighbor] == 0:
queue.append(neighbor)
return order if len(order) == numNodes else []var mostStonesRemoved = function(numNodes, edges) {
const adj = Array.from({ length: numNodes }, () => []);
const indegree = new Array(numNodes).fill(0);
for (const [u, v] of edges) {
adj[u].push(v);
indegree[v]++;
}
const queue = [];
for (let i = 0; i < numNodes; i++) {
if (indegree[i] === 0) queue.push(i);
}
const order = [];
while (queue.length) {
const node = queue.shift();
order.push(node);
for (const next of adj[node]) {
indegree[next]--;
if (indegree[next] === 0) queue.push(next);
}
}
return order.length === numNodes ? order : [];
};Edge Cases to Watch
- Key recognition trigger: "Remove stones sharing row or column".
- Constraint condition: Union-Find on (row, col) — each group keeps one stone.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n α(n)) time budget without stack overflow or time limit exceeded.
Master the Union Find Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Union Find pattern.
How to Recognize This Pattern
- "Remove stones sharing row or column"
- Union-Find on (row, col) — each group keeps one stone
Complexity Analysis
Time Complexity
O(n α(n))
Space Complexity
O(n)