Remove Stones to Minimize the Total
A medium Graph interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Graph
- Sheets
- 1
- Core for
- 12 roles
The problem
Given an m x n grid where 1 represents a stone and 0 represents empty space, return the maximum number of stones that can be removed. Two stones are connected if they share the same row or column.
Example 1
- Input
- stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
- Output
- 5
- Why
- All stones are connected, so 5 can be removed leaving 1.
Example 2
- Input
- stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
- Output
- 3
- Why
- Three stones can be removed, leaving two isolated stones.
Constraints
- 1 <= stones.length <= 1000
- stones[i].length == 2
- 0 <= stones[i][j] <= 10^4
How to think about it
Updated 2026-09-29Every connected cluster of stones sharing rows or columns can be reduced down to exactly one surviving stone: pick a spanning tree of the cluster and remove stones from leaf to root. Therefore, the maximum removable stones equals `total stones - number of connected components`. Unioning rows with column indices directly forms the components.
Approaches, worst first
DSU merging rows with columns
time O(N * α(N)) · space O(N)
Treat each row r and column c as graph nodes. For stone `(r, c)`, union node `r` with `c + 10001` to prevent coordinate collisions. Count unique roots across all stones: the answer is `stones.length - uniqueRoots`.
Stone-to-stone adjacency DFSWrite this one
time O(N^2) · space O(N^2)
Build an adjacency list connecting stone indices that share rows or columns. Run DFS to count connected components C, returning `stones.length - C`.
Where people lose marks · 3
- Colliding row index r with column index c (e.g. row 0 and column 0 are distinct lines in the 2D plane); offset column indices by an amount greater than max row coordinate.
- Attempting an actual grid simulation matrix: coordinates up to 10^4 mean a 10000x10000 matrix will allocate 100 million entries and run out of memory.
- Counting total roots in the parent map instead of roots specifically belonging to stones that actually exist.
Full solution
C++
#include <bits/stdc++.h>
using namespace std;
int minStoneSum(vector<int>& piles, int k) {
priority_queue<int> pq(piles.begin(), piles.end());
for (int i = 0; i < k; i++) {
if (pq.empty()) break;
int val = pq.top();
pq.pop();
int remove = val / 2;
val -= remove;
pq.push(val);
}
int total = 0;
while (!pq.empty()) {
total += pq.top();
pq.pop();
}
return total;
}
Java
import java.util.*;
public class Main {
static int minStoneSum(int[] piles, int k) {
PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
for (int x : piles) pq.offer(x);
for (int i = 0; i < k; i++) {
if (pq.isEmpty()) break;
int val = pq.poll();
val -= val / 2;
pq.offer(val);
}
int total = 0;
while (!pq.isEmpty()) {
total += pq.poll();
}
return total;
}
}
Python
import heapq
import math
def min_stone_sum(piles, k):
pq = [-x for x in piles]
heapq.heapify(pq)
for _ in range(k):
val = -heapq.heappop(pq)
remove = val // 2
val -= remove
heapq.heappush(pq, -val)
return -sum(pq)
JavaScript
function minStoneSum(piles, k) {
let maxVal = 0;
for (const p of piles) {
if (p > maxVal) maxVal = p;
}
const count = Array(maxVal + 1).fill(0);
for (const p of piles) {
count[p]++;
}
let curr = maxVal;
while (k > 0 && curr > 0) {
if (count[curr] === 0) {
curr--;
continue;
}
const take = Math.min(k, count[curr]);
const nextVal = curr - Math.floor(curr / 2);
count[curr] -= take;
count[nextVal] += take;
k -= take;
}
let total = 0;
for (let v = 1; v <= maxVal; v++) {
total += v * count[v];
}
return total;
}
The theory behind it
Graph — the ground this problem stands on. All Graph problems
What Graph is
A graph is a network of individual points, called vertices or nodes, connected by lines called edges. Think of a subway transit map, an electrical circuit, or a web of social friends. Unlike a tree, a graph has no designated top node and no parent-child hierarchy. Connections can run one-way or both ways, and paths can loop back on themselves to form closed cycles.
When to reach for it
Reach for graph algorithms when inputs describe relationships, networks, flights between cities, course prerequisites, or clone networks. Signals include finding the shortest route across unweighted connections, ordering tasks that depend on earlier tasks, counting isolated clusters, or checking whether a path contains an infinite loop. Whenever problems present pairs of related entities and ask for reachability, distances, or dependencies, graph representations apply.
How the pattern works
First convert edge lists into an adjacency list, mapping each node to an array of its neighbors. Choose your exploration strategy based on the goal: use a queue and breadth-first search to find the shortest path in unweighted networks, or use recursion and depth-first search to explore full paths and detect cycles. Because graphs can have loops, always track visited nodes in a set or boolean array. Add nodes to the visited set at the moment they enter the queue so they are never visited twice.
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