Algorithm
Problem Name: 947. Most Stones Removed with Same Row or Column
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] = [xi, yi] represents the location of the ith 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 <= 10000 <= xi, yi <= 104- No two stones are at the same coordinate point.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
void dfs(int **stones, int sz, int p, int *visited, int *n) {
int r, c, i;
if (visited[p]) return;
visited[p] = 1;
(*n) ++;
r = stones[p][0];
c = stones[p][1];
// remove stones on same row
for (i = 0; i < sz; i ++) {
if (stones[i][0] == r) dfs(stones, sz, i, visited, n);
}
// remove stones on same col
for (i = 0; i < sz; i ++) {
if (stones[i][1] == c) dfs(stones, sz, i, visited, n);
}
//visited[p] = 0;
}
int removeStones(int** stones, int stonesSize, int* stonesColSize){
int n, ans = 0;
int *visited;
int i;
visited = calloc(stonesSize, sizeof(int));
//assert(visited);
for (i = 0; i < stonesSize; i ++) {
n = 0;
dfs(stones, stonesSize, i, visited, &n);
if (n) ans += n - 1;
//printf("%d\n", n);
}
free(visited);
return ans;
}
Input
stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]
Output
5
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int removeStones(int[][] stones) {
Set<int[]> visited = new HashSet<>();
int numOfIslands = 0;
for (int[] stone : stones) {
if (!visited.contains(stone)) {
dfs(stones, visited, stone);
numOfIslands++;
}
}
return stones.length - numOfIslands;
}
private void dfs(int[][] stones, Set < int[]> visited, int[] stone) {
visited.add(stone);
for (int[] st : stones) {
if ((st[0] == stone[0] || st[1] == stone[1]) && !visited.contains(st)) {
dfs(stones, visited, st);
}
}
}
}
Input
stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]
Output
5
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const removeStones = function(stones) {
const f = new Map()
let islands = 0
for (let i = 0; i < stones.length; i++) {
union(stones[i][0], ~stones[i][1]) // row, col
}
return stones.length - islands
function find(x) {
if (!f.has(x)) {
islands++
f.set(x, x)
}
if (x != f.get(x)) {
f.set(x, find(f.get(x)))
}
return f.get(x)
}
function union(x, y) {
x = find(x)
y = find(y)
if (x !== y) {
f.set(x, y)
islands--
}
}
}
Input
stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output
3
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def removeStones(self, stones):
def dfs(i, j):
points.discard((i, j))
for y in rows[i]:
if (i, y) in points:
dfs(i, y)
for x in cols[j]:
if (x, j) in points:
dfs(x, j)
points, island, rows, cols = {(i, j) for i, j in stones}, 0, collections.defaultdict(list), collections.defaultdict(list)
for i, j in stones:
rows[i].append(j)
cols[j].append(i)
for i, j in stones:
if (i, j) in points:
dfs(i, j)
island += 1
return len(stones) - island
Input
stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output
3
#5 Code Example with C# Programming
Code -
C# Programming
using System.Collections.Generic;
namespace LeetCode
{
public class _0947_MostStonesRemovedWithSameRoworColumn
{
private const int BOARD_SIZE = 10000;
public int RemoveStones(int[][] stones)
{
var uf = new UnionFind(BOARD_SIZE * 2);
foreach (var stone in stones)
uf.Union(stone[0], stone[1] + BOARD_SIZE);
var set = new HashSet < int>();
foreach (var stone in stones)
set.Add(uf.FindParent(stone[0]));
return stones.Length - set.Count;
}
public class UnionFind
{
private int[] parents;
public UnionFind(int size)
{
parents = new int[size];
for (int i = 0; i < size; i++)
parents[i] = i;
}
public int FindParent(int x)
{
while (parents[x] != x)
x = parents[x];
return parents[x];
}
public void Union(int x, int y)
{
parents[FindParent(x)] = FindParent(y);
}
}
}
}
Input
stones = [[0,0]]