# Most Stones Removed With Same Row Or Column Problem

## Description

LeetCode Problem 947.

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:

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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:

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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:

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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 <= 10^4
- No two stones are at the same coordinate point.

## Sample C++ Code using Union Find

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class Solution {
public:
int _find(int p, vector<int>& nums) {
while (p != nums[p])
p = nums[p];
return p;
}
void _union(int i, int j, vector<int>& nums) {
int r1 = _find(i, nums);
int r2 = _find(j, nums);
nums[r1] = r2;
}
int removeStones(vector<vector<int>>& stones) {
int n = stones.size();
vector<int> parent(n);
for (int i = 0; i < n; i ++) parent[i] = i;
for (int i = 0; i < n; i ++) {
for (int j = 0; j < n; j ++) {
if (stones[i][0] == stones[j][0] ||
stones[i][1] == stones[j][1])
_union(i, j, parent);
}
}
unordered_map<int, int> ht;
for (int i = 0; i < n; i ++) {
ht[_find(i, parent)] = 1;
}
return n - ht.size();
}
};