947. Most Stones Removed with Same Row or Column

Description

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 <= 1000
• 0 <= xi, yi <= 104
• No two stones are at the same coordinate point.

 

Solutions

Solution: Union Find

• Time complexity: O(n²)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[][]} stones
 * @return {number}
 */
const removeStones = function (stones) {
  const n = stones.length;
  const stoneGroup = new Array(n)
    .fill('')
    .map((_, index) => index);

  const unionFind = x => {
    return stoneGroup[x] === x ? x : unionFind(stoneGroup[x]);
  };

  for (let a = 0; a < n; a++) {
    for (let b = a + 1; b < n; b++) {
      const [x1, y1] = stones[a];
      const [x2, y2] = stones[b];

      if (x1 !== x2 && y1 !== y2) continue;
      const groupA = unionFind(a);
      const groupB = unionFind(b);

      stoneGroup[groupB] = groupA;
    }
  }
  let result = n;

  for (let index = 0; index < n; index++) {
    if (index !== stoneGroup[index]) continue;
    result -= 1;
  }
  return result;
};