README.md

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}
 */
var removeStones = function(stones) {
    const n = stones.length;
    const stoneGroup = 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;
};