LeetCode in Kotlin

947. Most Stones Removed with Same Row or Column

Medium

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.

Solution

class Solution {
    private val roots = IntArray(20002)

    fun removeStones(stones: Array<IntArray>): Int {
        for (stone in stones) {
            init(stone[0] + 1, roots)
            init(stone[1] + 10000, roots)
            union(stone[0] + 1, stone[1] + 10000)
        }
        val set: HashSet<Int> = HashSet()
        for (n in roots) {
            if (n == 0) {
                continue
            }
            set.add(find(n))
        }
        return stones.size - set.size
    }

    private fun init(i: Int, roots: IntArray) {
        if (roots[i] != 0) {
            return
        }
        roots[i] = i
    }

    private fun union(i: Int, j: Int) {
        val ri = find(i)
        val rj = find(j)
        if (ri == rj) {
            return
        }
        roots[ri] = rj
    }

    private fun find(i: Int): Int {
        var cur = i
        while (cur != roots[cur]) {
            cur = roots[roots[cur]]
        }
        return cur
    }
}

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