LeetCode in Kotlin

934. Shortest Bridge

Medium

You are given an n x n binary matrix grid where 1 represents land and 0 represents water.

An island is a 4-directionally connected group of 1’s not connected to any other 1’s. There are exactly two islands in grid.

You may change 0’s to 1’s to connect the two islands to form one island.

Return the smallest number of 0’s you must flip to connect the two islands.

Example 1:

Input: grid = [[0,1],[1,0]]

Output: 1

Example 2:

Input: grid = [[0,1,0],[0,0,0],[0,0,1]]

Output: 2

Example 3:

Input: grid = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]

Output: 1

Constraints:

• n == grid.length == grid[i].length
• 2 <= n <= 100
• grid[i][j] is either 0 or 1.
• There are exactly two islands in grid.

Solution

class Solution {
    private class Pair(var x: Int, var y: Int)

    private val dirs = arrayOf(intArrayOf(1, 0), intArrayOf(0, 1), intArrayOf(-1, 0), intArrayOf(0, -1))

    fun shortestBridge(grid: Array<IntArray>): Int {
        val q: ArrayDeque<Pair> = ArrayDeque()
        var flag = false
        val visited = Array(grid.size) {
            BooleanArray(
                grid[0].size
            )
        }
        var i = 0
        while (i < grid.size && !flag) {
            var j = 0
            while (j < grid[i].size && !flag) {
                if (grid[i][j] == 1) {
                    dfs(grid, i, j, visited, q)
                    flag = true
                }
                j++
            }
            i++
        }
        var level = -1
        while (q.isNotEmpty()) {
            var size: Int = q.size
            level++
            while (size-- > 0) {
                val rem = q.removeFirst()
                for (dir in dirs) {
                    val newrow = rem.x + dir[0]
                    val newcol = rem.y + dir[1]
                    if (newrow >= 0 && newcol >= 0 && newrow < grid.size && newcol < grid[0].size &&
                        !visited[newrow][newcol]
                    ) {
                        if (grid[newrow][newcol] == 1) {
                            return level
                        }
                        q.add(Pair(newrow, newcol))
                        visited[newrow][newcol] = true
                    }
                }
            }
        }
        return -1
    }

    private fun dfs(grid: Array<IntArray>, row: Int, col: Int, visited: Array<BooleanArray>, q: ArrayDeque<Pair>) {
        visited[row][col] = true
        q.add(Pair(row, col))
        for (dir in dirs) {
            val newrow = row + dir[0]
            val newcol = col + dir[1]
            if (newrow >= 0 && newcol >= 0 && newrow < grid.size && newcol < grid[0].size &&
                !visited[newrow][newcol] && grid[newrow][newcol] == 1
            ) {
                dfs(grid, newrow, newcol, visited, q)
            }
        }
    }
}

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