# 934. Shortest Bridge

## Description

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.

## Solutions

DFS & BFS.

• class Solution {
private int[] dirs = {-1, 0, 1, 0, -1};
private Deque<int[]> q = new ArrayDeque<>();
private int[][] grid;
private int n;

public int shortestBridge(int[][] grid) {
this.grid = grid;
n = grid.length;
for (int i = 0, x = 1; i < n && x == 1; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
dfs(i, j);
x = 0;
break;
}
}
}
int ans = 0;
while (true) {
for (int i = q.size(); i > 0; --i) {
var p = q.pollFirst();
for (int k = 0; k < 4; ++k) {
int x = p[0] + dirs[k], y = p[1] + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n) {
if (grid[x][y] == 1) {
return ans;
}
if (grid[x][y] == 0) {
grid[x][y] = 2;
q.offer(new int[] {x, y});
}
}
}
}
++ans;
}
}

private void dfs(int i, int j) {
grid[i][j] = 2;
q.offer(new int[] {i, j});
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1) {
dfs(x, y);
}
}
}
}

• class Solution {
public:
const static inline vector<int> dirs = {-1, 0, 1, 0, -1};

int shortestBridge(vector<vector<int>>& grid) {
int n = grid.size();
queue<pair<int, int>> q;
function<void(int, int)> dfs = [&](int i, int j) {
grid[i][j] = 2;
q.emplace(i, j);
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1) {
dfs(x, y);
}
}
};
for (int i = 0, x = 1; i < n && x; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j]) {
dfs(i, j);
x = 0;
break;
}
}
}
int ans = 0;
while (1) {
for (int h = q.size(); h; --h) {
auto [i, j] = q.front();
q.pop();
for (int k = 0; k < 4; ++k) {
int x = i + dirs[k], y = j + dirs[k + 1];
if (x >= 0 && x < n && y >= 0 && y < n) {
if (grid[x][y] == 1) return ans;
if (grid[x][y] == 0) {
grid[x][y] = 2;
q.emplace(x, y);
}
}
}
}
++ans;
}
}
};

• class Solution:
def shortestBridge(self, grid: List[List[int]]) -> int:
def dfs(i, j):
q.append((i, j))
grid[i][j] = 2
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n and grid[x][y] == 1:
dfs(x, y)

n = len(grid)
dirs = (-1, 0, 1, 0, -1)
q = deque()
i, j = next((i, j) for i in range(n) for j in range(n) if grid[i][j])
dfs(i, j)
ans = 0
while 1:
for _ in range(len(q)):
i, j = q.popleft()
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < n and 0 <= y < n:
if grid[x][y] == 1:
return ans
if grid[x][y] == 0:
grid[x][y] = 2
q.append((x, y))
ans += 1


• func shortestBridge(grid [][]int) (ans int) {
n := len(grid)
dirs := []int{-1, 0, 1, 0, -1}
type pair struct{ i, j int }
q := []pair{}
var dfs func(int, int)
dfs = func(i, j int) {
grid[i][j] = 2
q = append(q, pair{i, j})
for k := 0; k < 4; k++ {
x, y := i+dirs[k], j+dirs[k+1]
if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 {
dfs(x, y)
}
}
}
for i, x := 0, 1; i < n && x == 1; i++ {
for j := 0; j < n; j++ {
if grid[i][j] == 1 {
dfs(i, j)
x = 0
break
}
}
}
for {
for i := len(q); i > 0; i-- {
p := q[0]
q = q[1:]
for k := 0; k < 4; k++ {
x, y := p.i+dirs[k], p.j+dirs[k+1]
if x >= 0 && x < n && y >= 0 && y < n {
if grid[x][y] == 1 {
return
}
if grid[x][y] == 0 {
grid[x][y] = 2
q = append(q, pair{x, y})
}
}
}
}
ans++
}
}