LeetCode #934 — MEDIUM

Shortest Bridge

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[0,1],[1,0]]

Example 2

[[0,1,0],[0,0,0],[0,0,1]]

Example 3

[[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]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #934: Shortest Bridge
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);
            }
        }
    }
}

// Accepted solution for LeetCode #934: Shortest Bridge
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++
	}
}

# Accepted solution for LeetCode #934: Shortest Bridge
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

// Accepted solution for LeetCode #934: Shortest Bridge
struct Solution;
use std::collections::VecDeque;

impl Solution {
    fn shortest_bridge(mut a: Vec<Vec<i32>>) -> i32 {
        let n = a.len();
        let m = a[0].len();
        let mut queue: VecDeque<(usize, usize, i32)> = VecDeque::new();
        let mut found = false;
        for i in 0..n {
            if found {
                break;
            }
            for j in 0..m {
                if a[i][j] == 1 {
                    Self::dfs(i, j, &mut queue, &mut a, n, m);
                    found = true;
                    break;
                }
            }
        }
        while let Some((i, j, d)) = queue.pop_front() {
            match a[i][j] {
                0 | 2 => {
                    a[i][j] = 3;
                    if i > 0 && a[i - 1][j] < 2 {
                        queue.push_back((i - 1, j, d + 1));
                    }
                    if j > 0 && a[i][j - 1] < 2 {
                        queue.push_back((i, j - 1, d + 1));
                    }
                    if i + 1 < n && a[i + 1][j] < 2 {
                        queue.push_back((i + 1, j, d + 1));
                    }
                    if j + 1 < m && a[i][j + 1] < 2 {
                        queue.push_back((i, j + 1, d + 1));
                    }
                }
                1 => {
                    return d - 1;
                }
                _ => {}
            }
        }
        0
    }

    fn dfs(
        i: usize,
        j: usize,
        queue: &mut VecDeque<(usize, usize, i32)>,
        a: &mut Vec<Vec<i32>>,
        n: usize,
        m: usize,
    ) {
        if a[i][j] == 1 {
            a[i][j] = 2;
            queue.push_back((i, j, 0));
            if i > 0 {
                Self::dfs(i - 1, j, queue, a, n, m);
            }
            if j > 0 {
                Self::dfs(i, j - 1, queue, a, n, m);
            }
            if i + 1 < n {
                Self::dfs(i + 1, j, queue, a, n, m);
            }
            if j + 1 < m {
                Self::dfs(i, j + 1, queue, a, n, m);
            }
        }
    }
}

#[test]
fn test() {
    let a = vec_vec_i32![[0, 1], [1, 0]];
    let res = 1;
    assert_eq!(Solution::shortest_bridge(a), res);
    let a = vec_vec_i32![[0, 1, 0], [0, 0, 0], [0, 0, 1]];
    let res = 2;
    assert_eq!(Solution::shortest_bridge(a), res);
    let a = vec_vec_i32![
        [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]
    ];
    let res = 1;
    assert_eq!(Solution::shortest_bridge(a), res);
}

// Accepted solution for LeetCode #934: Shortest Bridge
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #934: Shortest Bridge
// 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);
//             }
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.