LeetCode #749 — HARD

Contain Virus

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

A virus is spreading rapidly, and your task is to quarantine the infected area by installing walls.

The world is modeled as an m x n binary grid isInfected, where isInfected[i][j] == 0 represents uninfected cells, and isInfected[i][j] == 1 represents cells contaminated with the virus. A wall (and only one wall) can be installed between any two 4-directionally adjacent cells, on the shared boundary.

Every night, the virus spreads to all neighboring cells in all four directions unless blocked by a wall. Resources are limited. Each day, you can install walls around only one region (i.e., the affected area (continuous block of infected cells) that threatens the most uninfected cells the following night). There will never be a tie.

Return the number of walls used to quarantine all the infected regions. If the world will become fully infected, return the number of walls used.

Example 1:

Input: isInfected = [[0,1,0,0,0,0,0,1],[0,1,0,0,0,0,0,1],[0,0,0,0,0,0,0,1],[0,0,0,0,0,0,0,0]]
Output: 10
Explanation: There are 2 contaminated regions.
On the first day, add 5 walls to quarantine the viral region on the left. The board after the virus spreads is:

On the second day, add 5 walls to quarantine the viral region on the right. The virus is fully contained.

Example 2:

Input: isInfected = [[1,1,1],[1,0,1],[1,1,1]]
Output: 4
Explanation: Even though there is only one cell saved, there are 4 walls built.
Notice that walls are only built on the shared boundary of two different cells.

Example 3:

Input: isInfected = [[1,1,1,0,0,0,0,0,0],[1,0,1,0,1,1,1,1,1],[1,1,1,0,0,0,0,0,0]]
Output: 13
Explanation: The region on the left only builds two new walls.

Constraints:

  • m == isInfected.length
  • n == isInfected[i].length
  • 1 <= m, n <= 50
  • isInfected[i][j] is either 0 or 1.
  • There is always a contiguous viral region throughout the described process that will infect strictly more uncontaminated squares in the next round.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: A virus is spreading rapidly, and your task is to quarantine the infected area by installing walls. The world is modeled as an m x n binary grid isInfected, where isInfected[i][j] == 0 represents uninfected cells, and isInfected[i][j] == 1 represents cells contaminated with the virus. A wall (and only one wall) can be installed between any two 4-directionally adjacent cells, on the shared boundary. Every night, the virus spreads to all neighboring cells in all four directions unless blocked by a wall. Resources are limited. Each day, you can install walls around only one region (i.e., the affected area (continuous block of infected cells) that threatens the most uninfected cells the following night). There will never be a tie. Return the number of walls used to quarantine all the infected regions. If the world will become fully infected, return the number of walls used.

Baseline thinking

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

Pattern signal: Array

Example 1

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

Example 2

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

Example 3

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

Related Problems

  • Count the Number of Infection Sequences (count-the-number-of-infection-sequences)
Step 02

Core Insight

What unlocks the optimal approach

  • The implementation is long - we want to perfrom the following steps: * Find all viral regions (connected components), additionally for each region keeping track of the frontier (neighboring uncontaminated cells), and the perimeter of the region. * Disinfect the most viral region, adding it's perimeter to the answer. * Spread the virus in the remaining regions outward by 1 square.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #749: Contain Virus
class Solution {
    private static final int[] DIRS = {-1, 0, 1, 0, -1};
    private List<Integer> c = new ArrayList<>();
    private List<List<Integer>> areas = new ArrayList<>();
    private List<Set<Integer>> boundaries = new ArrayList<>();
    private int[][] infected;
    private boolean[][] vis;
    private int m;
    private int n;

    public int containVirus(int[][] isInfected) {
        infected = isInfected;
        m = infected.length;
        n = infected[0].length;
        vis = new boolean[m][n];
        int ans = 0;
        while (true) {
            for (boolean[] row : vis) {
                Arrays.fill(row, false);
            }
            c.clear();
            areas.clear();
            boundaries.clear();
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (infected[i][j] == 1 && !vis[i][j]) {
                        c.add(0);
                        areas.add(new ArrayList<>());
                        boundaries.add(new HashSet<>());
                        dfs(i, j);
                    }
                }
            }
            if (areas.isEmpty()) {
                break;
            }
            int idx = max(boundaries);
            ans += c.get(idx);
            for (int t = 0; t < areas.size(); ++t) {
                if (t == idx) {
                    for (int v : areas.get(t)) {
                        int i = v / n, j = v % n;
                        infected[i][j] = -1;
                    }
                } else {
                    for (int v : areas.get(t)) {
                        int i = v / n, j = v % n;
                        for (int k = 0; k < 4; ++k) {
                            int x = i + DIRS[k], y = j + DIRS[k + 1];
                            if (x >= 0 && x < m && y >= 0 && y < n && infected[x][y] == 0) {
                                infected[x][y] = 1;
                            }
                        }
                    }
                }
            }
        }
        return ans;
    }

    private int max(List<Set<Integer>> boundaries) {
        int idx = 0;
        int mx = boundaries.get(0).size();
        for (int i = 1; i < boundaries.size(); ++i) {
            int t = boundaries.get(i).size();
            if (mx < t) {
                mx = t;
                idx = i;
            }
        }
        return idx;
    }

    private void dfs(int i, int j) {
        vis[i][j] = true;
        int idx = areas.size() - 1;
        areas.get(idx).add(i * n + j);
        for (int k = 0; k < 4; ++k) {
            int x = i + DIRS[k], y = j + DIRS[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n) {
                if (infected[x][y] == 1 && !vis[x][y]) {
                    dfs(x, y);
                } else if (infected[x][y] == 0) {
                    c.set(idx, c.get(idx) + 1);
                    boundaries.get(idx).add(x * n + y);
                }
            }
        }
    }
}
Step 06

Interactive Study Demo

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

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.

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