827. Making A Large Island
You are given an n x n binary matrix grid. You are allowed to change at most one 0 to be 1.
Return the size of the largest island in grid after applying this operation.
An island is a 4-directionally connected group of 1s.
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| class Solution {
int getArea(vector<vector<int>>& grid, int sy, int sx, int island) {
int dx[4] = {0,1,0,-1}, dy[4] = {-1,0,1,0}, n = grid.size();
queue<pair<int, int>> q;
q.push({sy,sx});
int counter = 1;
grid[sy][sx] = island;
while(!q.empty()) {
auto [y, x] = q.front(); q.pop();
for(int i = 0; i < 4; i++) {
int ny = y + dy[i], nx = x + dx[i];
if(0 <= ny and ny < n and 0 <= nx and nx < n and grid[ny][nx] == 1) {
counter++;
grid[ny][nx] = island;
q.push({ny,nx});
}
}
}
return counter;
}
public:
int largestIsland(vector<vector<int>>& grid) {
vector<int> area(2);
int island = 2, n = grid.size(), res = 0;
int dx[4] = {0,1,0,-1}, dy[4] = {-1,0,1,0};
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == 1) {
int count = getArea(grid,i,j,island);
area.push_back(count);
island++;
res = max(res, count);
}
}
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
if(grid[i][j] == 0) {
int sumArea = 1;
unordered_set<int> counted;
for(int k = 0; k < 4; k++) {
int ny = i + dy[k], nx = j + dx[k];
if(0 <= ny and ny < n and 0 <= nx and nx < n and !counted.count(grid[ny][nx])) {
counted.insert(grid[ny][nx]);
sumArea += area[grid[ny][nx]];
}
}
res = max(res, sumArea);
}
}
}
return res;
}
};
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