Flood fill
827. Making A Large Island, Medium (if aware of flood fill) else Hard
- Problem necessitates Flood fill concepts and coloring the graphs.
- Approach
- We first traverse accross components that are not connected
- We give them some color (0, 1, 2 ,3 … ) and store their size.
- We then in another loop traverse only at points of no Land i.e. where grid[i][j] == 0
- and check what is the colour of the point located to its (left /right/ down/ up).
-
code
BFS Implementation
```cpp
int dr[] = {1, -1, 0, 0};
int dc[] = {0, 0, -1, 1};
int largestIsland(vector<vector> &grid) {
int n = grid.size();
int m = grid[0].size();
vector<vector> used(n, vector(m, false));
queue<pair<int, int>> qu;
map<int, int> mp;
int color = 1;
mp[0] = 0;
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1 and !used[i][j]) {
int Size = 1;
qu.push({i, j});
used[i][j] = color;
while (!qu.empty()) {
auto [r, c] = qu.front();
qu.pop();
for (int k = 0; k < 4; k++) {
int rr = r + dr[k];
int cc = c + dc[k];
if (rr < 0 or cc < 0 or rr >= n or cc >= n or used[rr][cc] or grid[rr][cc] != 1)
continue;
qu.push({rr, cc});
used[rr][cc] = color;
Size++;
}
}
ans = max(ans, Size);
mp[color] = Size;
color++;
}
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 0) {
int sum = 0;
map<int, bool> usedColor;
for (int k = 0; k < 4; k++) {
int rr = i + dr[k];
int cc = j + dc[k];
if (rr < 0 or cc < 0 or rr >= n or cc >= n)
continue;
int currentColor = used[rr][cc];
if (usedColor.count(currentColor))
continue;
sum += (mp[currentColor]);
if (currentColor != 0) {
usedColor[currentColor] = true;
}
}
ans = max(ans, sum + 1);
}
}
}
return ans;
```
</details>