Description
You are given a two-dimensional integer matrix containing 0s and 1s where 0 represents water and 1 represents land. Return a new matrix of the same dimensions where we increase the height of every land cell as much as possible given that:
- The height of any water cell staysÂ
0 - Any cell can differ by at mostÂ
1Â unit of height between neighboring cells (up, down, left, right)
You can assume there is at least one water cell.
Constraints:
-
n, m ≤ 250 whereÂn andÂm are the number of rows and columns inÂmatrix
Example 1
Input
matrix = [
[0, 1, 0],
[1, 1, 1],
[1, 1, 1]
]
Output
[
[0, 1, 0],
[1, 2, 1],
[2, 3, 2]
]
Intuition
Do multi source BFS from all the water positions i.e 0's by inserting all such cells into a queue
Implementation
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
vector<vector<int>> solve(vector<vector<int>>& matrix) {
int rows = matrix.size(), cols = matrix[0].size();
vector<vector<bool>> visited(rows, vector<bool>(cols, false));
queue<pair<int, int>> q;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (matrix[i][j] == 0) q.push({i, j});
}
}
int val = 0;
while (!q.empty()) {
int size = q.size();
for (int i = 0; i < size; i++) {
auto point = q.front();
q.pop();
if (visited[point.first][point.second]) continue;
visited[point.first][point.second] = true;
matrix[point.first][point.second] = val;
for (int j = 0; j < 4; j++) {
int x = point.first + dx[j], y = point.second + dy[j];
if (x >= 0 && x < rows && y >= 0 && y < cols && !visited[x][y]) q.push({x, y});
}
}
val++;
}
return matrix;
}
Time Complexity
- Time: O(n)
- Space: O(n)
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