2282. Number of People That Can Be Seen in a Grid
Explanation:
To solve this problem, we can iterate through the grid and for each cell, check if it can see any other person. A person in a cell can see another person if there are no obstacles in a straight line along the row or column. We can maintain four arrays to store the maximum indices of people seen in each direction for each cell. By iterating through the grid and updating these arrays, we can count the total number of people that can be seen.
- Initialize four arrays to store the maximum indices of people seen in each direction:
left,right,up, anddown. - Iterate through the grid and update the above arrays for each cell.
- For each cell, count the number of people that can be seen based on the maximum indices stored in the arrays.
Time Complexity: O(NM), where N is the number of rows and M is the number of columns in the grid. Space Complexity: O(NM) for storing the maximum indices arrays.
: :
class Solution {
public int numPeopleSeen(int[][] grid) {
int n = grid.length;
int m = grid[0].length;
int[] left = new int[m];
int[] right = new int[m];
int[] up = new int[n];
int[] down = new int[n];
int count = 0;
for (int i = 0; i < n; i++) {
int maxRight = -1;
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) {
left[j] = Math.max(left[j], i);
right[j] = Math.max(right[j], maxRight);
count++;
} else {
maxRight = j;
}
}
}
for (int j = 0; j < m; j++) {
int maxDown = -1;
for (int i = 0; i < n; i++) {
if (grid[i][j] == 1) {
up[i] = Math.max(up[i], j);
down[i] = Math.max(down[i], maxDown);
} else {
maxDown = i;
}
}
}
return count;
}
}Code Editor (Testing phase)
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