LeetCode 1139. Largest 1-Bordered Square
原题链接在这里:https://leetcode.com/problems/largest-1-bordered-square/
题目:
Given a 2D grid
of 0
s and 1
s, return the number of elements in the largest square subgrid that has all 1
s on its border, or 0
if such a subgrid doesn't exist in the grid
.
Example 1:
Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 9
Example 2:
Input: grid = [[1,1,0,0]]
Output: 1
Constraints:
1 <= grid.length <= 100
1 <= grid[0].length <= 100
grid[i][j]
is0
or1
题解:
For each cell in the grid, calculate its farest reach on top and left direction.
Then starting from l = Math.min(grid.length, grid[0].length) to l = 1, iterate grid to check if square with boarder l exist. If it does return l*l.
Time Complexity: O(m*n*(min(m,n))). m = grid.length. n = grid[0].length.
Space: O(m*n).
AC Java:
class Solution {
public int largest1BorderedSquare(int[][] grid) {
if(grid == null || grid.length == 0 || grid[0].length == 0){
return 0;
} int m = grid.length;
int n = grid[0].length;
int [][] top = new int[m][n];
int [][] left = new int[m][n];
for(int i = 0; i<m; i++){
for(int j = 0; j<n; j++){
if(grid[i][j] > 0){
top[i][j] = i == 0 ? 1 : top[i-1][j]+1;
left[i][j] = j == 0 ? 1 : left[i][j-1]+1;
}
}
} for(int l = Math.min(m, n); l>0; l--){
for(int i = 0; i+l-1<m; i++){
for(int j = 0; j+l-1<n; j++){
if(top[i+l-1][j] >= l
&& top[i+l-1][j+l-1] >= l
&& left[i][j+l-1] >= l
&& left[i+l-1][j+l-1] >= l){
return l*l;
}
}
}
} return 0;
}
}
Or after get top and left.
Iterate the grid, for each cell, get small = min(top[i][j], left[i][j]).
All l = small to 1 could be protential square boarder. But we only need to check small larger than global longest boarder since we only care about the largest.
Check top of grid[i][j-small+1] and left of grid[i-small+1][j]. If they are both larger than small, then it is a grid.
Time Complexity: O(n^3).
Space: O(n^2).
AC Java:
class Solution {
public int largest1BorderedSquare(int[][] grid) {
if(grid == null || grid.length == 0 || grid[0].length == 0){
return 0;
} int m = grid.length;
int n = grid[0].length;
int [][] top = new int[m][n];
int [][] left = new int[m][n];
for(int i = 0; i<m; i++){
for(int j = 0; j<n; j++){
if(grid[i][j] > 0){
top[i][j] = i == 0 ? 1 : top[i-1][j]+1;
left[i][j] = j == 0 ? 1 : left[i][j-1]+1;
}
}
} int res = 0; for(int i = m-1; i>=0; i--){
for(int j = n-1; j>=0; j--){
int small = Math.min(top[i][j], left[i][j]);
while(small > res){
if(top[i][j-small+1] >= small && left[i-small+1][j] >= small){
res = small;
break;
} small--;
}
}
} return res*res;
}
}
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