221. Maximal Square
Description
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Constraints
Approach
Links
GeeksforGeeks
ProgramCreek
YouTube
Examples
Input:
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Output: 4
Solutions
/**
* Time complexity : O((m*n)^2)
* Space complexity : O(1)
*/
class Solution {
public int maximalSquare(char[][] matrix) {
if(matrix == null || matrix.length == 0) return 0;
int rows = matrix.length,
cols = matrix[0].length;
int maxsqlen = 0;
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++) {
if(matrix[i][j] == '1') {
int sqlen = 1;
boolean flag = true;
while(sqlen + i < rows && sqlen + j < cols && flag) {
for(int k = j; k <= sqlen + j; k++) {
if (matrix[i + sqlen][k] == '0') {
flag = false;
break;
}
}
for(int k = i; k <= sqlen + i; k++) {
if (matrix[k][j + sqlen] == '0') {
flag = false;
break;
}
}
if(flag) sqlen++;
}
if(maxsqlen < sqlen) {
maxsqlen = sqlen;
}
}
}
}
return maxsqlen * maxsqlen;
}
}/**
* Time complexity : O(m*n)
* Space complexity : O(m*n)
*/
class Solution {
public int maximalSquare(char[][] matrix) {
if(matrix == null || matrix.length == 0) return 0;
int row = matrix.length,
col = matrix[0].length;
int[][] dp = new int[row+1][col+1];
int sqlen = 0;
for(int i = 1; i <= row; i++) {
for(int j = 1; j <= col; j++) {
if(matrix[i-1][j-1] == '1') {
dp[i][j] = 1 + Math.min(dp[i-1][j-1],
Math.min(dp[i][j-1], dp[i-1][j]));
sqlen = Math.max(sqlen, dp[i][j]);
}
}
}
return sqlen * sqlen;
}
}/**
* Time complexity : O(m*n), where m = row length, n = col length
* Space complexity : O(n), where n = col length
*/
class Solution {
public int maximalSquare(char[][] matrix) {
if(matrix == null || matrix.length == 0) return 0;
int row = matrix.length,
col = matrix[0].length;
int[] dp = new int[col+1];
int sqlen = 0, prev = 0;
for(int i = 1; i <= row; i++) {
for(int j = 1; j <= col; j++) {
int tmp = dp[j];
if(matrix[i-1][j-1] == '1') {
dp[j] = 1 + Math.min(dp[j], Math.min(dp[j-1], prev));
sqlen = Math.max(sqlen, dp[j]);
} else {
dp[j] = 0;
}
prev = tmp;
}
}
return sqlen * sqlen;
}
}Follow up
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