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Maximal Square

Medium Dynamic Programming Squares Original on LeetCode

Maximal Square is a medium dynamic programming problem solved with the squares pattern. The best approach, optimal (dp on the bottom-right corner), runs in O(m · n) time and O(n) space. Below are 2 approaches in Java, from brute force (grow squares) up.

Problem

In a binary matrix, find the largest square containing only 1s and return its area.

Examples

Example 1

Input
matrix = [[1,0,1,0,0],[1,0,1,1,1],[1,1,1,1,1],[1,0,0,1,0]]
Output
4
Why
A 2 × 2 square of 1s.

Example 2

Input
matrix = [[0]]
Output
0

Constraints

  • 1 <= m, n <= 300; cells are '0' or '1'.
  • Return the area.

Animated walkthrough

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Concepts first, then the problem and every approach, step by step.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
Brute force (grow squares)O((m · n)²)O(1)
Optimal (DP on the bottom-right corner)O(m · n)O(n)

1Brute force (grow squares)

TimeO((m · n)²)
SpaceO(1)

From every 1, try to grow the square one layer at a time, checking the new row and column.

  1. For each cell, increase the size while the next ring is all 1s.
Java
class Solution {
    public int maximalSquare(char[][] matrix) {
        int m = matrix.length, n = matrix[0].length, best = 0;
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++) {
                int size = 0;
                grow:
                while (i + size < m && j + size < n) {
                    for (int k = 0; k <= size; k++)
                        if (matrix[i + size][j + k] != '1' || matrix[i + k][j + size] != '1') break grow;
                    size++;
                }
                best = Math.max(best, size);
            }
        return best * best;
    }
}

2Optimal (DP on the bottom-right corner)

TimeO(m · n)
SpaceO(n)

A square ending at (i, j) can only be as big as the smallest of the three squares ending at its top, left and top-left neighbours, plus one. Keep one row of the table plus the diagonal value.

  1. If the cell is '1': dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]).
  2. Track the largest side; return side².
Java
class Solution {
    public int maximalSquare(char[][] matrix) {
        int n = matrix[0].length, best = 0;
        int[] dp = new int[n + 1];
        for (char[] row : matrix) {
            int diag = 0;
            for (int j = 1; j <= n; j++) {
                int up = dp[j];
                dp[j] = row[j - 1] == '1' ? 1 + Math.min(Math.min(dp[j], dp[j - 1]), diag) : 0;
                diag = up;
                best = Math.max(best, dp[j]);
            }
        }
        return best * best;
    }
}

Edge cases to test

  • No 1s at all
  • Single row or column

Hints

Hint 1

dp[i][j] = side of the largest square whose bottom-right corner is (i, j) = 1 + min(top, left, top-left).

FAQ

What is the best time complexity for Maximal Square?

Optimal (DP on the bottom-right corner) runs in O(m · n) time and O(n) extra space.

Which pattern does Maximal Square use?

It is a dynamic programming problem that uses the squares pattern. Other problems with the same pattern: Count Square Submatrices with All Ones, Largest 1-Bordered Square.

Is there a brute force solution for Maximal Square?

Yes. Brute force (grow squares) takes O((m · n)²) time and O(1) space. From every 1, try to grow the square one layer at a time, checking the new row and column.

Which edge cases should I test for Maximal Square?

No 1s at all; Single row or column.