How do you approach Maximal Square?

dp[i][j] is the side of the largest square whose bottom-right corner is (i, j).

What are common mistakes on Maximal Square?

Iterating the dimensions in an order that reads a cell before it has been written. Getting the base case wrong. It carries the whole recurrence, and the error stays invisible until a large input. Rolling down to one dimension without reversing the inner loop, which lets a 0/1 item be taken twice.

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Maximal Square

MediumAlgorithms

Description

Given an `m x n` binary `matrix` of `'0'`s and `'1'`s, find the largest square containing only `'1'`s and return its **area**.

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

Explanation: A 2x2 square of 1s has area 4.

Constraints

- m == matrix.length - n == matrix[i].length - 1 <= m, n <= 300 - matrix[i][j] is '0' or '1'

Maximal Square: approach, complexity and pitfalls

A medium dynamic programming problem, graded against 4 test cases (2 of them hidden).

The pattern: Dynamic Programming

Overlapping subproblems solved once and reused - the pattern candidates fear most.

Reach for it when you see: "Number of ways", "minimum/maximum cost", or a recursion that recomputes the same state.

More Dynamic Programmingproblems →

If you are stuck

Show the 2-step hint ladder

dp[i][j] is the side of the largest square whose bottom-right corner is (i, j).
On a '1', dp[i][j] = 1 + min of the three neighbors (up, left, up-left).

The optimal approach

Show the approach: The approach

Square-side DP. For a `'1'` cell, the largest square ending there is limited by its top, left, and top-left neighbors: `dp[i][j] = min(up, left, up-left) + 1`. Track the max side; the area is side².

Time: O(m·n). Space: O(m·n).

The full reference solution in every supported language stays in the editor above - reveal it there once you have had a real attempt.

What the tests check

Read off this problem's own test suite, so these are the cases a submission actually has to survive.

Single element. A one-element case is included. Two-pointer and window solutions frequently return the wrong thing when the two indices start out equal.
Duplicate values. Repeated values appear in the input, so any logic that assumes distinct elements - or that skips over equal neighbours - is exercised.

Common Dynamic Programming mistakes

These apply to the pattern as a whole, not just this problem.

Iterating the dimensions in an order that reads a cell before it has been written.
Getting the base case wrong. It carries the whole recurrence, and the error stays invisible until a large input.
Rolling down to one dimension without reversing the inner loop, which lets a 0/1 item be taken twice.