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Community-reported · source · updated 2026-10-08. ● asked in the last 6 months.
Go step by step: try it yourself before the video and the visual. Progress saves on this device.
0 / 130 pointsSolved (Medium) 0/40Reviews 0/50Mastered 0/25Practice steps 0/15
Step 1
Know these before you start
Practise each one here, or tick what you already know
Concept videoBuilds on: Recursion: the base case and the call stack
Step 2
Try it on paper
5–10 minutes, no code yet
In the wild · Spell check and diff: “Did you mean…?” compares words with edit distance, and diff tools align two files with longest common subsequence. Both fill a DP table.
About this problem
Break a problem into overlapping subproblems and reuse their answers.
Read the full statement and examples on LeetCode, try it for 20–30 minutes, then come back for the walkthrough.
Examples
Full statement on LeetCode ↗Example 1
- Input
- matrix = [["0", "1"], ["1", "0"]]
- Output
- 1
Example 2
1×1 grid- Input
- matrix = [["0"]]
- Output
- 0
Example 3
All the same- Input
- matrix = [["1", "1", "1"],["1", "1", "1"],["1", "1", "1"]]
- Output
- 9
Code it here
Tab indents · ⌘/Ctrl+Enter runs · Esc leaves the editor
Step 6
Get it accepted
All test cases pass
Step 7
Review it so it sticks
Re-solve from memory after
The step that matters most: without reviews most of a solved problem fades within weeks. Three short re-solves, each just as you start to forget, are what make it yours.
Your notes
Three lines in your own words. They come back with this problem's reviews, so you can check yourself after recalling the solution.
Solve it yourself before you look
Reading the solution first feels like progress, but it makes the next similar problem — and the interview version — much harder, because you skipped the part where you figure it out. Give it an honest 20–30 minutes. Stuck? Re-read the pattern, watch the concept video, or try the brute force first.
Hidden: optimal solution
Solution code
Optimal solutiondp[r][c] is the side of the biggest all-1 square whose bottom-right corner is (r, c): one more than the smallest of the squares above, to the left and diagonally up-left. The answer is the biggest side, squared.
class Solution:
def maximalSquare(self, matrix: List[List[str]]) -> int:
cols = len(matrix[0])
side = [0] * (cols + 1) # previous row, shifted by one
best = 0
for row in matrix:
diag = 0 # side[c] from the row above, one to the left
for c in range(1, cols + 1):
above = side[c]
side[c] = min(side[c], side[c - 1], diag) + 1 if row[c - 1] == "1" else 0
best = max(best, side[c])
diag = above
return best * bestReference solution, tested in Python, Java and C++.
Other ways to solve it
- Thought of another approach (slower, or cleverer)? Code it here: the same tests check any approach.
- Community solutions on LeetCode ↗ for many more versions, in many languages. Read them after you've solved it.