Maximum Rectangle in Binary Matrix
A hard Dynamic Programming problem included in Love Babbar 450. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
- Platform
- LeetCode
The problem
Given a 2D binary matrix, find the area of the largest rectangle containing only ones.
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
- 6
- Why
- The maximal rectangle spans rows 1-2 and columns 2-4, area = 6.
Example 2
- Input
- matrix = [["0"]]
- Output
- 0
- Why
- No rectangle of ones exists.
Example 3
- Input
- matrix = [["1"]]
- Output
- 1
- Why
- Single cell with value 1, rectangle area is 1.
Constraints
- rows == matrix.length
- cols == matrix[i].length
- 1 <= rows, cols <= 200
- matrix[i][j] is '0' or '1'
How to think about it
Updated 2026-09-09Every row can be converted into a histogram of continuous 1s extending upward. Finding the largest rectangle in the binary matrix then reduces to running the Largest Rectangle in Histogram algorithm across each of the m rows.
Approaches, worst first
Exhaustive coordinate bounds
time O((m * n)^3) · space O(1)
Iterate over every pair of top-left (r1, c1) and bottom-right (r2, c2) coordinates, verifying that all cells within the rectangle contain '1'. Extremely slow.
Row-wise width expansion
time O(m^2 * n) · space O(m * n)
For each cell, compute the number of consecutive ones to its left. For each cell (r, c), sweep upward row by row updating the minimum width seen to calculate candidate areas.
Histogram heights with monotonic stackWrite this one
time O(m * n) · space O(n)
Maintain an array `heights` of size n. For each row, increment heights[j] if matrix[i][j] == '1', otherwise reset to 0. Find the largest rectangle in this histogram using a monotonic increasing stack.
Where people lose marks · 3
- Failing to reset height to 0 when encountering '0': continuous columns break immediately on a zero.
- Input matrix contains character digits '0' and '1', not integers 0 and 1; comparing with integer 1 evaluates to false.
- Single-row or single-column matrices: stack processing must push sentinel zero or flush remaining elements to clear the stack.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
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