Count Square Submatrices with All Ones
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
The problem
Given a 2D binary matrix, count the total number of square submatrices with all ones.
Example 1
- Input
- matrix = [[1,1,0],[1,1,1],[0,1,1]]
- Output
- 9
- Why
- Seven single cells plus the two 2x2 squares (bottom-right corners at row 1 col 1 and row 2 col 2) make 9.
Example 2
- Input
- matrix = [[0,1],[1,1],[1,0]]
- Output
- 4
- Why
- Four 1x1 squares at positions with value 1. No 2x2 all-ones squares exist.
Example 3
- Input
- matrix = [[0,0],[0,0]]
- Output
- 0
- Why
- No squares of ones exist.
Constraints
- 1 <= matrix.length, matrix[i].length <= 300
- matrix[i][j] is 0 or 1
How to think about it
Updated 2026-09-29The side length of the largest all-ones square whose bottom-right corner rests at (i, j) is exactly equal to the number of square submatrices anchored at that corner. Because every larger square contains smaller squares inside it, summing dp[i][j] across all cells gives the total count.
Approaches, worst first
Exhaustive corner expansion
time O(m * n * min(m, n)) · space O(1)
For every cell with value 1, test possible side lengths k from 1 upward, checking if all cells in the k x k square are 1. Inefficient due to repetitive cell scans.
Corner-anchored dynamic programming
time O(m * n) · space O(m * n)
Let dp[i][j] be the side length of the largest square ending at (i, j). If matrix[i][j] == 1, dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) for i, j > 0. Accumulate all dp[i][j] values.
1D rolling array optimizationWrite this one
time O(m * n) · space O(n)
Compress the DP table to a single row array plus a variable storing the top-left diagonal cell. Reduces space to linear in column count while accumulating the running total.
Where people lose marks · 3
- First row and first column handling: cells with matrix[i][j] == 1 on row 0 or col 0 can only ever form squares of size 1 (dp[i][j] = 1).
- Confusing max square area with total square count: the task asks for the sum of all valid square sizes, not the single maximum square.
- Overwriting diagonal predecessor state: during 1D row compression, the diagonal value dp[i-1][j-1] must be saved in a temporary variable before dp[j] is overwritten.
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.
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