How do you solve Count Square Submatrices with All Ones on LeetCode?

Count Square Submatrices with All Ones (LeetCode #1277) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Dynamic programming allows us to build solutions incrementally, reducing redundant checks.

What is the time complexity of Count Square Submatrices with All Ones?

The optimal solution for Count Square Submatrices with All Ones runs in O(m * n) time and uses O(m * n) space. The complexity is O(m * n) because we iterate through each cell once and perform constant time operations for each cell.

What is the brute force solution for Count Square Submatrices with All Ones?

The brute force approach for Count Square Submatrices with All Ones is Brute Force, with O(n^3) time complexity and O(1) space complexity. The brute-force approach checks every possible square submatrix in the given matrix. For each possible starting point, it verifies if all elements in the square are ones, counting them if they are. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Count Square Submatrices with All Ones?

Count Square Submatrices with All Ones uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Count Square Submatrices with All Ones?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process while coding to demonstrate your understanding. Practice writing clean and efficient code, as readability matters in interviews.

What are common mistakes when solving Count Square Submatrices with All Ones?

Not considering edge cases like empty matrices or matrices filled with zeros. Failing to correctly update the DP table based on previous values.

#1277

Count Square Submatrices with All Ones

Medium

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^3)	O(m * n)
Space	O(1)	O(m * n)

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Intuition

Time O(m * n)Space O(m * n)

The optimal approach uses dynamic programming to build a table that keeps track of the size of the largest square submatrix ending at each cell. This allows us to calculate the total count efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table of the same size as the input matrix initialized to zero.
2Step 2: Iterate through each cell in the matrix. If the cell is 1, update the DP table based on the minimum of the three neighboring cells (top, left, top-left diagonal).
3Step 3: Sum the values in the DP table to get the total count of square submatrices.

solution.py17 lines

1# Full working Python code
2
3def countSquares(matrix):
4    if not matrix:
5        return 0
6    count = 0
7    m, n = len(matrix), len(matrix[0])
8    dp = [[0] * n for _ in range(m)]
9    for i in range(m):
10        for j in range(n):
11            if matrix[i][j] == 1:
12                dp[i][j] = 1
13                if i > 0 and j > 0:
14                    dp[i][j] += min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
15                count += dp[i][j]
16    return count
17

ℹ

Complexity note: The complexity is O(m * n) because we iterate through each cell once and perform constant time operations for each cell.

1Dynamic programming allows us to build solutions incrementally, reducing redundant checks.
2Understanding how to relate subproblems is crucial for optimizing solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.