How do you solve Count Submatrices With All Ones on LeetCode?

Count Submatrices With All Ones (LeetCode #1504) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n²) time complexity and O(n) space complexity. Key insight: Understanding how to represent the problem in terms of heights of consecutive ones is crucial.

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

The brute force approach for Count Submatrices With All Ones is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible submatrix in the given binary matrix. For each submatrix, we verify if all elements are 1, which can be slow but is straightforward to understand. The optimal Optimal Solution approach improves this to O(m * n²).

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

Count Submatrices With All Ones uses the following concepts: Array, Dynamic Programming, Stack, Matrix, Monotonic Stack. The recommended patterns to study are: Dynamic Programming, Monotonic Stack.

What are the interview tips for Count Submatrices With All Ones?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches before diving into coding. Practice dry runs of your code to ensure correctness.

What are common mistakes when solving Count Submatrices With All Ones?

Not considering edge cases such as empty matrices. Failing to reset the height array correctly for each row.

#1504

Count Submatrices With All Ones

Medium

ArrayDynamic ProgrammingStackMatrixMonotonic StackDynamic ProgrammingMonotonic Stack

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n²)
Space	O(1)	O(n)

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Intuition

Time O(m * n²)Space O(n)

The optimal solution leverages dynamic programming to count the number of submatrices with all ones by calculating the height of consecutive ones for each column and using that to find rectangles efficiently.

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Algorithm

3 steps

1Step 1: Create a height array to store the number of consecutive 1s for each column.
2Step 2: For each row, update the height array based on the current row's values.
3Step 3: For each column in the current row, calculate the number of rectangles that can be formed using the height array.

solution.py17 lines

1def countSubmatrices(mat):
2    if not mat:
3        return 0
4    m, n = len(mat), len(mat[0])
5    height = [0] * n
6    count = 0
7    for r in range(m):
8        for c in range(n):
9            height[c] = height[c] + 1 if mat[r][c] == 1 else 0
10        for c in range(n):
11            minHeight = height[c]
12            for k in range(c, n):
13                if height[k] == 0:
14                    break
15                minHeight = min(minHeight, height[k])
16                count += minHeight
17    return count

ℹ

Complexity note: The time complexity is O(m * n²) because we iterate through each row and for each column in that row, we check the heights of columns to count rectangles, which can take linear time for each column.

1Understanding how to represent the problem in terms of heights of consecutive ones is crucial.
2Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.

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