Leetcode #1504: Count Submatrices With All Ones
In this guide, we solve Leetcode #1504 Count Submatrices With All Ones in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an m x n binary matrix mat, return the number of submatrices that have all ones. Example 1: Input: mat = [[1,0,1],[1,1,0],[1,1,0]] Output: 13 Explanation: There are 6 rectangles of side 1x1.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Array, Dynamic Programming, Matrix, Monotonic Stack
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: mat = [[1,0,1],[1,1,0],[1,1,0]]
Output: 13
Explanation:
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2.
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.
Python Solution
class Solution:
def numSubmat(self, mat: List[List[int]]) -> int:
m, n = len(mat), len(mat[0])
g = [[0] * n for _ in range(m)]
for i in range(m):
for j in range(n):
if mat[i][j]:
g[i][j] = 1 if j == 0 else 1 + g[i][j - 1]
ans = 0
for i in range(m):
for j in range(n):
col = inf
for k in range(i, -1, -1):
col = min(col, g[k][j])
ans += col
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.