Solution Explanation: Counting Submatrices with All Ones

This problem asks to count the number of submatrices within a given binary matrix that contain only ones. The optimal solution leverages a combination of prefix sum calculation and iterative counting.

Approach

1. Prefix Sum Calculation: We first preprocess the input matrix mat to create a new matrix g. g[i][j] stores the width of the largest submatrix containing only ones that ends at position (i, j). This is essentially a horizontal prefix sum. If mat[i][j] is 0, g[i][j] is 0; otherwise, it's 1 plus the value to its left (g[i][j-1]).

2. Iterative Counting: We iterate through each cell (i, j) in the matrix g. For each cell, we consider it as the bottom-right corner of potential submatrices. We iterate upwards from row i to row 0. For each row k, we find the minimum width among all the cells in column j from row k to row i. This minimum width represents the width of the submatrix with (i,j) as its bottom-right corner and spanning from row k to i. We add this width to the final count (ans).

Time and Space Complexity Analysis

• Time Complexity: The preprocessing step (creating g) takes O(mn) time, where 'm' is the number of rows and 'n' is the number of columns. The nested loops for counting submatrices also take O(mn*m) time in the worst case (innermost loop iterates up to 'm' rows). Therefore, the overall time complexity is O(m²n).

• Space Complexity: We use an auxiliary matrix g of size mn, resulting in a space complexity of **O(mn)**.

Code Implementation (Python)

class Solution:
    def numSubmat(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        g = [[0] * n for _ in range(m)] #Initialize the prefix sum matrix
        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] #Calculate prefix sum
 
        ans = 0
        for i in range(m):
            for j in range(n):
                col = float('inf') #Initialize minimum column width
                for k in range(i, -1, -1):
                    col = min(col, g[k][j]) #Find minimum width
                    ans += col #Add to count
        return ans
 

The implementations in other languages (Java, C++, Go, TypeScript) follow the same logic, only differing in syntax and data structure handling. They all achieve the same time and space complexity as the Python solution.