Suppose we have m x n binary matrix, we have to find how many submatrices have all ones.
So, if the input is like
| 1 | 0 | 1 |
| 0 | 1 | 1 |
| 0 | 1 | 1 |
then the output will be 13 as there are 6 (1x1) matrices, 3 (2,1) matrices, 2 (1x2) matrices, 1 (3x1) matrix and 1 (4x4) matrix.
To solve this, we will follow these steps −
m := row count of matrix
n := column count of matrix
dp := a zero matrix of same size m x n
for i in range 0 to m - 1, do
for j in range 0 to n - 1, do
if i is same as 0 and matrix[i, j], then
dp[i, j] := 1
otherwise when matrix[i][j] is non-zero, then
dp[i, j] := dp[i-1, j] + 1
total := 0
for i in range 0 to m - 1, do
for j in range 0 to n - 1, do
for k in range j+1 to n, do
total := total + minimum of dp[i,j] to dp[i,k]
return total
Let us see the following implementation to get better understanding −
Example
def solve(matrix): m = len(matrix) n = len(matrix[0]) dp = [[0] * n for _ in range(m)] for i in range(m): for j in range(n): if i == 0 and matrix[i][j]: dp[i][j] = 1 elif matrix[i][j]: dp[i][j] = dp[i-1][j] + 1 total = 0 for i in range(m): for j in range(n): for k in range(j+1, n+1): total += min(dp[i][j:k]) return total matrix = [[1,0,1],[0,1,1],[0,1,1]] print(solve(matrix))
Input
[4,6,7,8], 11
Output
13