Count of square submatrices with average at least K

Given a matrix arr[][] of size NxM and an integer K, the task is to find the count of square submatrices in the given matrix with the average of elements greater than or equal to K.

Examples:

Input: K = 4, arr[][] = {{2, 2, 3}, {3, 4, 5}, {4, 5, 5}}
Output: 7
Explanation: 
The following square submatrices have an average greater than or equal to K:

1. Square submatrices of dimension (1×1), formed by taking the elements at positions {(2, 2)}. The average of the submatrix is equal to 4.
2. Square submatrices of dimension (1×1), formed by taking the elements at positions {(2, 3)}. The average of the submatrix is equal to 5.
3. Square submatrices of dimension (1×1), formed by taking the elements at positions {(3, 1)}. The average of the submatrix is equal to 4.
4. Square submatrices of dimension (1×1), formed by taking the elements at positions {(3, 2)}. The average of the submatrix is equal to 5.
5. Square submatrices of dimension (1×1), formed by taking the elements at positions {(3, 3)}. The average of the submatrix is equal to 5.
6. Square submatrices of dimension (2×2), formed by taking the elements at positions {(2, 1), (2, 2), (3, 1), (3, 2)}. The average of the submatrix is equal to (3+4+4+5 = 16)/4 = 4.
7. Square submatrices of dimension (2×2), formed by taking the elements at positions {(2, 2), (2, 3), (3, 2), (3, 3)}. The average of the submatrix is equal to (4+5+5+5 = 19)/4 = 4.75.

Therefore, there are totals of 7 square submatrices with an average greater than or equals to K.

Input: K = 3, arr[][] = {{1, 1, 1}, {1, 1, 1}}
Output: 0

Naive Approach: The simplest approach is to generate all possible square submatrices and check sum of all the elements of the sub-square is greater than or equals to K multiplied by the size of the submatrix.

Time Complexity: O(N³ * M³) 
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using the prefix sum matrix which results in constant time calculation of the sum of a submatrix. Follow the steps below to solve the problem:

• Initialize a variable, say count as 0 to store the count of submatrices with an average greater than or equal to K.
• Calculate the prefix sum of the matrix arr[][] and store it in a vector of vectors say pre[][].
• Traverse over every element of the matrix using the variables i and j and perform the following steps:
  • Initialize two variables, say l as i and r as j.
  • Iterate until l and r are greater than 0 and in each iteration perform the following steps:
    • Calculate the sum of the square submatrix with the bottom right vertex as (i, j) and the top left vertex as (l, r) and store it in a variable, say sum, i.e sum = pre[i][j] - pre[l-1][r] - pre[l][r-1] + pre[l-1][r-1].
      • Now if the value of K*(i-l+1)*(j-r+1) is equal to the sum, then increment the count by 1.
      • Decrement l and r by 1.
• Finally, after completing the above steps, print the value of the count as the result.

Below is the implementation of the above approach:

c

7

Time Complexity: O(M * N * (min(N, M))
Auxiliary Space: O(M * N)