Count of square submatrices with average at least K

• Difficulty Level : Easy
• Last Updated : 26 Jul, 2021

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++

 // C++ program for the above approach #include using namespace std;   #define MAX 1000   // Function to count submatrixes with // average greater than or equals to K int cntMatrices(vector > arr, int N, int M,                 int K) {       // Stores count of submatrices     int cnt = 0;       // Stores the prefix sum of matrix     vector > pre(N + 1, vector(M + 1, 0));       // Iterate over the range [1, N]     for (int i = 1; i <= N; i++) {           // Iterate over the range         // [1, M]         for (int j = 1; j <= M; j++) {               // Update the prefix sum             pre[i][j] = arr[i - 1][j - 1] + pre[i - 1][j]                         + pre[i][j - 1] - pre[i - 1][j - 1];         }     }       // Iterate over the range [1, N]     for (int i = 1; i <= N; i++) {           // Iterate over the range         // [1, M]         for (int j = 1; j <= M; j++) {               // Iterate until l and r             // are greater than 0             for (int l = i, r = j; l > 0 && r > 0;                  l--, r--) {                   // Update count                 int sum1 = (K * (i - l + 1) * (i - r + 1));                   // Stores sum of submatrix                 // with bottom right corner                 // as (i, j) and top left                 // corner as (l, r)                 int sum2 = pre[i][j] - pre[l - 1][r]                            - pre[l][r - 1]                            + pre[l - 1][r - 1];                   // If sum1 is less than or                 // equal to sum2                 if (sum1 <= sum2)                       // Increment cnt by 1                     cnt++;             }         }     }       // Return cnt as the answer     return cnt; }   // Driver Code int main() {     // Given Input     vector > arr         = { { 2, 2, 3 }, { 3, 4, 5 }, { 4, 5, 5 } };     int K = 4;     int N = arr.size();     int M = arr.size();       // Function Call     cout << cntMatrices(arr, N, M, K);       return 0; }

Java

 // Java program for the above approach import java.util.*;   class GFG{       static int MAX = 1000;   // Function to count submatrixes with // average greater than or equals to K static int cntMatrices(int[][] arr, int N,                     int M, int K) {           // Stores count of submatrices     int cnt = 0;       // Stores the prefix sum of matrix     int[][] pre = new int[N + 1][M + 1];       // Iterate over the range [1, N]     for(int i = 1; i <= N; i++)     {                   // Iterate over the range         // [1, M]         for(int j = 1; j <= M; j++)         {                           // Update the prefix sum             pre[i][j] = arr[i - 1][j - 1] + pre[i - 1][j] +                             pre[i][j - 1] - pre[i - 1][j - 1];         }     }       // Iterate over the range [1, N]     for(int i = 1; i <= N; i++)     {                   // Iterate over the range         // [1, M]         for(int j = 1; j <= M; j++)         {                           // Iterate until l and r             // are greater than 0             for(int l = i, r = j;                     l > 0 && r > 0; l--, r--)             {                                   // Update count                 int sum1 = (K * (i - l + 1) *                                 (i - r + 1));                   // Stores sum of submatrix                 // with bottom right corner                 // as (i, j) and top left                 // corner as (l, r)                 int sum2 = pre[i][j] - pre[l - 1][r] -                        pre[l][r - 1] + pre[l - 1][r - 1];                   // If sum1 is less than or                 // equal to sum2                 if (sum1 <= sum2)                       // Increment cnt by 1                     cnt++;             }         }     }       // Return cnt as the answer     return cnt; }   // Driver Code public static void main(String args[]) {           // Given Input     int[][] arr = { { 2, 2, 3 },                     { 3, 4, 5 },                     { 4, 5, 5 } };     int K = 4;     int N = arr.length;     int M = arr.length;       // Function Call     System.out.println( cntMatrices(arr, N, M, K)); } }   // This code is contributed by avijitmondal1998

Python3

 # Python3 program for the above approach # define MAX 1000   # Function to count submatrixes with # average greater than or equals to K def cntMatrices(arr, N, M, K):       # Stores count of submatrices     cnt = 0       # Stores the prefix sum of matrix     pre = [[0 for i in range(M + 1)]               for i in range(N + 1)]                     # Iterate over the range [1, N]     for i in range(1, N + 1):                   # Iterate over the range         # [1, M]         for j in range(1, M + 1):                           # Update the prefix sum             pre[i][j] = (arr[i - 1][j - 1] +                          pre[i - 1][j] +                          pre[i][j - 1] -                          pre[i - 1][j - 1])       # Iterate over the range [1, N]     for i in range(1, N + 1):                   # Iterate over the range         # [1, M]         for j in range(1, M + 1):                           # Iterate until l and r             # are greater than 0             l, r = i, j             while l > 0 and r > 0:                                   # Update count                 sum1 = (K * (i - l + 1) * (i - r + 1))                   # Stores sum of submatrix                 # with bottom right corner                 # as (i, j) and top left                 # corner as (l, r)                 sum2 = (pre[i][j] -                         pre[l - 1][r] - pre[l][r - 1] +                         pre[l - 1][r - 1])                   # If sum1 is less than or                 # equal to sum2                 if (sum1 <= sum2):                                           # Increment cnt by 1                     cnt += 1                                       l -= 1                 r -= 1       # Return cnt as the answer     return cnt   # Driver Code if __name__ == '__main__':           # Given Input     arr = [ [ 2, 2, 3 ],             [ 3, 4, 5 ],             [ 4, 5, 5 ] ]     K = 4     N = len(arr)     M = len(arr)       # Function Call     print(cntMatrices(arr, N, M, K))   # This code is contributed by mohit kumar 29

C#

 // C# program for the above approach using System; class GFG {     static int MAX = 1000;     // Function to count submatrixes with   // average greater than or equals to K   static int cntMatrices(int[,] arr, int N,                          int M, int K)   {       // Stores count of submatrices     int cnt = 0;       // Stores the prefix sum of matrix     int[,] pre = new int[N + 1, M + 1];       // Iterate over the range [1, N]     for(int i = 1; i <= N; i++)     {         // Iterate over the range       // [1, M]       for(int j = 1; j <= M; j++)       {           // Update the prefix sum         pre[i, j] = arr[i - 1, j - 1] + pre[i - 1, j] +           pre[i, j - 1] - pre[i - 1, j - 1];       }     }       // Iterate over the range [1, N]     for(int i = 1; i <= N; i++)     {         // Iterate over the range       // [1, M]       for(int j = 1; j <= M; j++)       {           // Iterate until l and r         // are greater than 0         for(int l = i, r = j;             l > 0 && r > 0; l--, r--)         {             // Update count           int sum1 = (K * (i - l + 1) *                       (i - r + 1));             // Stores sum of submatrix           // with bottom right corner           // as (i, j) and top left           // corner as (l, r)           int sum2 = pre[i, j] - pre[l - 1, r] -             pre[l, r - 1] + pre[l - 1, r - 1];             // If sum1 is less than or           // equal to sum2           if (sum1 <= sum2)               // Increment cnt by 1             cnt++;         }       }     }       // Return cnt as the answer     return cnt;   }     // Driver code   public static void Main(string[] args)   {       // Given Input     int[,] arr = { { 2, 2, 3 },                   { 3, 4, 5 },                   { 4, 5, 5 } };     int K = 4;     int N = arr.GetLength(0);     int M = arr.GetLength(0);       // Function Call     Console.WriteLine( cntMatrices(arr, N, M, K));   } }   // This code is contributed by sanjoy_62.

Javascript



Output

7

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

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