Given an n x n square matrix, find sum of all sub-squares of size k x k

Given an n x n square matrix, find sum of all sub-squares of size k x k where k is smaller than or equal to n.

Examples

Input:
n = 5, k = 3
arr[][] = { {1, 1, 1, 1, 1},
            {2, 2, 2, 2, 2},
            {3, 3, 3, 3, 3},
            {4, 4, 4, 4, 4},
            {5, 5, 5, 5, 5},
         };
Output:
       18  18  18
       27  27  27
       36  36  36


Input:
n = 3, k = 2
arr[][] = { {1, 2, 3},
            {4, 5, 6},
            {7, 8, 9},
         };
Output:
       12  16
       24  28

A Simple Solution is to one by one pick starting point (leftmost-topmost corner) of all possible sub-squares. Once the starting point is picked, calculate sum of sub-square starting with the picked starting point.

Following is C++ implementation of this idea.

// A simple C++ program to find sum of all subsquares of size k x k
#include <iostream>
using namespace std;

// Size of given matrix
#define n 5

// A simple function to find sum of all sub-squares of size k x k
// in a given square matrix of size n x n
void printSumSimple(int mat[][n], int k)
{
   // k must be smaller than or equal to n
   if (k > n) return;

   // row number of first cell in current sub-square of size k x k
   for (int i=0; i<n-k+1; i++)
   {
      // column of first cell in current sub-square of size k x k
      for (int j=0; j<n-k+1; j++)
      {
          // Calculate and print sum of current sub-square
          int sum = 0;
          for (int p=i; p<k+i; p++)
             for (int q=j; q<k+j; q++)
                 sum += mat[p][q];
           cout << sum << "  ";
      }

      // Line separator for sub-squares starting with next row
      cout << endl;
   }
}

// Driver program to test above function
int main()
{
    int mat[n][n] = {{1, 1, 1, 1, 1},
                     {2, 2, 2, 2, 2},
                     {3, 3, 3, 3, 3},
                     {4, 4, 4, 4, 4},
                     {5, 5, 5, 5, 5},
                    };
    int k = 3;
    printSumSimple(mat, k);
    return 0;
}

Output:

  18  18  18
  27  27  27
  36  36  36

Time complexity of above solution is O(k2n2). We can solve this problem in O(n2) time using a Tricky Solution. The idea is to preprocess the given square matrix. In the preprocessing step, calculate sum of all vertical strips of size k x 1 in a temporary square matrix stripSum[][]. Once we have sum of all vertical strips, we can calculate sum of first sub-square in a row as sum of first k strips in that row, and for remaining sub-squares, we can calculate sum in O(1) time by removing the leftmost strip of previous subsquare and adding the rightmost strip of new square.

Following is C++ implementation of this idea.

// An efficient C++ program to find sum of all subsquares of size k x k
#include <iostream>
using namespace std;

// Size of given matrix
#define n 5

// A O(n^2) function to find sum of all sub-squares of size k x k
// in a given square matrix of size n x n
void printSumTricky(int mat[][n], int k)
{
   // k must be smaller than or equal to n
   if (k > n) return;

   // 1: PREPROCESSING
   // To store sums of all strips of size k x 1
   int stripSum[n][n];

   // Go column by column
   for (int j=0; j<n; j++)
   {
       // Calculate sum of first k x 1 rectangle in this column
       int sum = 0;
       for (int i=0; i<k; i++)
          sum += mat[i][j];
       stripSum[0][j] = sum;

       // Calculate sum of remaining rectangles
       for (int i=1; i<n-k+1; i++)
       {
            sum += (mat[i+k-1][j] - mat[i-1][j]);
            stripSum[i][j] = sum;
       }
   }

   // 2: CALCULATE SUM of Sub-Squares using stripSum[][]
   for (int i=0; i<n-k+1; i++)
   {
      // Calculate and print sum of first subsquare in this row
      int sum = 0;
      for (int j = 0; j<k; j++)
           sum += stripSum[i][j];
      cout << sum << "  ";

      // Calculate sum of remaining squares in current row by
      // removing the leftmost strip of previous sub-square and
      // adding a new strip
      for (int j=1; j<n-k+1; j++)
      {
         sum += (stripSum[i][j+k-1] - stripSum[i][j-1]);
         cout << sum << "  ";
      }

      cout << endl;
   }
}

// Driver program to test above function
int main()
{
    int mat[n][n] = {{1, 1, 1, 1, 1},
                     {2, 2, 2, 2, 2},
                     {3, 3, 3, 3, 3},
                     {4, 4, 4, 4, 4},
                     {5, 5, 5, 5, 5},
                    };
    int k = 3;
    printSumTricky(mat, k);
    return 0;
}

Output:

  18  18  18
  27  27  27
  36  36  36

This article is contributed by Rahul Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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