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

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

We can solve this problem in O(n2) time using aTricky 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.

// 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 j=0; j<k; k++)

          sum += mat[k][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;

   }

}

-- O(K^2*N^2)

// 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;

   }

}

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