Smallest element from all square submatrices of size K from a given Matrix

Given a matrix arr[][] and an integer K, the task is to find the smallest element from all possible square submatrices of size K from the given matrix.

Examples:

Input: K = 2, arr[][] ={ {1, 2, 3}, {4, 5, 6},  {7, 8, 9} }
Output: 
1 2 
4 5
Explanation:
Smallest elements from all possible square submatrices of size 2 are as follows:
{ {1, 2}, {4, 5} } -> 1
{ {2, 3}, {5, 6} } -> 2
{ {4, 5}, {7, 8} } -> 4
{ {5, 6}, {8, 9} } -> 5 

Input: K = 3,  
arr[][] = { {-1, 5, 4, 1, -3}, 
{4, 3, 1, 1, 6}, 
{2, -2, 5, 3, 1}, 
{8, 5, 1, 9, -4}, 
{12, 3, 5, 8, 1} }
Output: 
-2 -2 -3
-2 -2 -4
-2 -2 -4

Naive Approach: The simplest approach to solve the problem is to generate all possible square submatrices of size K from the given matrix and print the smallest element from each such submatrices.
Time Complexity: O(N * M * K2)
Auxiliary Space: O(1)

Efficient Approach: Follow the steps below to optimize the above approach:



Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program for
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to returns a smallest
// elements of all KxK submatrices
// of a given NxM matrix
vector<vector<int> > matrixMinimum(
       vector<vector<int> > nums, int K)
{
  // Stores the dimensions
  // of the given matrix
  int N = nums.size();
  int M = nums[0].size();
 
  // Stores the required
  // smallest elements
  vector<vector<int> > res(N - K + 1,
                           vector<int>(M - K + 1));
 
  // Update the smallest elements row-wise
  for (int i = 0; i < N; i++)
  {
    for (int j = 0; j < M - K + 1; j++)
    {
      int mini = INT_MAX;
      for (int k = j; k < j + K; k++)
      {
        mini = min(mini, nums[i][k]);
      }
      nums[i][j] = mini;
    }
  }
 
  // Update the minimum column-wise
  for (int j = 0; j < M; j++)
  {
    for (int i = 0; i < N - K + 1; i++)
    {
      int mini = INT_MAX;
      for (int k = i; k < i + K; k++)
      {
        mini = min(mini, nums[k][j]);
      }
      nums[i][j] = mini;
    }
  }
 
  // Store the final submatrix with
  // required minimum values
  for (int i = 0; i < N - K + 1; i++)
    for (int j = 0; j < M - K + 1; j++)
      res[i][j] = nums[i][j];
 
  // Return the resultant matrix
  return res;
}
 
void smallestinKsubmatrices(vector<vector<int> > arr,
                            int K)
{
  // Function Call
  vector<vector<int> > res = matrixMinimum(arr, K);
 
  // Print resultant matrix with the
  // minimum values of KxK sub-matrix
  for (int i = 0; i < res.size(); i++)
  {
    for (int j = 0; j < res[0].size(); j++)
    {
      cout << res[i][j] << " ";
    }
    cout << endl;
  }
}
 
// Driver Code
int main()
{
 
  // Given matrix
  vector<vector<int> > arr = {{-1, 5, 4, 1, -3},
                              {4, 3, 1, 1, 6},
                              {2, -2, 5, 3, 1},
                              {8, 5, 1, 9, -4},
                              {12, 3, 5, 8, 1}};
 
  // Given K
  int K = 3;
 
  smallestinKsubmatrices(arr, K);
}
 
// This code is contributed by Chitranayal
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program for the above approach
 
import java.util.*;
import java.lang.*;
 
class GFG {
 
    // Function to returns a smallest
    // elements of all KxK submatrices
    // of a given NxM matrix
    public static int[][] matrixMinimum(
        int[][] nums, int K)
    {
        // Stores the dimensions
        // of the given matrix
        int N = nums.length;
        int M = nums[0].length;
 
        // Stores the required
        // smallest elements
        int[][] res
            = new int[N - K + 1][M - K + 1];
 
        // Update the smallest elements row-wise
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < M - K + 1; j++) {
 
                int min = Integer.MAX_VALUE;
                for (int k = j; k < j + K; k++) {
                    min = Math.min(min, nums[i][k]);
                }
                nums[i][j] = min;
            }
        }
 
        // Update the minimum column-wise
        for (int j = 0; j < M; j++) {
            for (int i = 0; i < N - K + 1; i++) {
                int min = Integer.MAX_VALUE;
                for (int k = i; k < i + K; k++) {
                    min = Math.min(min, nums[k][j]);
                }
                nums[i][j] = min;
            }
        }
 
        // Store the final submatrix with
        // required minimum values
        for (int i = 0; i < N - K + 1; i++)
            for (int j = 0; j < M - K + 1; j++)
                res[i][j] = nums[i][j];
 
        // Return the resultant matrix
        return res;
    }
 
    public static void smallestinKsubmatrices(
        int arr[][], int K)
    {
        // Function Call
        int[][] res = matrixMinimum(arr, K);
 
        // Print resultant matrix with the
        // minimum values of KxK sub-matrix
        for (int i = 0; i < res.length; i++) {
            for (int j = 0; j < res[0].length; j++) {
                System.out.print(res[i][j] + " ");
            }
            System.out.println();
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        // Given matrix
        int[][] arr = { { -1, 5, 4, 1, -3 },
                        { 4, 3, 1, 1, 6 },
                        { 2, -2, 5, 3, 1 },
                        { 8, 5, 1, 9, -4 },
                        { 12, 3, 5, 8, 1 } };
 
        // Given K
        int K = 3;
 
        smallestinKsubmatrices(arr, K);
    }
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
import sys
 
# Function to returns a smallest
# elements of all KxK submatrices
# of a given NxM matrix
def matrixMinimum(nums, K):
 
    # Stores the dimensions
    # of the given matrix
    N = len(nums)
    M = len(nums[0])
 
    # Stores the required
    # smallest elements
    res = [[0 for x in range(M - K + 1)]
              for y in range(N - K + 1)]
 
    # Update the smallest elements row-wise
    for i in range(N):
        for j in range(M - K + 1):
            mn = sys.maxsize
             
            for k in range(j, j + K):
                mn = min(mn, nums[i][k])
 
            nums[i][j] = mn
 
    # Update the minimum column-wise
    for j in range(M):
        for i in range(N - K + 1):
            mn = sys.maxsize
             
            for k in range(i, i + K):
                mn = min(mn, nums[k][j])
 
            nums[i][j] = mn
 
    # Store the final submatrix with
    # required minimum values
    for i in range(N - K + 1):
        for j in range(M - K + 1):
            res[i][j] = nums[i][j]
 
    # Return the resultant matrix
    return res
 
def smallestinKsubmatrices(arr, K):
 
    # Function call
    res = matrixMinimum(arr, K)
 
    # Print resultant matrix with the
    # minimum values of KxK sub-matrix
    for i in range(len(res)):
        for j in range(len(res[0])):
            print(res[i][j], end = " ")
             
        print()
 
# Driver Code
 
# Given matrix
arr = [ [ -1, 5, 4, 1, -3 ],
        [ 4, 3, 1, 1, 6 ],
        [ 2, -2, 5, 3, 1 ],
        [ 8, 5, 1, 9, -4 ],
        [ 12, 3, 5, 8, 1 ] ]
 
# Given K
K = 3
 
# Function call
smallestinKsubmatrices(arr, K)
 
# This code is contributed by Shivam Singh
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
 
class GFG{
 
// Function to returns a smallest
// elements of all KxK submatrices
// of a given NxM matrix
public static int[,] matrixMinimum(int[,] nums,
                                   int K)
{
     
    // Stores the dimensions
    // of the given matrix
    int N = nums.GetLength(0);
    int M = nums.GetLength(1);
 
    // Stores the required
    // smallest elements
    int[,] res = new int[N - K + 1,
                         M - K + 1];
 
    // Update the smallest elements row-wise
    for(int i = 0; i < N; i++)
    {
        for(int j = 0; j < M - K + 1; j++)
        {
            int min = int.MaxValue;
            for(int k = j; k < j + K; k++)
            {
                min = Math.Min(min, nums[i, k]);
            }
            nums[i, j] = min;
        }
    }
 
    // Update the minimum column-wise
    for(int j = 0; j < M; j++)
    {
        for(int i = 0; i < N - K + 1; i++)
        {
            int min = int.MaxValue;
            for(int k = i; k < i + K; k++)
            {
                min = Math.Min(min, nums[k, j]);
            }
            nums[i, j] = min;
        }
    }
 
    // Store the readonly submatrix with
    // required minimum values
    for(int i = 0; i < N - K + 1; i++)
        for(int j = 0; j < M - K + 1; j++)
            res[i, j] = nums[i, j];
 
    // Return the resultant matrix
    return res;
}
 
public static void smallestinKsubmatrices(int [,]arr,
                                          int K)
{
     
    // Function call
    int[,] res = matrixMinimum(arr, K);
 
    // Print resultant matrix with the
    // minimum values of KxK sub-matrix
    for(int i = 0; i < res.GetLength(0); i++)
    {
        for(int j = 0; j < res.GetLength(1); j++)
        {
            Console.Write(res[i, j] + " ");
        }
        Console.WriteLine();
    }
}
 
// Driver Code
public static void Main(String[] args)
{
 
    // Given matrix
    int[,] arr = { { -1, 5, 4, 1, -3 },
                   { 4, 3, 1, 1, 6 },
                   { 2, -2, 5, 3, 1 },
                   { 8, 5, 1, 9, -4 },
                   { 12, 3, 5, 8, 1 } };
 
    // Given K
    int K = 3;
 
    smallestinKsubmatrices(arr, K);
}
}
 
// This code is contributed by Amit Katiyar
chevron_right

Output: 
-2 -2 -3 
-2 -2 -4 
-2 -2 -4



Time Complexity: O( max(N, M)3 )
Auxiliary Space: O( (N-K+1)*(M-K+1) ) 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.





Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :