Data Modelling & AI Data Structure & Algorithm

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

23 June 2025

0

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.

C++

#include <iostream>
#include <vector>
#include <climits> // for INT_MAX
using namespace std;
 
// Function to return the smallest elements of all KxK submatrices of a given NxM matrix
void print_smallest_elements(const vector<vector<int>> &arr, int K) {
    // Get the number of rows and columns in the matrix
    int rows = arr.size();
    int cols = arr[0].size();
     
    // Loop through the matrix, creating submatrices of size KxK
    for (int row = 0; row <= rows - K; row++) {
        for (int col = 0; col <= cols - K; col++) {
            // Create a submatrix by extracting elements from the original matrix
            vector<vector<int>> subarr;
            for (int i = row; i < row + K; i++) {
                vector<int> subcol;
                for (int j = col; j < col + K; j++) {
                    subcol.push_back(arr[i][j]);
                }
                subarr.push_back(subcol);
            }
             
            // Find the minimum value in the submatrix
            int minimum = INT_MAX;
            for (const auto &x : subarr) {
                for (const auto &y : x) {
                    minimum = min(minimum, y);
                }
            }
             
            // Print the minimum value of the submatrix
            cout << minimum << " ";
        }
        cout << endl;
    }
}
 
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}};
    int K = 3;
     
    // Call the function to print the smallest elements of all KxK submatrices
    print_smallest_elements(arr, K);
     
    return 0;
}

Java

import java.util.ArrayList;
 
public class Main { // Function to return the smallest
                    // elements of all KxK submatrices of a
                    // given NxM matrix
    static void printSmallestElements(
        ArrayList<ArrayList<Integer> > arr, int K)
    {
        // Get the number of rows and columns in the matrix
        int rows = arr.size();
        int cols = arr.get(0).size();
 
        // Loop through the matrix, creating submatrices of
        // size KxK
        for (int row = 0; row <= rows - K; row++) {
            for (int col = 0; col <= cols - K; col++) {
                // Create a submatrix by extracting elements
                // from the original matrix
                ArrayList<ArrayList<Integer> > subarr
                    = new ArrayList<ArrayList<Integer> >();
                for (int i = row; i < row + K; i++) {
                    ArrayList<Integer> subcol
                        = new ArrayList<Integer>();
                    for (int j = col; j < col + K; j++) {
                        subcol.add(arr.get(i).get(j));
                    }
                    subarr.add(subcol);
                }
 
                // Find the minimum value in the submatrix
                int minimum = Integer.MAX_VALUE;
                for (ArrayList<Integer> x : subarr) {
                    for (int y : x) {
                        minimum = Math.min(minimum, y);
                    }
                }
 
                // Print the minimum value of the submatrix
                System.out.print(minimum + " ");
            }
            System.out.println();
        }
    }
 
    public static void main(String[] args)
    {
        // Given matrix
        ArrayList<ArrayList<Integer> > arr
            = new ArrayList<ArrayList<Integer> >();
        arr.add(new ArrayList<Integer>() {
            {
                add(-1);
                add(5);
                add(4);
                add(1);
                add(-3);
            }
        });
        arr.add(new ArrayList<Integer>() {
            {
                add(4);
                add(3);
                add(1);
                add(1);
                add(6);
            }
        });
        arr.add(new ArrayList<Integer>() {
            {
                add(2);
                add(-2);
                add(5);
                add(3);
                add(1);
            }
        });
        arr.add(new ArrayList<Integer>() {
            {
                add(8);
                add(5);
                add(1);
                add(9);
                add(-4);
            }
        });
        arr.add(new ArrayList<Integer>() {
            {
                add(12);
                add(3);
                add(5);
                add(8);
                add(1);
            }
        });
        int K = 3;
 
        // Call the function to print the smallest elements
        // of all KxK submatrices
        printSmallestElements(arr, K);
    }
}

Python3

# Python3 program for the above approach
 
# Function to returns a smallest
# elements of all KxK submatrices
# of a given NxM matrix
 
 
def print_smallest_elements(arr, K):
    rows = len(arr)
    cols = len(arr[0])
    for row in range(rows-K+1):
        for col in range(cols-K+1):
            subarr = [row[col:col+K] for row in arr[row:row+K]]
            print(min([min(x) for x in subarr]), 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
print_smallest_elements(arr, K)
 
# This code is contributed by Aman Kumar

C#

using System;
using System.Collections.Generic;
 
public class Mainn
{
    // Function to return the smallest elements of all KxK submatrices of a
    // given NxM matrix
    static void PrintSmallestElements(List<List<int>> arr, int K)
    {
        // Get the number of rows and columns in the matrix
        int rows = arr.Count;
        int cols = arr[0].Count;
 
        // Loop through the matrix, creating submatrices of size KxK
        for (int row = 0; row <= rows - K; row++)
        {
            for (int col = 0; col <= cols - K; col++)
            {
                // Create a submatrix by extracting elements from the original matrix
                List<List<int>> subarr = new List<List<int>>();
                for (int i = row; i < row + K; i++)
                {
                    List<int> subcol = new List<int>();
                    for (int j = col; j < col + K; j++)
                    {
                        subcol.Add(arr[i][j]);
                    }
                    subarr.Add(subcol);
                }
 
                // Find the minimum value in the submatrix
                int minimum = int.MaxValue;
                foreach (List<int> x in subarr)
                {
                    foreach (int y in x)
                    {
                        minimum = Math.Min(minimum, y);
                    }
                }
 
                // Print the minimum value of the submatrix
                Console.Write(minimum + " ");
            }
            Console.WriteLine();
        }
    }
 
    public static void Main()
    {
        // Given matrix
        List<List<int>> arr = new List<List<int>>()
        {
            new List<int>() {-1, 5, 4, 1, -3},
            new List<int>() {4, 3, 1, 1, 6},
            new List<int>() {2, -2, 5, 3, 1},
            new List<int>() {8, 5, 1, 9, -4},
            new List<int>() {12, 3, 5, 8, 1}
        };
        int K = 3;
 
        // Call the function to print the smallest elements of all KxK submatrices
        PrintSmallestElements(arr, K);
    }
}

Javascript

// Function to return the smallest elements of all KxK submatrices of a given NxM matrix
function print_smallest_elements(arr, K) {
    // Get the number of rows and columns in the matrix
    let rows = arr.length;
    let cols = arr[0].length;
 
    // Loop through the matrix, creating submatrices of size KxK
    for (let row = 0; row <= rows - K; row++) {
        for (let col = 0; col <= cols - K; col++) {
            // Create a submatrix by extracting elements from the original matrix
            let subarr = [];
            for (let i = row; i < row + K; i++) {
                let subcol = [];
                for (let j = col; j < col + K; j++) {
                    subcol.push(arr[i][j]);
                }
                subarr.push(subcol);
            }
 
            // Find the minimum value in the submatrix
            let minimum = Number.MAX_SAFE_INTEGER;
            subarr.forEach(x => {
                x.forEach(y => {
                    minimum = Math.min(minimum, y);
                })
            })
 
            // Print the minimum value of the submatrix
            process.stdout.write(minimum + " ");
        }
        console.log();
    }
}
 
// Given matrix
let 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]];
let K = 3;
 
// Call the function to print the smallest elements of all KxK submatrices
print_smallest_elements(arr, K);

Output

-2 -2 -3 
-2 -2 -4 
-2 -2 -4

Time Complexity: O(N * M * K²)
Auxiliary Space: O(K*K)

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

Traverse over each row of the matrix and for every arr[i][j], update in-place the smallest element present between indices arr[i][j] to arr[i][j + K – 1] (0 <= j < M – K + 1).
Similarly, traverse over each column of the matrix and for every arr[i][j], update in-place the smallest element present between indices arr[i][j] to arr[i+K-1][j] (0 <= i < N – K + 1).
After performing the above operations, the top-left submatrix of size (N – K + 1)*(M – K + 1) of the matrix arr[][] consists of all the smallest elements of all the K x K sub-matrices of the given matrix.
Therefore, print the submatrix of size (N – K + 1)*(M – K + 1) as the required output.

Below is the implementation of the above approach:

C++

// 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

Java

// 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);
    }
}

Python3

# 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

C#

// 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

Javascript

<script>
// javascript program for the
// above approach
 
// Function to returns a smallest
    // elements of all KxK submatrices
    // of a given NxM matrix
    function matrixMinimum(
        nums, K)
    {
        // Stores the dimensions
        // of the given matrix
        let N = nums.length;
        let M = nums[0].length;
  
        // Stores the required
        // smallest elements
        let res
            = new Array(N - K + 1); 
        for (var i = 0; i < res.length; i++) {
            res[i] = new Array(2);
        }
  
        // Update the smallest elements row-wise
        for (let i = 0; i < N; i++) {
            for (let j = 0; j < M - K + 1; j++) {
  
                let min = Number.MAX_VALUE;
                for (let k = j; k < j + K; k++) {
                    min = Math.min(min, nums[i][k]);
                }
                nums[i][j] = min;
            }
        }
  
        // Update the minimum column-wise
        for (let j = 0; j < M; j++) {
            for (let i = 0; i < N - K + 1; i++) {
                let min = Number.MAX_VALUE;
                for (let 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 (let i = 0; i < N - K + 1; i++)
            for (let j = 0; j < M - K + 1; j++)
                res[i][j] = nums[i][j];
  
        // Return the resultant matrix
        return res;
    }
  
    function smallestinKsubmatrices(arr, K)
    {
        // Function Call
        let res = matrixMinimum(arr, K);
  
        // Print resultant matrix with the
        // minimum values of KxK sub-matrix
        for (let i = 0; i < res.length; i++) {
            for (let j = 0; j < res[0].length; j++) {
                document.write(res[i][j] + " ");
            }
            document.write("<br/>");
        }
    }
  
// Driver Code
 
     // Given matrix
        let 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
        let K = 3;
  
        smallestinKsubmatrices(arr, K);
              
</script>

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) )

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

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