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# Find the Submatrix which holds the given co-ordinate

Given a matrix mat of N*N (N is a perfect square|) and two points x and y, the task is to return all the elements of the submatrix in which the element A[x][y] lies.

Note: The matrix is divided into N equal submatrix each of size K*K (where K is the square root of N)

Examples:

Input: N = 9, x = 4, y = 4
mat = {{1, 2, 3, 9, 8, 7, 1, 2, 1}, {4, 5, 6, 1, 2, 3, 7, 9, 8}, {7, 8, 9, 2, 2, 0, 1, 5, 7},
{0, 2, 9, 1, 2, 3, 4, 5, 3}, {9, 8, 7, 4, 5, 6, 7, 4, 1}, {1, 4, 7, 7, 8, 9, 9, 8, 7},
{5, 6, 1, 9, 8, 7, 5, 2, 3}, {4, 5, 1, 6, 5, 4, 9, 7, 4}, {8, 7, 9, 3, 2, 1, 9, 4, 9}},
Output: {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
Explanation: Given x = 4, y = 4 that element lies in the middle grid.

Input Matrix

Input: N = 9, x = 6, y= 4
mat = {{1, 2, 3, 9, 8, 7, 1, 2, 1}, {4, 5, 6, 1, 2, 3, 7, 9, 8}, {7, 8, 9, 2, 2, 0, 1, 5, 7},
{0, 2, 9, 1, 2, 3, 4, 5, 3}, {9, 8, 7, 4, 5, 6, 7, 4, 1}, {1, 4, 7, 7, 8, 9, 9, 8, 7},
{5, 6, 1, 9, 8, 7, 5, 2, 3}, {4, 5, 1, 6, 5, 4, 9, 7, 4}, {8, 7, 9, 3, 2, 1, 9, 4, 9}}
Output: {{9, 8, 7}, {6, 5, 4}, {3, 2, 1}}
Explanation: Given x =6, y = 4 that element lies in the grid shown below.

Input Matrix

Approach: The problem can be solved based on the following observation:

An element at index (x, y) in a square matrix of perfect square length, lies in submatrix[n*(x/n), (n*(y/n)], where each value shows the positioning with respect to other submatrices. So the idea is to just print that submatrix.

Follow the steps mentioned below to implement the idea:

• Find square root on N.
• Store the submatrix where the coordinate (x, y) lies in a new matrix.
• Return the new array.

Below is the implementation of the above approach.

## C++

```// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to print submatrix
vector<vector<int> >
getSubGrid(int N, vector<vector<int> >& matrix, int x,
int y)
{
int n = sqrt(N);
vector<vector<int> > subGrid(n, vector<int>(n));
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
subGrid[i][j]
= matrix[n * (x / n) + (j / n) + i]
[n * (y / n) + j % n];
}
}

// Return the submatrix
return subGrid;
}

// Driver Code
int main()
{
int N = 9;
vector<vector<int> > matrix
= { { 1, 2, 3, 9, 8, 7, 1, 2, 1 },
{ 4, 5, 6, 1, 2, 3, 7, 9, 8 },
{ 7, 8, 9, 2, 2, 0, 1, 5, 7 },
{ 0, 2, 9, 1, 2, 3, 4, 5, 3 },
{ 9, 8, 7, 4, 5, 6, 7, 4, 1 },
{ 1, 4, 7, 7, 8, 9, 9, 8, 7 },
{ 5, 6, 1, 9, 8, 7, 5, 2, 3 },
{ 4, 5, 1, 6, 5, 4, 9, 7, 4 },
{ 8, 7, 9, 3, 2, 1, 9, 4, 9 } };
int x = 4;
int y = 4;

// Function call
vector<vector<int> > subGrid
= getSubGrid(N, matrix, x, y);
for (int i = 0; i < subGrid.size(); i++) {
cout << "[";
int j = 0;
for (; j < subGrid.size() - 1; j++) {
cout << subGrid[i][j] << ", ";
}
cout << subGrid[i][j] << "] ";
}

return 0;
}

// This code is contributed by Rohit Pradhan```

## Java

```// Java code to implement the approach

import java.io.*;
import java.util.*;

class GFG {

// Function to print submatrix
static int[][] getSubGrid(int N, int[][] matrix,
int x, int y)
{
int n = (int)Math.sqrt(N);
int subGrid[][] = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
subGrid[i][j]
= matrix[n * (x / n) + (j / n) + i]
[n * (y / n) + j % n];
}
}

// Return the submatrix
return subGrid;
}

// Driver code
public static void main(String[] args)
{
int N = 9;
int matrix[][] = { { 1, 2, 3, 9, 8, 7, 1, 2, 1 },
{ 4, 5, 6, 1, 2, 3, 7, 9, 8 },
{ 7, 8, 9, 2, 2, 0, 1, 5, 7 },
{ 0, 2, 9, 1, 2, 3, 4, 5, 3 },
{ 9, 8, 7, 4, 5, 6, 7, 4, 1 },
{ 1, 4, 7, 7, 8, 9, 9, 8, 7 },
{ 5, 6, 1, 9, 8, 7, 5, 2, 3 },
{ 4, 5, 1, 6, 5, 4, 9, 7, 4 },
{ 8, 7, 9, 3, 2, 1, 9, 4, 9 } };
int x = 4;
int y = 4;

// Function call
int subGrid[][] = getSubGrid(N, matrix, x, y);

System.out.println(Arrays.deepToString(subGrid));
}
}```

## Python3

```# python3 code to implement the approach
import math

# Function to print submatrix
def getSubGrid(N, matrix, x, y):

n = int(math.sqrt(N))
subGrid = [[0 for _ in range(n)] for _ in range(n)]
for i in range(0, n):
for j in range(0, n):
subGrid[i][j] = matrix[n *
(x // n) + (j // n) + i][n * (y // n) + j % n]

# Return the submatrix
return subGrid

# Driver Code
if __name__ == "__main__":

N = 9
matrix = [[1, 2, 3, 9, 8, 7, 1, 2, 1],
[4, 5, 6, 1, 2, 3, 7, 9, 8],
[7, 8, 9, 2, 2, 0, 1, 5, 7],
[0, 2, 9, 1, 2, 3, 4, 5, 3],
[9, 8, 7, 4, 5, 6, 7, 4, 1],
[1, 4, 7, 7, 8, 9, 9, 8, 7],
[5, 6, 1, 9, 8, 7, 5, 2, 3],
[4, 5, 1, 6, 5, 4, 9, 7, 4],
[8, 7, 9, 3, 2, 1, 9, 4, 9]]
x = 4
y = 4

# Function call
subGrid = getSubGrid(N, matrix, x, y)
for i in range(0, len(subGrid)):
print("[", end="")
j = 0
while(j < len(subGrid) - 1):
print(subGrid[i][j], end=", ")
j += 1

print(subGrid[i][j], end="] ")

# This code is contributed by rakeshsahni
```

## C#

```// C# code to implement the approach

using System;

class GFG {

// Function to print submatrix
static int[,] getSubGrid(int N, int[,] matrix,
int x, int y)
{
int n = (int)Math.Sqrt(N);
int [,]subGrid = new int[n,n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
subGrid[i,j] = matrix[n * (x / n) + (j / n) + i,n * (y / n) + j % n];
}
}

// Return the submatrix
return subGrid;
}

// Driver code
public static void Main(string[] args)
{
int N = 9;
int [,]matrix = { { 1, 2, 3, 9, 8, 7, 1, 2, 1 },
{ 4, 5, 6, 1, 2, 3, 7, 9, 8 },
{ 7, 8, 9, 2, 2, 0, 1, 5, 7 },
{ 0, 2, 9, 1, 2, 3, 4, 5, 3 },
{ 9, 8, 7, 4, 5, 6, 7, 4, 1 },
{ 1, 4, 7, 7, 8, 9, 9, 8, 7 },
{ 5, 6, 1, 9, 8, 7, 5, 2, 3 },
{ 4, 5, 1, 6, 5, 4, 9, 7, 4 },
{ 8, 7, 9, 3, 2, 1, 9, 4, 9 } };
int x = 4;
int y = 4;

// Function call
int [,]subGrid = getSubGrid(N, matrix, x, y);
Console.Write("[") ;
for (int i = 0; i< subGrid.GetLength(0); i++){
Console.Write("[") ;
for (int j = 0; j < subGrid.GetLength(1) ; j++)
Console.Write(subGrid[i,j] + " ,");
Console.Write("] ");
}
Console.Write("]") ;

}
}

// This code is contributed by AnkThon```

## Javascript

```<script>
// javascript code to implement the approach
// Function to print submatrix
function getSubGrid(N, matrix,
x , y)
{
var n = parseInt(Math.sqrt(N));
var subGrid = Array(n).fill(0).map(x => Array(n).fill(0));
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++) {
subGrid[i][j]
= matrix[n * parseInt(x / n) + parseInt(j / n) + i]
[n * parseInt(y / n) + j % n];
}
}

// Return the submatrix
return subGrid;
}

// Driver code

var N = 9;
var matrix = [ [ 1, 2, 3, 9, 8, 7, 1, 2, 1 ],
[ 4, 5, 6, 1, 2, 3, 7, 9, 8 ],
[ 7, 8, 9, 2, 2, 0, 1, 5, 7 ],
[ 0, 2, 9, 1, 2, 3, 4, 5, 3 ],
[ 9, 8, 7, 4, 5, 6, 7, 4, 1 ],
[ 1, 4, 7, 7, 8, 9, 9, 8, 7 ],
[ 5, 6, 1, 9, 8, 7, 5, 2, 3 ],
[ 4, 5, 1, 6, 5, 4, 9, 7, 4 ],
[ 8, 7, 9, 3, 2, 1, 9, 4, 9 ] ];
var x = 4;
var y = 4;

// Function call
var subGrid = getSubGrid(N, matrix, x, y);
document.write("[");
for (let i = 0; i < subGrid.length; i++) {
document.write("[");
for (let j = 0; j < subGrid[i].length; j++) {
document.write(subGrid[i][j]+", ");
}
document.write("]");
}
document.write("]");

// This code contributed by shikhasingrajput
</script>
```
Output

`[[1, 2, 3], [4, 5, 6], [7, 8, 9]]`

Time Complexity: O(N)
Auxiliary Space: O(1)

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