Given a matrix of integers mat[][] of size N * M. The task is to find the maximum perimeter of a square in the matrix. The perimeter of a square is defined as the sum of all the values lying on the sides of the square.
Examples:
Input: mat[][] = {
{-3, -2, 7},
{-4, 6, 0},
{-4, 8, 2}}
Output: 16
The maximum perimeter square is
{6, 0}
{8, 2}
Input: mat[][] = {
{1, 1, 0},
{1, 1, 1},
{0, 1, 1}}
Output: 6
Naive approach: A simple solution is to generate all the square that can be possible inside the given matrix mat[][] and then find their perimeter and take the maximum out of them.
Efficient approach:
- To find the perimeter of the square, the length of the sides should be known.
- Here the length is described as the sum of elements on that particular row and column.
- Two matrices of size N * M will be created and that will store the prefix sum of the rows of the original matrix and the prefix sum of the columns so that the length of sides can be calculated in constant time.
- There will be two nested loops one from 1 to N and the other from 1 to M to find the left corner of the square (Top-Left corner) and one loop will be min(N – i, M – j) that will tell us the size of the square and make sure that the index will not exceed the size of the matrix.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to calculate the prefix sum of the // rows and the columns of the given matrix void prefix_calculate(vector<vector< int > >& A,
vector<vector< int > >& row,
vector<vector< int > >& col)
{ // Number of rows and cols
int n = ( int )A.size();
int m = ( int )A[0].size();
// First column of the row prefix array
for ( int i = 0; i < n; ++i) {
row[i][0] = A[i][0];
}
// Update the prefix sum for the rows
for ( int i = 0; i < n; ++i) {
for ( int j = 1; j < m; ++j) {
row[i][j] = row[i][j - 1]
+ A[i][j];
}
}
// First row of the column prefix array
for ( int i = 0; i < m; ++i) {
col[0][i] = A[0][i];
}
// Update the prefix sum for the columns
for ( int i = 0; i < m; ++i) {
for ( int j = 1; j < n; ++j) {
col[j][i] = A[j][i]
+ col[j - 1][i];
}
}
} // Function to return the perimeter // of the square having top-left corner // at (i, j) and size k int perimeter( int i, int j, int k,
vector<vector< int > >& row,
vector<vector< int > >& col,
vector<vector< int > >& A)
{ // i and j represent the top left
// corner of the square and
// k is the size
int row_s, col_s;
// Get the upper row sum
if (j == 0)
row_s = 0;
else
row_s = row[i][j - 1];
// Get the left column sum
if (i == 0)
col_s = 0;
else
col_s = col[i - 1][j];
int upper_row = row[i][j + k] - row_s;
int left_col = col[i + k][j] - col_s;
// At the distance of k in
// both direction
if (j == 0)
row_s = 0;
else
row_s = row[i + k][j - 1];
if (i == 0)
col_s = 0;
else
col_s = col[i - 1][j + k];
int lower_row = row[i + k][j + k] - row_s;
int right_col = col[i + k][j + k] - col_s;
// The perimeter will be
// sum of all the values
int sum = upper_row
+ lower_row
+ left_col
+ right_col;
// Since all the corners are
// included twice, they need to
// be subtract from the sum
sum -= (A[i][j]
+ A[i + k][j]
+ A[i][j + k]
+ A[i + k][j + k]);
return sum;
} // Function to return the maximum perimeter // of a square in the given matrix int maxPerimeter(vector<vector< int > >& A)
{ // Number of rows and cols
int n = ( int )A.size();
int m = ( int )A[0].size();
vector<vector< int > > row(n, vector< int >(m, 0));
vector<vector< int > > col(n, vector< int >(m, 0));
// Function call to calculate
// the prefix sum of rows and cols
prefix_calculate(A, row, col);
// To store the maximum perimeter
int maxPer = 0;
// Nested loops to choose the top-left
// corner of the square
for ( int i = 0; i < n; ++i) {
for ( int j = 0; j < m; ++j) {
// Loop for the size of the square
for ( int k = 0; k < min(n - i, m - j); ++k) {
// Get the perimeter of the current square
int perimtr = perimeter(i, j, k, row, col, A);
// Update the maximum perimeter so far
maxPer = max(maxPer, perimtr);
}
}
}
return maxPer;
} // Driver code int main()
{ vector<vector< int > > A = {
{ 1, 1, 0 },
{ 1, 1, 1 },
{ 0, 1, 1 }
};
cout << maxPerimeter(A);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ // Function to calculate the prefix sum of the // rows and the columns of the given matrix static void prefix_calculate( int [][] A,
int [][] row,
int [][] col)
{ // Number of rows and cols
int n = ( int )A.length;
int m = ( int )A[ 0 ].length;
// First column of the row prefix array
for ( int i = 0 ; i < n; ++i)
{
row[i][ 0 ] = A[i][ 0 ];
}
// Update the prefix sum for the rows
for ( int i = 0 ; i < n; ++i)
{
for ( int j = 1 ; j < m; ++j)
{
row[i][j] = row[i][j - 1 ]
+ A[i][j];
}
}
// First row of the column prefix array
for ( int i = 0 ; i < m; ++i)
{
col[ 0 ][i] = A[ 0 ][i];
}
// Update the prefix sum for the columns
for ( int i = 0 ; i < m; ++i)
{
for ( int j = 1 ; j < n; ++j)
{
col[j][i] = A[j][i]
+ col[j - 1 ][i];
}
}
} // Function to return the perimeter // of the square having top-left corner // at (i, j) and size k static int perimeter( int i, int j, int k,
int [][] row, int [][] col,
int [][] A)
{ // i and j represent the top left
// corner of the square and
// k is the size
int row_s, col_s;
// Get the upper row sum
if (j == 0 )
row_s = 0 ;
else
row_s = row[i][j - 1 ];
// Get the left column sum
if (i == 0 )
col_s = 0 ;
else
col_s = col[i - 1 ][j];
int upper_row = row[i][j + k] - row_s;
int left_col = col[i + k][j] - col_s;
// At the distance of k in
// both direction
if (j == 0 )
row_s = 0 ;
else
row_s = row[i + k][j - 1 ];
if (i == 0 )
col_s = 0 ;
else
col_s = col[i - 1 ][j + k];
int lower_row = row[i + k][j + k] - row_s;
int right_col = col[i + k][j + k] - col_s;
// The perimeter will be
// sum of all the values
int sum = upper_row + lower_row +
left_col + right_col;
// Since all the corners are
// included twice, they need to
// be subtract from the sum
sum -= (A[i][j] + A[i + k][j] +
A[i][j + k] + A[i + k][j + k]);
return sum;
} // Function to return the maximum perimeter // of a square in the given matrix static int maxPerimeter( int [][] A)
{ // Number of rows and cols
int n = ( int )A.length;
int m = ( int )A[ 0 ].length;
int [][] row = new int [n][m];
int [][] col = new int [n][m];
// Function call to calculate
// the prefix sum of rows and cols
prefix_calculate(A, row, col);
// To store the maximum perimeter
int maxPer = 0 ;
// Nested loops to choose the top-left
// corner of the square
for ( int i = 0 ; i < n; ++i)
{
for ( int j = 0 ; j < m; ++j)
{
// Loop for the size of the square
for ( int k = 0 ; k < Math.min(n - i, m - j); ++k)
{
// Get the perimeter of the current square
int perimtr = perimeter(i, j, k,
row, col, A);
// Update the maximum perimeter so far
maxPer = Math.max(maxPer, perimtr);
}
}
}
return maxPer;
} // Driver code public static void main(String[] args)
{ int [][] A = {
{ 1 , 1 , 0 },
{ 1 , 1 , 1 },
{ 0 , 1 , 1 }
};
System.out.print(maxPerimeter(A));
} } // This code is contributed by PrinciRaj1992 |
# Python3 implementation of the approach # Function to calculate the prefix sum of the # rows and the columns of the given matrix def prefix_calculate(A, row, col):
# Number of rows and cols
n = len (A)
m = len (A[ 0 ])
# First column of the row prefix array
for i in range (n):
row[i][ 0 ] = A[i][ 0 ]
# Update the prefix sum for the rows
for i in range (n):
for j in range ( 1 , m):
row[i][j] = row[i][j - 1 ] + A[i][j]
# First row of the column prefix array
for i in range (m):
col[ 0 ][i] = A[ 0 ][i]
# Update the prefix sum for the columns
for i in range (m):
for j in range ( 1 , m):
col[j][i] = A[j][i] + col[j - 1 ][i]
# Function to return the perimeter # of the square having top-left corner # at (i, j) and size k def perimeter(i, j, k, row, col, A):
# i and j represent the top left
# corner of the square and
# k is the size
row_s, col_s = 0 , 0
# Get the upper row sum
if (j = = 0 ):
row_s = 0
else :
row_s = row[i][j - 1 ]
# Get the left column sum
if (i = = 0 ):
col_s = 0
else :
col_s = col[i - 1 ][j]
upper_row = row[i][j + k] - row_s
left_col = col[i + k][j] - col_s
# At the distance of k in
# both direction
if (j = = 0 ):
row_s = 0
else :
row_s = row[i + k][j - 1 ]
if (i = = 0 ):
col_s = 0
else :
col_s = col[i - 1 ][j + k]
lower_row = row[i + k][j + k] - row_s
right_col = col[i + k][j + k] - col_s
# The perimeter will be
# sum of all the values
sum = upper_row + lower_row + \
left_col + right_col
# Since all the corners are
# included twice, they need to
# be subtract from the sum
sum - = (A[i][j] + A[i + k][j] + \
A[i][j + k] + A[i + k][j + k])
return sum
# Function to return the maximum perimeter # of a square in the given matrix def maxPerimeter(A):
# Number of rows and cols
n = len (A)
m = len (A[ 0 ])
row = [[ 0 for i in range (m)]
for i in range (n)]
col = [[ 0 for i in range (m)]
for i in range (n)]
# Function call to calculate
# the prefix sum of rows and cols
prefix_calculate(A, row, col)
# To store the maximum perimeter
maxPer = 0
# Nested loops to choose the top-left
# corner of the square
for i in range (n):
for j in range (m):
# Loop for the size of the square
for k in range ( min (n - i, m - j)):
# Get the perimeter of the current square
perimtr = perimeter(i, j, k,
row, col, A)
# Update the maximum perimeter so far
maxPer = max (maxPer, perimtr)
return maxPer
# Driver code A = [[ 1 , 1 , 0 ],
[ 1 , 1 , 1 ],
[ 0 , 1 , 1 ]]
print (maxPerimeter(A))
# This code is contributed by Mohit Kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to calculate the prefix sum of the // rows and the columns of the given matrix static void prefix_calculate( int [,] A,
int [,] row,
int [,] col)
{ // Number of rows and cols
int n = ( int )A.GetLength(0);
int m = ( int )A.GetLength(1);
// First column of the row prefix array
for ( int i = 0; i < n; ++i)
{
row[i, 0] = A[i, 0];
}
// Update the prefix sum for the rows
for ( int i = 0; i < n; ++i)
{
for ( int j = 1; j < m; ++j)
{
row[i, j] = row[i, j - 1]
+ A[i, j];
}
}
// First row of the column prefix array
for ( int i = 0; i < m; ++i)
{
col[0, i] = A[0, i];
}
// Update the prefix sum for the columns
for ( int i = 0; i < m; ++i)
{
for ( int j = 1; j < n; ++j)
{
col[j, i] = A[j, i]
+ col[j - 1, i];
}
}
} // Function to return the perimeter // of the square having top-left corner // at (i, j) and size k static int perimeter( int i, int j, int k,
int [,] row, int [,] col,
int [,] A)
{ // i and j represent the top left
// corner of the square and
// k is the size
int row_s, col_s;
// Get the upper row sum
if (j == 0)
row_s = 0;
else
row_s = row[i, j - 1];
// Get the left column sum
if (i == 0)
col_s = 0;
else
col_s = col[i - 1, j];
int upper_row = row[i, j + k] - row_s;
int left_col = col[i + k, j] - col_s;
// At the distance of k in
// both direction
if (j == 0)
row_s = 0;
else
row_s = row[i + k, j - 1];
if (i == 0)
col_s = 0;
else
col_s = col[i - 1, j + k];
int lower_row = row[i + k, j + k] - row_s;
int right_col = col[i + k, j + k] - col_s;
// The perimeter will be
// sum of all the values
int sum = upper_row + lower_row +
left_col + right_col;
// Since all the corners are
// included twice, they need to
// be subtract from the sum
sum -= (A[i, j] + A[i + k, j] +
A[i, j + k] + A[i + k, j + k]);
return sum;
} // Function to return the maximum perimeter // of a square in the given matrix static int maxPerimeter( int [,] A)
{ // Number of rows and cols
int n = ( int )A.GetLength(0);
int m = ( int )A.GetLength(1);
int [,] row = new int [n, m];
int [,] col = new int [n, m];
// Function call to calculate
// the prefix sum of rows and cols
prefix_calculate(A, row, col);
// To store the maximum perimeter
int maxPer = 0;
// Nested loops to choose the top-left
// corner of the square
for ( int i = 0; i < n; ++i)
{
for ( int j = 0; j < m; ++j)
{
// Loop for the size of the square
for ( int k = 0; k < Math.Min(n - i, m - j); ++k)
{
// Get the perimeter of the current square
int perimtr = perimeter(i, j, k,
row, col, A);
// Update the maximum perimeter so far
maxPer = Math.Max(maxPer, perimtr);
}
}
}
return maxPer;
} // Driver code public static void Main(String[] args)
{ int [,] A = {{ 1, 1, 0 },
{ 1, 1, 1 },
{ 0, 1, 1 }};
Console.Write(maxPerimeter(A));
} } // This code is contributed by PrinciRaj1992 |
<script> // Javascript implementation of the approach // Function to calculate the prefix sum of the // rows and the columns of the given matrix function prefix_calculate(A, row, col)
{ // Number of rows and cols
let n = A.length;
let m = A[0].length;
// First column of the row prefix array
for (let i = 0; i < n; ++i)
{
row[i][0] = A[i][0];
}
// Update the prefix sum for the rows
for (let i = 0; i < n; ++i)
{
for (let j = 1; j < m; ++j)
{
row[i][j] = row[i][j - 1]
+ A[i][j];
}
}
// First row of the column prefix array
for (let i = 0; i < m; ++i)
{
col[0][i] = A[0][i];
}
// Update the prefix sum for the columns
for (let i = 0; i < m; ++i)
{
for (let j = 1; j < n; ++j)
{
col[j][i] = A[j][i]
+ col[j - 1][i];
}
}
} // Function to return the perimeter // of the square having top-left corner // at (i, j) and size k function perimeter(i,j,k,row,col,A)
{ // i and j represent the top left
// corner of the square and
// k is the size
let row_s, col_s;
// Get the upper row sum
if (j == 0)
row_s = 0;
else
row_s = row[i][j - 1];
// Get the left column sum
if (i == 0)
col_s = 0;
else
col_s = col[i - 1][j];
let upper_row = row[i][j + k] - row_s;
let left_col = col[i + k][j] - col_s;
// At the distance of k in
// both direction
if (j == 0)
row_s = 0;
else
row_s = row[i + k][j - 1];
if (i == 0)
col_s = 0;
else
col_s = col[i - 1][j + k];
let lower_row = row[i + k][j + k] - row_s;
let right_col = col[i + k][j + k] - col_s;
// The perimeter will be
// sum of all the values
let sum = upper_row + lower_row +
left_col + right_col;
// Since all the corners are
// included twice, they need to
// be subtract from the sum
sum -= (A[i][j] + A[i + k][j] +
A[i][j + k] + A[i + k][j + k]);
return sum;
} // Function to return the maximum perimeter // of a square in the given matrix function maxPerimeter(A)
{ // Number of rows and cols
let n = A.length;
let m = A[0].length;
let row = new Array(n);
let col = new Array(n);
for (let i=0;i<n;i++)
{
row[i]= new Array(m);
col[i]= new Array(m);
}
// Function call to calculate
// the prefix sum of rows and cols
prefix_calculate(A, row, col);
// To store the maximum perimeter
let maxPer = 0;
// Nested loops to choose the top-left
// corner of the square
for (let i = 0; i < n; ++i)
{
for (let j = 0; j < m; ++j)
{
// Loop for the size of the square
for (let k = 0; k < Math.min(n - i, m - j); ++k)
{
// Get the perimeter of the current square
let perimtr = perimeter(i, j, k,
row, col, A);
// Update the maximum perimeter so far
maxPer = Math.max(maxPer, perimtr);
}
}
}
return maxPer;
} // Driver code let A = [[1, 1, 0 ],[1, 1, 1],[0, 1, 1]]; document.write(maxPerimeter(A)); // This code is contributed by unknown2108 </script> |
6
Time Complexity: O(N * M * min(N, M))
Auxiliary Space: O(N*M)