Given an n x n square matrix, find sum of all sub-squares of size k x k
Last Updated :
12 Dec, 2022
Given an n x n square matrix, find sum of all sub-squares of size k x k where k is smaller than or equal to n.
Examples :
Input:
n = 5, k = 3
arr[][] = { {1, 1, 1, 1, 1},
{2, 2, 2, 2, 2},
{3, 3, 3, 3, 3},
{4, 4, 4, 4, 4},
{5, 5, 5, 5, 5},
};
Output:
18 18 18
27 27 27
36 36 36
Input:
n = 3, k = 2
arr[][] = { {1, 2, 3},
{4, 5, 6},
{7, 8, 9},
};
Output:
12 16
24 28
A Simple Solution is to one by one pick starting point (leftmost-topmost corner) of all possible sub-squares. Once the starting point is picked, calculate sum of sub-square starting with the picked starting point.
Following is the implementation of this idea.
C++
// A simple C++ program to find sum of all subsquares of
// size k x k
#include <iostream>
using namespace std;
// Size of given matrix
#define n 5
// 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;
}
}
// Driver program to test above function
int main()
{
int mat[n][n] = {
{ 1, 1, 1, 1, 1 }, { 2, 2, 2, 2, 2 },
{ 3, 3, 3, 3, 3 }, { 4, 4, 4, 4, 4 },
{ 5, 5, 5, 5, 5 },
};
int k = 3;
printSumSimple(mat, k);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// A simple C program to find sum of all subsquares of
// size k x k
#include <stdio.h>
// Size of given matrix
#define n 5
// 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];
printf("%d ", sum);
}
// Line separator for sub-squares starting with next
// row
printf("\n");
}
}
// Driver program to test above function
int main()
{
int mat[n][n] = {
{ 1, 1, 1, 1, 1 }, { 2, 2, 2, 2, 2 },
{ 3, 3, 3, 3, 3 }, { 4, 4, 4, 4, 4 },
{ 5, 5, 5, 5, 5 },
};
int k = 3;
printSumSimple(mat, k);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// A simple Java program to find sum of all
// subsquares of size k x k
import java.io.*;
class GFG {
// Size of given matrix
static final int n = 5;
// 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
static void printSumSimple(int mat[][], 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];
System.out.print(sum + " ");
}
// Line separator for sub-squares starting with
// next row
System.out.println();
}
}
// Driver Program to test above function
public static void main(String arg[])
{
int mat[][] = { { 1, 1, 1, 1, 1 },
{ 2, 2, 2, 2, 2 },
{ 3, 3, 3, 3, 3 },
{ 4, 4, 4, 4, 4 },
{ 5, 5, 5, 5, 5 } };
int k = 3;
printSumSimple(mat, k);
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
# A simple Python 3 program to find sum
# of all subsquares of size k x k
# Size of given matrix
n = 5
# 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
def printSumSimple(mat, 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 i in range(n - k + 1):
# column of first cell in current
# sub-square of size k x k
for j in range(n - k + 1):
# Calculate and print sum of
# current sub-square
sum = 0
for p in range(i, k + i):
for q in range(j, k + j):
sum += mat[p][q]
print(sum, end = " ")
# Line separator for sub-squares
# starting with next row
print()
# Driver Code
if __name__ == "__main__":
mat = [[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3],
[4, 4, 4, 4, 4],
[5, 5, 5, 5, 5]]
k = 3
printSumSimple(mat, k)
# This code is contributed by ita_c
// A simple C# program to find sum of all
// subsquares of size k x k
using System;
class GFG
{
// Size of given matrix
static int n = 5;
// 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
static void printSumSimple(int [,]mat, 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];
Console.Write(sum+ " ");
}
// Line separator for sub-squares
// starting with next row
Console.WriteLine();
}
}
// Driver Program to test above function
public static void Main()
{
int [,]mat = {{1, 1, 1, 1, 1},
{2, 2, 2, 2, 2},
{3, 3, 3, 3, 3},
{4, 4, 4, 4, 4},
{5, 5, 5, 5, 5}};
int k = 3;
printSumSimple(mat, k);
}
}
// This code is contributed by Sam007
<?php
// A simple PHP program to find
// sum of all subsquares of size
// k x k
// Size of given matrix
$n = 5;
// function to find sum of all sub -
// squares of size k x k in a given
// square matrix of size n x n
function printSumSimple( $mat, $k)
{
global $n;
// k must be smaller than
// or equal to n
if ($k > $n) return;
// row number of first cell in
// current sub-square of size
// k x k
for($i = 0; $i < $n - $k + 1; $i++)
{
// column of first cell in
// current sub-square of size
// k x k
for($j = 0; $j < $n - $k + 1; $j++)
{
// Calculate and print sum of
// current sub-square
$sum = 0;
for ($p = $i; $p < $k + $i; $p++)
for ($q = $j; $q < $k + $j; $q++)
$sum += $mat[$p][$q];
echo $sum , " ";
}
// Line separator for sub-squares
// starting with next row
echo "\n";
}
}
// Driver Code
$mat = array(array(1, 1, 1, 1, 1),
array(2, 2, 2, 2, 2,),
array(3, 3, 3, 3, 3,),
array(4, 4, 4, 4, 4,),
array(5, 5, 5, 5, 5));
$k = 3;
printSumSimple($mat, $k);
// This code is contributed by anuj_67.
?>
<script>
// A simple Javascript program to find sum of all
// subsquares of size k x k
// Size of given matrix
let n = 5;
// 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
function printSumSimple(mat,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 (let i = 0; i < n-k+1; i++)
{
// column of first cell in current
// sub-square of size k x k
for (let j = 0; j < n-k+1; j++)
{
// Calculate and print sum of
// current sub-square
let sum = 0;
for (let p = i; p < k+i; p++)
for (let q = j; q < k+j; q++)
sum += mat[p][q];
document.write(sum+ " ");
}
// Line separator for sub-squares
// starting with next row
document.write("<br>");
}
}
// Driver Program to test above function
let mat=[[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3],
[4, 4, 4, 4, 4],
[5, 5, 5, 5, 5]]
let k = 3;
printSumSimple(mat, k);
// This code is contributed by avanitrachhadiya2155
</script>
Output18 18 18
27 27 27
36 36 36
Time complexity: O(k2n2).
Auxiliary Space: O(1)
We can solve this problem in O(n2) time using a Tricky 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.
Following is the implementation of this idea.
C++
// An efficient C++ program to find sum of all subsquares of
// size k x k
#include <iostream>
using namespace std;
// Size of given matrix
#define n 5
// 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 i = 0; i < k; i++)
sum += mat[i][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;
}
}
// Driver program to test above function
int main()
{
int mat[n][n] = {
{ 1, 1, 1, 1, 1 }, { 2, 2, 2, 2, 2 },
{ 3, 3, 3, 3, 3 }, { 4, 4, 4, 4, 4 },
{ 5, 5, 5, 5, 5 },
};
int k = 3;
printSumTricky(mat, k);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// An efficient C program to find sum of all subsquares of
// size k x k
#include <stdio.h>
// Size of given matrix
#define n 5
// 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 i = 0; i < k; i++)
sum += mat[i][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];
printf("%d ", 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]);
printf("%d ", sum);
}
printf("\n");
}
}
// Driver program to test above function
int main()
{
int mat[n][n] = {
{ 1, 1, 1, 1, 1 }, { 2, 2, 2, 2, 2 },
{ 3, 3, 3, 3, 3 }, { 4, 4, 4, 4, 4 },
{ 5, 5, 5, 5, 5 },
};
int k = 3;
printSumTricky(mat, k);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// An efficient Java program to find sum of all subsquares
// of size k x k
import java.io.*;
class GFG {
// Size of given matrix
static int n = 5;
// 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
static void printSumTricky(int mat[][], 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[][] = new int[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 i = 0; i < k; i++)
sum += mat[i][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];
System.out.print(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]);
System.out.print(sum + " ");
}
System.out.println();
}
}
// Driver program to test above function
public static void main(String[] args)
{
int mat[][] = {
{ 1, 1, 1, 1, 1 }, { 2, 2, 2, 2, 2 },
{ 3, 3, 3, 3, 3 }, { 4, 4, 4, 4, 4 },
{ 5, 5, 5, 5, 5 },
};
int k = 3;
printSumTricky(mat, k);
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
# An efficient Python3 program to find sum
# of all subsquares of size k x k
# 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
def printSumTricky(mat, k):
global n
# 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
stripSum = [[None] * n for i in range(n)]
# Go column by column
for j in range(n):
# Calculate sum of first k x 1
# rectangle in this column
Sum = 0
for i in range(k):
Sum += mat[i][j]
stripSum[0][j] = Sum
# Calculate sum of remaining rectangles
for i in range(1, n - k + 1):
Sum += (mat[i + k - 1][j] -
mat[i - 1][j])
stripSum[i][j] = Sum
# 2: CALCULATE SUM of Sub-Squares
# using stripSum[][]
for i in range(n - k + 1):
# Calculate and print sum of first
# subsquare in this row
Sum = 0
for j in range(k):
Sum += stripSum[i][j]
print(Sum, end = " ")
# Calculate sum of remaining squares
# in current row by removing the leftmost
# strip of previous sub-square and adding
# a new strip
for j in range(1, n - k + 1):
Sum += (stripSum[i][j + k - 1] -
stripSum[i][j - 1])
print(Sum, end = " ")
print()
# Driver Code
n = 5
mat = [[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3],
[4, 4, 4, 4, 4],
[5, 5, 5, 5, 5]]
k = 3
printSumTricky(mat, k)
# This code is contributed by PranchalK
// An efficient C# program to find
// sum of all subsquares of size k x k
using System;
class GFG {
// Size of given matrix
static int n = 5;
// 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
static void printSumTricky(int [,]mat, 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 = new int[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 i = 0; i < k; i++)
sum += mat[i,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];
Console.Write(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]);
Console.Write(sum + " ");
}
Console.WriteLine();
}
}
// Driver program to test above function
public static void Main()
{
int [,]mat = {{1, 1, 1, 1, 1},
{2, 2, 2, 2, 2},
{3, 3, 3, 3, 3},
{4, 4, 4, 4, 4},
{5, 5, 5, 5, 5},
};
int k = 3;
printSumTricky(mat, k);
}
}
// This code is contributed by nitin mittal.
<?php
// An efficient PHP program to
// find sum of all subsquares
// of size k x k
// Size of given matrix
$n = 5;
// A O(n^2) function to find
// sum of all sub-squares of
// size k x k in a given
// square matrix of size n x n
function printSumTricky($mat, $k)
{
global $n;
// k must be smaller
// than or equal to n
if ($k > $n) return;
// 1: PREPROCESSING
// To store sums of all
// strips of size k x 1
$stripSum = array(array());
// Go column by column
for ($j = 0; $j < $n; $j++)
{
// Calculate sum of first
// k x 1 rectangle in this column
$sum = 0;
for ($i = 0; $i < $k; $i++)
$sum += $mat[$i][$j];
$stripSum[0][$j] = $sum;
// Calculate sum of
// remaining rectangles
for ($i = 1; $i < $n - $k + 1; $i++)
{
$sum += ($mat[$i + $k - 1][$j] -
$mat[$i - 1][$j]);
$stripSum[$i][$j] = $sum;
}
}
// 2: CALCULATE SUM of
// Sub-Squares using stripSum[][]
for ($i = 0; $i < $n - $k + 1; $i++)
{
// Calculate and print sum of
// first subsquare in this row
$sum = 0;
for ($j = 0; $j < $k; $j++)
$sum += $stripSum[$i][$j];
echo $sum , " ";
// Calculate sum of remaining
// squares in current row by
// removing the leftmost strip
// of previous sub-square and
// adding a new strip
for ($j = 1; $j < $n - $k + 1; $j++)
{
$sum += ($stripSum[$i][$j + $k - 1] -
$stripSum[$i][$j - 1]);
echo $sum , " ";
}
echo "\n";
}
}
// Driver Code
$mat = array(array(1, 1, 1, 1, 1),
array(2, 2, 2, 2, 2),
array(3, 3, 3, 3, 3),
array(4, 4, 4, 4, 4),
array(5, 5, 5, 5, 5));
$k = 3;
printSumTricky($mat, $k);
// This code is contributed by anuj_67.
?>
<script>
// An efficient Javascript program to find
// sum of all subsquares of size k x k
// Size of given matrix
let n = 5;
// 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
function printSumTricky(mat, 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
let stripSum = new Array(n);
for(let i = 0; i < n; i++)
{
stripSum[i] = new Array(n);
}
for(let i = 0; i < n; i++)
{
for(let j = 0; j < n; j++)
{
stripSum[i][j] = 0;
}
}
// Go column by column
for(let j = 0; j < n; j++)
{
// Calculate sum of first k x 1
// rectangle in this column
let sum = 0;
for(let i = 0; i < k; i++)
sum += mat[i][j];
stripSum[0][j] = sum;
// Calculate sum of remaining rectangles
for(let 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(let i = 0; i < n - k + 1; i++)
{
// Calculate and print sum of first
// subsquare in this row
let sum = 0;
for (let j = 0; j < k; j++)
sum += stripSum[i][j];
document.write(sum + " ");
// Calculate sum of remaining squares
// in current row by removing the
// leftmost strip of previous sub-square
// and adding a new strip
for(let j = 1; j < n - k + 1; j++)
{
sum += (stripSum[i][j + k - 1] -
stripSum[i][j - 1]);
document.write(sum + " ");
}
document.write("<br>");
}
}
// Driver code
let mat = [ [ 1, 1, 1, 1, 1 ],
[ 2, 2, 2, 2, 2 ],
[ 3, 3, 3, 3, 3 ],
[ 4, 4, 4, 4, 4 ],
[ 5, 5, 5, 5, 5 ] ];
let k = 3;
printSumTricky(mat, k);
// This code is contributed by rag2127
</script>
Output18 18 18
27 27 27
36 36 36
Time complexity: O(n2).
Auxiliary Space: O(n2).
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