Dada una array cuadrada de nxn, encuentre la suma de todos los subcuadrados de tamaño kxk donde k es menor o igual que n.
Ejemplos:
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
Una solución simple es elegir uno por uno el punto de inicio (la esquina superior izquierda) de todos los subcuadrados posibles. Una vez que se elige el punto de inicio, calcule la suma de los subcuadrados que comienzan con el punto de inicio seleccionado.
A continuación se muestra la implementación de esta 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)
C
// 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)
Java
// A simple Java program to find sum of all
// subsquares of size k x k
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)
Python3
# 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
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
<?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.
?>
Javascript
<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>
Producción
18 18 18 27 27 27 36 36 36
La complejidad temporal de la solución anterior es O(k 2 n 2 ). Podemos resolver este problema en O(n 2 ) usando una Tricky Solution .
La idea es preprocesar la array cuadrada dada. En el paso de preprocesamiento, calcule la suma de todas las franjas verticales de tamaño kx 1 en una array cuadrada temporal stripSum[][]. Una vez que tenemos la suma de todas las tiras verticales, podemos calcular la suma del primer subcuadrado en una fila como la suma de las primeras k tiras en esa fila, y para los subcuadrados restantes, podemos calcular la suma en tiempo O(1) eliminando la franja más a la izquierda del subcuadrado anterior y agregando la franja más a la derecha del nuevo cuadrado.
A continuación se muestra la implementación de esta 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)
C
// 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)
Java
// 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)
Python3
# 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
C#
// 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
<?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.
?>
Javascript
<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>
Producción
18 18 18 27 27 27 36 36 36
Complejidad temporal: O(n 2 ).
Espacio Auxiliar: O(n 2 ).
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA