Dada una array N x M de enteros y un entero K , la tarea es encontrar el tamaño de la subarray cuadrada máxima (S x S) , tal que todas las subarrays cuadradas de la array dada de ese tamaño tengan una suma menos que k .
Ejemplos:
Input: K = 30
mat[N][M] = {{1, 2, 3, 4, 6},
{5, 3, 8, 1, 2},
{4, 6, 7, 5, 5},
{2, 4, 8, 9, 4} };
Output: 2
Explanation:
All Sub-matrices of size 2 x 2
have sum less than 30
Input : K = 100
mat[N][M] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
Output: 3
Explanation:
All Sub-matrices of size 3 x 3
have sum less than 100
Enfoque ingenuo La solución básica es elegir el tamaño S de la subarray y encontrar todas las subarrays de ese tamaño y verificar que la suma de todas las subarrays sea menor que la suma dada, mientras que esto se puede mejorar calculando la suma de las array utilizando este enfoque. Por lo tanto, la tarea será elegir el tamaño máximo posible y la posición inicial y final de todas las subarrays posibles. Debido a lo cual la complejidad temporal total será O(N 3 ).
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to find the
// maximum size square submatrix
// such that their sum is less than K
#include <bits/stdc++.h>
using namespace std;
// Size of matrix
#define N 4
#define M 5
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
void preProcess(int mat[N][M],
int aux[N][M])
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
int sumQuery(int aux[N][M], int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1][tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
int maximumSquareSize(int mat[N][M], int K)
{
int aux[N][M];
preProcess(mat, aux);
// Loop to choose the size of matrix
for (int i = min(N, M); i >= 1; i--) {
bool satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
int main()
{
int K = 30;
int mat[N][M] = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
cout << maximumSquareSize(mat, K);
return 0;
}
Java
// Java implementation to find the
// maximum size square submatrix
// such that their sum is less than K
class GFG{
// Size of matrix
static final int N = 4;
static final int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [][]mat,
int [][]aux)
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [][]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1][tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [][]mat, int K)
{
int [][]aux = new int[N][M];
preProcess(mat, aux);
// Loop to choose the size of matrix
for (int i = Math.min(N, M); i >= 1; i--) {
boolean satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
public static void main(String[] args)
{
int K = 30;
int mat[][] = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
System.out.print(maximumSquareSize(mat, K));
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 implementation to find the # maximum size square submatrix # such that their sum is less than K # Size of matrix N = 4 M = 5 # Function to preprocess the matrix # for computing the sum of every # possible matrix of the given size def preProcess(mat, aux): # Loop to copy the first row # of the matrix into the aux matrix for i in range (M): aux[0][i] = mat[0][i] # Computing the sum column-wise for i in range (1, N): for j in range (M): aux[i][j] = (mat[i][j] + aux[i - 1][j]) # Computing row wise sum for i in range (N): for j in range (1, M): aux[i][j] += aux[i][j - 1] # Function to find the sum of a # submatrix with the given indices def sumQuery(aux, tli, tlj, rbi, rbj): # Overall sum from the top to # right corner of matrix res = aux[rbi][rbj] # Removing the sum from the top # corer of the matrix if (tli > 0): res = res - aux[tli - 1][rbj] # Remove the overlapping sum if (tlj > 0): res = res - aux[rbi][tlj - 1] # Add the sum of top corner # which is subtracted twice if (tli > 0 and tlj > 0): res = (res + aux[tli - 1][tlj - 1]) return res # Function to find the maximum # square size possible with the # such that every submatrix have # sum less than the given sum def maximumSquareSize(mat, K): aux = [[0 for x in range (M)] for y in range (N)] preProcess(mat, aux) # Loop to choose the size of matrix for i in range (min(N, M), 0, -1): satisfies = True # Loop to find the sum of the # matrix of every possible submatrix for x in range (N): for y in range (M) : if (x + i - 1 <= N - 1 and y + i - 1 <= M - 1): if (sumQuery(aux, x, y, x + i - 1, y + i - 1) > K): satisfies = False if (satisfies == True): return (i) return 0 # Driver Code if __name__ == "__main__": K = 30 mat = [[1, 2, 3, 4, 6], [5, 3, 8, 1, 2], [4, 6, 7, 5, 5], [2, 4, 8, 9, 4]] print( maximumSquareSize(mat, K)) # This code is contributed by Chitranayal
C#
// C# implementation to find the
// maximum size square submatrix
// such that their sum is less than K
using System;
public class GFG{
// Size of matrix
static readonly int N = 4;
static readonly int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [,]mat,
int [,]aux)
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0,i] = mat[0,i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i,j] = mat[i,j] +
aux[i - 1,j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i,j] += aux[i,j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [,]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi,rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1,rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi,tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1,tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [,]mat, int K)
{
int [,]aux = new int[N,M];
preProcess(mat, aux);
// Loop to choose the size of matrix
for (int i = Math.Min(N, M); i >= 1; i--) {
bool satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
public static void Main(String[] args)
{
int K = 30;
int [,]mat = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
Console.Write(maximumSquareSize(mat, K));
}
}
// This code contributed by PrinciRaj1992
Javascript
<script>
// JavaScript implementation to find the
// maximum size square submatrix
// such that their sum is less than K
// Size of matrix
let N = 4;
let M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
function preProcess(mat,aux)
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (let i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (let i = 1; i < N; i++)
for (let j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (let i = 0; i < N; i++)
for (let j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
function sumQuery(aux,tli,tlj,rbi,rbj)
{
// Overall sum from the top to
// right corner of matrix
let res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res +
aux[tli - 1][tlj - 1];
return res;
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
function maximumSquareSize(mat,k)
{
let aux = new Array(N);
for(let i=0;i<N;i++)
{
aux[i]=new Array(M);
}
preProcess(mat, aux);
// Loop to choose the size of matrix
for (let i = Math.min(N, M); i >= 1; i--) {
let satisfies = true;
// Loop to find the sum of the
// matrix of every possible submatrix
for (let x = 0; x < N; x++) {
for (let y = 0; y < M; y++) {
if (x + i - 1 <= N - 1 &&
y + i - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + i - 1, y + i - 1) > K)
satisfies = false;
}
}
}
if (satisfies == true)
return (i);
}
return 0;
}
// Driver Code
let mat = [[1, 2, 3, 4, 6],
[5, 3, 8, 1, 2],
[4, 6, 7, 5, 5],
[2, 4, 8, 9, 4]];
let K = 30;
document.write(maximumSquareSize(mat, K));
// This code is contributed by avanitrachhadiya2155
</script>
Producción:
2
- Complejidad temporal: O(N 3 )
- Espacio Auxiliar: O(N 2 )
Enfoque eficiente: la observación clave es que, si un cuadrado de lado s tiene el tamaño máximo que satisface la condición, entonces todos los tamaños más pequeños satisfarán la condición. Usando esto, podemos reducir nuestro espacio de búsqueda en cada paso a la mitad, que es precisamente la idea de Binary Search . A continuación se muestra la ilustración de los pasos del enfoque:
- Espacio de búsqueda: El espacio de búsqueda para este problema será de [1, min(N, M)]. Ese es el espacio de búsqueda para la búsqueda binaria se define como –
low = 1 high = min(N, M)
- Siguiente espacio de búsqueda: en cada iteración, encuentre la mitad del espacio de búsqueda y luego, finalmente, verifique que todos los subarreglos de ese tamaño tengan una suma menor que K . Si todos los subarreglos de ese tamaño tienen una suma menor que K. Entonces, el siguiente espacio de búsqueda posible estará a la derecha del medio. De lo contrario, el siguiente espacio de búsqueda posible estará a la izquierda del medio. Eso es menos que el medio.
- Caso 1: Condición cuando todos los subarreglos de tamaño medio tienen una suma menor que K . Después –
if checkSubmatrix(mat, mid, K):
low = mid + 1
- Caso 2: Condición cuando todos los subarreglos de tamaño medio tienen una suma mayor que K . Después –
if not checkSubmatrix(mat, mid, K):
high = mid - 1
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to find the
// maximum size square submatrix
// such that their sum is less than K
#include <bits/stdc++.h>
using namespace std;
// Size of matrix
#define N 4
#define M 5
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
void preProcess(int mat[N][M],
int aux[N][M])
{
// Loop to copy the first row
// of the matrix into the aux matrix
for (int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for (int i = 1; i < N; i++)
for (int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for (int i = 0; i < N; i++)
for (int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
int sumQuery(int aux[N][M], int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1][tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
bool check(int mid, int aux[N][M],
int K)
{
bool satisfies = true;
// Iterating through all possible
// submatrices of given size
for (int x = 0; x < N; x++) {
for (int y = 0; y < M; y++) {
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1) {
if (sumQuery(aux, x, y,
x + mid - 1, y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
int maximumSquareSize(int mat[N][M],
int K)
{
int aux[N][M];
preProcess(mat, aux);
// Search space
int low = 1, high = min(N, M);
int mid;
// Binary search for size
while (high - low > 1) {
mid = (low + high) / 2;
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K)) {
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
int main()
{
int K = 30;
int mat[N][M] = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
cout << maximumSquareSize(mat, K);
return 0;
}
Java
// Java implementation to find the
// maximum size square submatrix
// such that their sum is less than K
class GFG{
// Size of matrix
static final int N = 4;
static final int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [][]mat,
int [][]aux)
{
// Loop to copy the first row of
// the matrix into the aux matrix
for(int i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for(int i = 1; i < N; i++)
for(int j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for(int i = 0; i < N; i++)
for(int j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [][]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1][tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
static boolean check(int mid, int [][]aux,
int K)
{
boolean satisfies = true;
// Iterating through all possible
// submatrices of given size
for(int x = 0; x < N; x++)
{
for(int y = 0; y < M; y++)
{
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1)
{
if (sumQuery(aux, x, y,
x + mid - 1,
y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [][]mat,
int K)
{
int [][]aux = new int[N][M];
preProcess(mat, aux);
// Search space
int low = 1, high = Math.min(N, M);
int mid;
// Binary search for size
while (high - low > 1)
{
mid = (low + high) / 2;
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K))
{
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
public static void main(String[] args)
{
int K = 30;
int [][]mat = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
System.out.print(maximumSquareSize(mat, K));
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation to find the # maximum size square submatrix # such that their sum is less than K # Function to preprocess the matrix # for computing the sum of every # possible matrix of the given size def preProcess(mat, aux): # Loop to copy the first row # of the matrix into the aux matrix for i in range(5): aux[0][i] = mat[0][i] # Computing the sum column-wise for i in range(1, 4): for j in range(5): aux[i][j] = (mat[i][j] + aux[i - 1][j]) # Computing row wise sum for i in range(4): for j in range(1, 5): aux[i][j] += aux[i][j - 1] return aux # Function to find the sum of a # submatrix with the given indices def sumQuery(aux, tli, tlj, rbi, rbj): # Overall sum from the top to # right corner of matrix res = aux[rbi][rbj] # Removing the sum from the top # corer of the matrix if (tli > 0): res = res - aux[tli - 1][rbj] # Remove the overlapping sum if (tlj > 0): res = res - aux[rbi][tlj - 1] # Add the sum of top corner # which is subtracted twice if (tli > 0 and tlj > 0): res = res + aux[tli - 1][tlj - 1] return res # Function to check whether square # sub matrices of size mid satisfy # the condition or not def check(mid, aux, K): satisfies = True # Iterating through all possible # submatrices of given size for x in range(4): for y in range(5): if (x + mid - 1 <= 4 - 1 and y + mid - 1 <= 5 - 1): if (sumQuery(aux, x, y, x + mid - 1, y + mid - 1) > K): satisfies = False return True if satisfies == True else False # Function to find the maximum # square size possible with the # such that every submatrix have # sum less than the given sum def maximumSquareSize(mat, K): aux = [[0 for i in range(5)] for i in range(4)] aux = preProcess(mat, aux) # Search space low , high = 1, min(4, 5) mid = 0 # Binary search for size while (high - low > 1): mid = (low + high) // 2 # Check if the mid satisfies # the given condition if (check(mid, aux, K)): low = mid else: high = mid if (check(high, aux, K)): return high return low # Driver Code if __name__ == '__main__': K = 30 mat = [ [ 1, 2, 3, 4, 6 ], [ 5, 3, 8, 1, 2 ], [ 4, 6, 7, 5, 5 ], [ 2, 4, 8, 9, 4 ] ] print(maximumSquareSize(mat, K)) # This code is contributed by mohit kumar 29
C#
// C# implementation to find the
// maximum size square submatrix
// such that their sum is less than K
using System;
class GFG{
// Size of matrix
static readonly int N = 4;
static readonly int M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
static void preProcess(int [,]mat,
int [,]aux)
{
// Loop to copy the first row of
// the matrix into the aux matrix
for(int i = 0; i < M; i++)
aux[0, i] = mat[0, i];
// Computing the sum column-wise
for(int i = 1; i < N; i++)
for(int j = 0; j < M; j++)
aux[i, j] = mat[i, j] +
aux[i - 1, j];
// Computing row wise sum
for(int i = 0; i < N; i++)
for(int j = 1; j < M; j++)
aux[i, j] += aux[i, j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
static int sumQuery(int [,]aux, int tli,
int tlj, int rbi, int rbj)
{
// Overall sum from the top to
// right corner of matrix
int res = aux[rbi, rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1, rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi, tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1, tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
static bool check(int mid, int [,]aux,
int K)
{
bool satisfies = true;
// Iterating through all possible
// submatrices of given size
for(int x = 0; x < N; x++)
{
for(int y = 0; y < M; y++)
{
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1)
{
if (sumQuery(aux, x, y,
x + mid - 1,
y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
static int maximumSquareSize(int [,]mat,
int K)
{
int [,]aux = new int[N, M];
preProcess(mat, aux);
// Search space
int low = 1, high = Math.Min(N, M);
int mid;
// Binary search for size
while (high - low > 1)
{
mid = (low + high) / 2;
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K))
{
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
public static void Main(String[] args)
{
int K = 30;
int [,]mat = { { 1, 2, 3, 4, 6 },
{ 5, 3, 8, 1, 2 },
{ 4, 6, 7, 5, 5 },
{ 2, 4, 8, 9, 4 } };
Console.Write(maximumSquareSize(mat, K));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript implementation to find the
// maximum size square submatrix
// such that their sum is less than K
// Size of matrix
let N = 4;
let M = 5;
// Function to preprocess the matrix
// for computing the sum of every
// possible matrix of the given size
function preProcess(mat,aux)
{
// Loop to copy the first row of
// the matrix into the aux matrix
for(let i = 0; i < M; i++)
aux[0][i] = mat[0][i];
// Computing the sum column-wise
for(let i = 1; i < N; i++)
for(let j = 0; j < M; j++)
aux[i][j] = mat[i][j] +
aux[i - 1][j];
// Computing row wise sum
for(let i = 0; i < N; i++)
for(let j = 1; j < M; j++)
aux[i][j] += aux[i][j - 1];
}
// Function to find the sum of a
// submatrix with the given indices
function sumQuery(aux,tli,tlj,rbi,rbj)
{
// Overall sum from the top to
// right corner of matrix
let res = aux[rbi][rbj];
// Removing the sum from the top
// corer of the matrix
if (tli > 0)
res = res - aux[tli - 1][rbj];
// Remove the overlapping sum
if (tlj > 0)
res = res - aux[rbi][tlj - 1];
// Add the sum of top corner
// which is subtracted twice
if (tli > 0 && tlj > 0)
res = res + aux[tli - 1][tlj - 1];
return res;
}
// Function to check whether square
// sub matrices of size mid satisfy
// the condition or not
function check(mid,aux,K)
{
let satisfies = true;
// Iterating through all possible
// submatrices of given size
for(let x = 0; x < N; x++)
{
for(let y = 0; y < M; y++)
{
if (x + mid - 1 <= N - 1 &&
y + mid - 1 <= M - 1)
{
if (sumQuery(aux, x, y,
x + mid - 1,
y + mid - 1) > K)
satisfies = false;
}
}
}
return (satisfies == true);
}
// Function to find the maximum
// square size possible with the
// such that every submatrix have
// sum less than the given sum
function maximumSquareSize(mat,K)
{
let aux = new Array(N);
for(let i=0;i<N;i++)
{
aux[i]=new Array(M);
}
preProcess(mat, aux);
// Search space
let low = 1, high = Math.min(N, M);
let mid;
// Binary search for size
while (high - low > 1)
{
mid = Math.floor((low + high) / 2);
// Check if the mid satisfies
// the given condition
if (check(mid, aux, K))
{
low = mid;
}
else
high = mid;
}
if (check(high, aux, K))
return high;
return low;
}
// Driver Code
let K = 30;
let mat = [[1, 2, 3, 4, 6 ],
[ 5, 3, 8, 1, 2 ],
[ 4, 6, 7, 5, 5 ],
[ 2, 4, 8, 9, 4 ] ];
document.write(maximumSquareSize(mat, K));
// This code is contributed by rag2127
</script>
Producción:
2
Análisis de rendimiento:
- Complejidad de tiempo: O(N 2 * log(N) )
- Espacio Auxiliar: O(N 2 )