Dada una array NxN , la tarea es encontrar la suma de AND bit a bit de todas sus subarrays rectangulares.
Ejemplos:
Input : arr[][] = {{1, 1, 1},
{1, 1, 1},
{1, 1, 1}}
Output : 36
Explanation: All the possible submatrices will have AND value 1.
Since, there are 36 submatrices in total, ans = 36
Input : arr[][] = {{9, 7, 4},
{8, 9, 2},
{11, 11, 5}}
Output : 135
Requisito previo : Número de subarrays rectangulares de una array binaria con todos 1s .
Solución ingenua: una solución simple es generar todas las subarrays y encontrar el AND requerido para cada una de ellas. La complejidad temporal de este enfoque será O(N 6 ).
Enfoque eficiente: en aras de una mejor comprensión, supongamos que cualquier parte de un elemento está representada por la variable ‘i’, y la variable ‘suma’ se usa para almacenar la suma final.
La idea aquí es que intentaremos encontrar el número de valores AND (subarrays con bits y (&)) con i -ésimo conjunto de bits. Supongamos que existe un número ‘S i ‘ de subarrays con i -ésimo conjunto de bits. Para el i -ésimo bit, la suma se puede actualizar como sum += (2 i * S i ) .
Para cada bit ‘i’, cree una array booleana set_bit que almacene ‘1’ en un índice (R, C) si i thse establece el bit de arr[R][C]. De lo contrario, almacena ‘0’. Entonces, para este arreglo booleano, tratamos de encontrar el número de subarrays rectangulares con todos 1s(S i ). Para, i -ésimo bit, la suma final se actualizará como:
sum += 2i * Si
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find sum of Bit-wise AND
// of all submatrices
#include <iostream>
#include <stack>
using namespace std;
#define n 3
// Function to find prefix-count for each row
// from right to left
void findPrefixCount(int p_arr[][n], bool set_bit[][n])
{
for (int i = 0; i < n; i++) {
for (int j = n - 1; j >= 0; j--) {
if (!set_bit[i][j])
continue;
if (j != n - 1)
p_arr[i][j] += p_arr[i][j + 1];
p_arr[i][j] += (int)set_bit[i][j];
}
}
}
// Function to find the number of submatrices
// with all 1s
int matrixAllOne(bool set_bit[][n])
{
// Array to store required prefix count of 1s from
// right to left for boolean array
int p_arr[n][n] = { 0 };
findPrefixCount(p_arr, set_bit);
// Variable to store the final answer
int ans = 0;
// For each index of a column, determine the number
// of sub-matrices starting from that index
// and has all 1s
for (int j = 0; j < n; j++) {
int i = n - 1;
// Stack to store elements and the count
// of the numbers they popped
// First part of pair is value of inserted element
// Second part is count of the number of elements
// pushed before with a greater value
stack<pair<int, int> > q;
// variable to store the number of submatrices
// with all 1s
int to_sum = 0;
while (i >= 0) {
int c = 0;
while (q.size() != 0 and q.top().first > p_arr[i][j]) {
to_sum -= (q.top().second + 1) * (q.top().first - p_arr[i][j]);
c += q.top().second + 1;
q.pop();
}
to_sum += p_arr[i][j];
ans += to_sum;
q.push({ p_arr[i][j], c });
i--;
}
}
return ans;
}
// Function to find the sum of Bitwise-AND
// of all submatrices
int sumAndMatrix(int arr[][n])
{
int sum = 0;
int mul = 1;
for (int i = 0; i < 30; i++) {
// matrix to store the status
// of ith bit of each element
// of matrix arr
bool set_bit[n][n];
for (int R = 0; R < n; R++)
for (int C = 0; C < n; C++)
set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
sum += (mul * matrixAllOne(set_bit));
mul *= 2;
}
return sum;
}
// Driver Code
int main()
{
int arr[][n] = { { 9, 7, 4 },
{ 8, 9, 2 },
{ 11, 11, 5 } };
cout << sumAndMatrix(arr);
return 0;
}
Java
// Java program to find sum of Bit-wise AND
// of all submatrices
import java.util.*;
class GFG
{
static int n = 3;
// Function to find prefix-count for
// each row from right to left
static void findPrefixCount(int p_arr[][],
boolean set_bit[][])
{
for (int i = 0; i < n; i++)
{
for (int j = n - 1; j >= 0; j--)
{
if (!set_bit[i][j])
continue;
if (j != n - 1)
p_arr[i][j] += p_arr[i][j + 1];
p_arr[i][j] += (set_bit[i][j]) ? 1 : 0;
}
}
}
static class pair
{
int first,second;
pair(){}
pair(int a, int b)
{
first = a;
second = b;
}
}
// Function to find the number of
// submatrices with all 1s
static int matrixAllOne(boolean set_bit[][])
{
// Array to store required prefix count of 1s from
// right to left for boolean array
int p_arr[][] = new int[n][n];
for(int i = 0; i < n; i++)
for(int j = 0; j < n;j++)
p_arr[i][j] = 0;
findPrefixCount(p_arr, set_bit);
// Variable to store the final answer
int ans = 0;
// For each index of a column, determine the number
// of sub-matrices starting from that index
// and has all 1s
for (int j = 0; j < n; j++)
{
int i = n - 1;
// Stack to store elements and the count
// of the numbers they popped
// First part of pair is value of inserted element
// Second part is count of the number of elements
// pushed before with a greater value
Stack<pair > q = new Stack<pair >();
// variable to store the number of submatrices
// with all 1s
int to_sum = 0;
while (i >= 0)
{
int c = 0;
while (q.size() != 0 &&
q.peek().first > p_arr[i][j])
{
to_sum -= (q.peek().second + 1) *
(q.peek().first - p_arr[i][j]);
c += q.peek().second + 1;
q.pop();
}
to_sum += p_arr[i][j];
ans += to_sum;
q.push(new pair( p_arr[i][j], c ));
i--;
}
}
return ans;
}
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumAndMatrix(int arr[][])
{
int sum = 0;
int mul = 1;
for (int i = 0; i < 30; i++)
{
// matrix to store the status
// of ith bit of each element
// of matrix arr
boolean set_bit[][] = new boolean[n][n];
for (int R = 0; R < n; R++)
for (int C = 0; C < n; C++)
set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
sum += (mul * matrixAllOne(set_bit));
mul *= 2;
}
return sum;
}
// Driver Code
public static void main(String args[])
{
int arr[][] = { { 9, 7, 4 },
{ 8, 9, 2 },
{ 11, 11, 5 } };
System.out.println( sumAndMatrix(arr));
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 program to find sum of # Bitwise-AND of all submatrices # Function to find prefix-count for # each row from right to left def findPrefixCount(p_arr, set_bit): for i in range(0, n): for j in range(n - 1, -1, -1): if not set_bit[i][j]: continue if j != n - 1: p_arr[i][j] += p_arr[i][j + 1] p_arr[i][j] += int(set_bit[i][j]) # Function to create a boolean matrix # set_bit which stores ‘1’ at an index # (R, C) if ith bit of arr[R][C] is set. def matrixAllOne(set_bit): # Array to store prefix count of zeros # from right to left for boolean array p_arr = [[0 for i in range(n)] for j in range(n)] findPrefixCount(p_arr, set_bit) # Variable to store the final answer ans = 0 # For each index of a column we # will try to determine the number # of sub-matrices starting from # that index and has all 1s for j in range(0, n): i = n - 1 # stack to store elements and the # count of the numbers they popped # First part of pair will be the # value of inserted element. # Second part will be the count # of the number of elements pushed # before with a greater value q = [] # Variable to store the number # of submatrices with all 0s to_sum = 0 while i >= 0: c = 0 while (len(q) != 0 and q[-1][0] > p_arr[i][j]): to_sum -= ((q[-1][1] + 1) * (q[-1][0] - p_arr[i][j])) c += q.pop()[1] + 1 to_sum += p_arr[i][j] ans += to_sum q.append((p_arr[i][j], c)) i -= 1 # Return the final answer return ans # Function to find sum of # Bitwise-AND of all submatrices def sumAndMatrix(arr): Sum, mul = 0, 1 for i in range(0, 30): # matrix to store the status # of ith bit of each element # of matrix arr set_bit = [[False for i in range(n)] for j in range(n)] for R in range(0, n): for C in range(0, n): set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0) Sum += (mul * matrixAllOne(set_bit)) mul *= 2 return Sum # Driver Code if __name__ == "__main__": n = 3 arr = [[9, 7, 4], [8, 9, 2], [11, 11, 5]] print(sumAndMatrix(arr)) # This code is contributed by Rituraj Jain
C#
// C# program to find sum of Bit-wise AND
// of all submatrices
using System;
using System.Collections.Generic;
class GFG
{
static int n = 3;
// Function to find prefix-count for
// each row from right to left
static void findPrefixCount(int [,]p_arr,
bool [,]set_bit)
{
for (int i = 0; i < n; i++)
{
for (int j = n - 1; j >= 0; j--)
{
if (!set_bit[i, j])
continue;
if (j != n - 1)
p_arr[i, j] += p_arr[i, j + 1];
p_arr[i, j] += (set_bit[i, j]) ? 1 : 0;
}
}
}
public class pair
{
public int first,second;
public pair(){}
public pair(int a, int b)
{
first = a;
second = b;
}
}
// Function to find the number of
// submatrices with all 1s
static int matrixAllOne(bool [,]set_bit)
{
// Array to store required prefix count of 1s from
// right to left for boolean array
int [,]p_arr = new int[n, n];
for(int i = 0; i < n; i++)
for(int j = 0; j < n;j++)
p_arr[i, j] = 0;
findPrefixCount(p_arr, set_bit);
// Variable to store the final answer
int ans = 0;
// For each index of a column, determine the number
// of sub-matrices starting from that index
// and has all 1s
for (int j = 0; j < n; j++)
{
int i = n - 1;
// Stack to store elements and the count
// of the numbers they popped
// First part of pair is value of inserted element
// Second part is count of the number of elements
// pushed before with a greater value
Stack<pair > q = new Stack<pair >();
// variable to store the number of submatrices
// with all 1s
int to_sum = 0;
while (i >= 0)
{
int c = 0;
while (q.Count != 0 &&
q.Peek().first > p_arr[i,j])
{
to_sum -= (q.Peek().second + 1) *
(q.Peek().first - p_arr[i,j]);
c += q.Peek().second + 1;
q.Pop();
}
to_sum += p_arr[i,j];
ans += to_sum;
q.Push(new pair( p_arr[i,j], c ));
i--;
}
}
return ans;
}
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumAndMatrix(int [,]arr)
{
int sum = 0;
int mul = 1;
for (int i = 0; i < 30; i++)
{
// matrix to store the status
// of ith bit of each element
// of matrix arr
bool [,]set_bit = new bool[n,n];
for (int R = 0; R < n; R++)
for (int C = 0; C < n; C++)
set_bit[R, C] = ((arr[R, C] & (1 << i)) != 0);
sum += (mul * matrixAllOne(set_bit));
mul *= 2;
}
return sum;
}
// Driver Code
public static void Main(String []args)
{
int [,]arr = { { 9, 7, 4 },
{ 8, 9, 2 },
{ 11, 11, 5 } };
Console.WriteLine(sumAndMatrix(arr));
}
}
// This code contributed by Rajput-Ji
Javascript
<script>
// Javascript program to find sum of Bit-wise AND
// of all submatrices
var n = 3;
// Function to find prefix-count for each row
// from right to left
function findPrefixCount(p_arr, set_bit)
{
for (var i = 0; i < n; i++) {
for (var j = n - 1; j >= 0; j--) {
if (!set_bit[i][j])
continue;
if (j != n - 1)
p_arr[i][j] += p_arr[i][j + 1];
p_arr[i][j] += set_bit[i][j];
}
}
}
// Function to find the number of submatrices
// with all 1s
function matrixAllOne(set_bit)
{
// Array to store required prefix count of 1s from
// right to left for boolean array
var p_arr = Array.from(Array(n), ()=> Array(n).fill(0));
findPrefixCount(p_arr, set_bit);
// Variable to store the final answer
var ans = 0;
// For each index of a column, determine the number
// of sub-matrices starting from that index
// and has all 1s
for (var j = 0; j < n; j++) {
var i = n - 1;
// Stack to store elements and the count
// of the numbers they popped
// First part of pair is value of inserted element
// Second part is count of the number of elements
// pushed before with a greater value
var q = [];
// variable to store the number of submatrices
// with all 1s
var to_sum = 0;
while (i >= 0) {
var c = 0;
while (q.length != 0 && q[q.length - 1][0] > p_arr[i][j]) {
to_sum -= (q[q.length - 1][1] + 1) * (q[q.length - 1][0] - p_arr[i][j]);
c += q[q.length - 1][1] + 1;
q.pop();
}
to_sum += p_arr[i][j];
ans += to_sum;
q.push([p_arr[i][j], c ]);
i--;
}
}
return ans;
}
// Function to find the sum of Bitwise-AND
// of all submatrices
function sumAndMatrix(arr)
{
var sum = 0;
var mul = 1;
for (var i = 0; i < 30; i++) {
// matrix to store the status
// of ith bit of each element
// of matrix arr
var set_bit = Array.from(Array(n), ()=> Array(n));
for (var R = 0; R < n; R++)
for (var C = 0; C < n; C++)
set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
sum += (mul * matrixAllOne(set_bit));
mul *= 2;
}
return sum;
}
// Driver Code
var arr = [ [ 9, 7, 4 ],
[ 8, 9, 2 ],
[ 11, 11, 5 ] ];
document.write( sumAndMatrix(arr));
</script>
Producción:
135
Complejidad de tiempo: O (N * N), ya que estamos usando bucles anidados para atravesar la array.
Espacio auxiliar: O(N*N), ya que estamos usando cualquier espacio adicional.
Publicación traducida automáticamente
Artículo escrito por DivyanshuShekhar1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA