Dada una array N * N mat[][] que consta de enteros no negativos y algunas consultas que consisten en la esquina superior izquierda e inferior derecha de la subarray, la tarea es encontrar el OR bit a bit de todos los elementos de la subarray dada en cada consulta.
Ejemplos:
Entrada: mat[][] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}},
q[] = {{1, 1, 1, 1}, { 1, 2, 2, 2}}
Salida:
5
15
Consulta 1: Solo el elemento en la subarray es 5.
Consulta 2: 6 O 9 = 15
Entrada: mat[][] = {
{12, 23, 13} ,
{41, 15, 46},
{75, 82, 123}},
q[] = {{0, 0, 2, 2}, {1, 1, 2, 1}}
Salida:
127
95
Enfoque ingenuo: iterar a través de la subarray y encontrar el OR bit a bit de todos los números en ese rango. Esto tomará O(n 2 ) tiempo para cada consulta en el peor de los casos.
Enfoque eficiente: si observamos los números enteros como un número binario, podemos ver fácilmente que la condición para que se establezca el i – ésimo bit de nuestra respuesta es que se establezca el i -ésimo bit de al menos un entero en la subarray.
Entonces, calcularemos el conteo de prefijos para cada bit. Usaremos esto para encontrar el número de enteros en la subarray con i -ésimo conjunto de bits. Si no es cero, también se establecerá el i -ésimo bit de nuestra respuesta.
Para esto, crearemos una array 3d, prefix_count[][][] donde prefix_count[i][x][y] almacenará el recuento de todos los elementos de la subarray con la esquina superior izquierda en {0, 0} y la esquina inferior derecha en {x, y} y el i -ésimo conjunto de bits. Consulte
este artículo para comprender prefix_count en caso de array.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
#define bitscount 32
#define n 3
using namespace std;
// Array to store bit-wise
// prefix count
int prefix_count[bitscount][n][n];
// Function to find the prefix sum
void findPrefixCount(int arr[][n])
{
// Loop for each bit
for (int i = 0; i < bitscount; i++) {
// Loop to find prefix-count
// for each row
for (int j = 0; j < n; j++) {
prefix_count[i][j][0] = ((arr[j][0] >> i) & 1);
for (int k = 1; k < n; k++) {
prefix_count[i][j][k] = ((arr[j][k] >> i) & 1);
prefix_count[i][j][k] += prefix_count[i][j][k - 1];
}
}
}
// Finding column-wise prefix
// count
for (int i = 0; i < bitscount; i++)
for (int j = 1; j < n; j++)
for (int k = 0; k < n; k++)
prefix_count[i][j][k] += prefix_count[i][j - 1][k];
}
// Function to return the result for a query
int rangeOr(int x1, int y1, int x2, int y2)
{
// To store the answer
int ans = 0;
// Loop for each bit
for (int i = 0; i < bitscount; i++) {
// To store the number of variables
// with ith bit set
int p;
if (x1 == 0 and y1 == 0)
p = prefix_count[i][x2][y2];
else if (x1 == 0)
p = prefix_count[i][x2][y2]
- prefix_count[i][x2][y1 - 1];
else if (y1 == 0)
p = prefix_count[i][x2][y2]
- prefix_count[i][x1 - 1][y2];
else
p = prefix_count[i][x2][y2]
- prefix_count[i][x1 - 1][y2]
- prefix_count[i][x2][y1 - 1]
+ prefix_count[i][x1 - 1][y1 - 1];
// If count of variables with ith bit
// set is greater than 0
if (p != 0)
ans = (ans | (1 << i));
}
return ans;
}
// Driver code
int main()
{
int arr[][n] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
findPrefixCount(arr);
int queries[][4] = { { 1, 1, 1, 1 }, { 1, 2, 2, 2 } };
int q = sizeof(queries) / sizeof(queries[0]);
for (int i = 0; i < q; i++)
cout << rangeOr(queries[i][0],
queries[i][1],
queries[i][2],
queries[i][3])
<< endl;
return 0;
}
Java
// Java implementation of the approach
class GFG
{
final static int bitscount = 32 ;
final static int n = 3 ;
// Array to store bit-wise
// prefix count
static int prefix_count[][][] = new int [bitscount][n][n];
// Function to find the prefix sum
static void findPrefixCount(int arr[][])
{
// Loop for each bit
for (int i = 0; i < bitscount; i++)
{
// Loop to find prefix-count
// for each row
for (int j = 0; j < n; j++)
{
prefix_count[i][j][0] = ((arr[j][0] >> i) & 1);
for (int k = 1; k < n; k++)
{
prefix_count[i][j][k] = ((arr[j][k] >> i) & 1);
prefix_count[i][j][k] += prefix_count[i][j][k - 1];
}
}
}
// Finding column-wise prefix
// count
for (int i = 0; i < bitscount; i++)
for (int j = 1; j < n; j++)
for (int k = 0; k < n; k++)
prefix_count[i][j][k] += prefix_count[i][j - 1][k];
}
// Function to return the result for a query
static int rangeOr(int x1, int y1, int x2, int y2)
{
// To store the answer
int ans = 0;
// Loop for each bit
for (int i = 0; i < bitscount; i++)
{
// To store the number of variables
// with ith bit set
int p;
if (x1 == 0 && y1 == 0)
p = prefix_count[i][x2][y2];
else if (x1 == 0)
p = prefix_count[i][x2][y2]
- prefix_count[i][x2][y1 - 1];
else if (y1 == 0)
p = prefix_count[i][x2][y2]
- prefix_count[i][x1 - 1][y2];
else
p = prefix_count[i][x2][y2]
- prefix_count[i][x1 - 1][y2]
- prefix_count[i][x2][y1 - 1]
+ prefix_count[i][x1 - 1][y1 - 1];
// If count of variables with ith bit
// set is greater than 0
if (p != 0)
ans = (ans | (1 << i));
}
return ans;
}
// Driver code
public static void main (String[] args)
{
int arr[][] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
findPrefixCount(arr);
int queries[][] = { { 1, 1, 1, 1 }, { 1, 2, 2, 2 } };
int q = queries.length;
for (int i = 0; i < q; i++)
System.out.println( rangeOr(queries[i][0],
queries[i][1],
queries[i][2],
queries[i][3]) );
}
}
// This code is contributed by AnkitRai
Python3
# Python 3 implementation of the approach bitscount = 32 n = 3 # Array to store bit-wise # prefix count prefix_count = [[[0 for i in range(n)] for j in range(n)] for k in range(bitscount)] # Function to find the prefix sum def findPrefixCount(arr): # Loop for each bit for i in range(bitscount): # Loop to find prefix-count # for each row for j in range(n): prefix_count[i][j][0] = ((arr[j][0] >> i) & 1) for k in range(1,n): prefix_count[i][j][k] = ((arr[j][k] >> i) & 1) prefix_count[i][j][k] += prefix_count[i][j][k - 1] # Finding column-wise prefix # count for i in range(bitscount): for j in range(1,n): for k in range(n): prefix_count[i][j][k] += prefix_count[i][j - 1][k] # Function to return the result for a query def rangeOr(x1, y1, x2, y2): # To store the answer ans = 0 # Loop for each bit for i in range(bitscount): # To store the number of variables # with ith bit set if (x1 == 0 and y1 == 0): p = prefix_count[i][x2][y2] elif (x1 == 0): p = prefix_count[i][x2][y2] - prefix_count[i][x2][y1 - 1] elif (y1 == 0): p = prefix_count[i][x2][y2] - prefix_count[i][x1 - 1][y2] else: p = prefix_count[i][x2][y2] - prefix_count[i][x1 - 1][y2] - prefix_count[i][x2][y1 - 1] + prefix_count[i][x1 - 1][y1 - 1]; # If count of variables with ith bit # set is greater than 0 if (p != 0): ans = (ans | (1 << i)) return ans # Driver code if __name__ == '__main__': arr = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] findPrefixCount(arr) queries = [[1, 1, 1, 1], [1, 2, 2, 2]] q = len(queries) for i in range(q): print(rangeOr(queries[i][0],queries[i][1],queries[i][2],queries[i][3])) # This code is contributed by # Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
static int bitscount = 32 ;
static int n = 3 ;
// Array to store bit-wise
// prefix count
static int [,,]prefix_count = new int [bitscount,n,n];
// Function to find the prefix sum
static void findPrefixCount(int [,]arr)
{
// Loop for each bit
for (int i = 0; i < bitscount; i++)
{
// Loop to find prefix-count
// for each row
for (int j = 0; j < n; j++)
{
prefix_count[i,j,0] = ((arr[j,0] >> i) & 1);
for (int k = 1; k < n; k++)
{
prefix_count[i, j, k] = ((arr[j, k] >> i) & 1);
prefix_count[i, j, k] += prefix_count[i, j, k - 1];
}
}
}
// Finding column-wise prefix
// count
for (int i = 0; i < bitscount; i++)
for (int j = 1; j < n; j++)
for (int k = 0; k < n; k++)
prefix_count[i, j, k] += prefix_count[i, j - 1, k];
}
// Function to return the result for a query
static int rangeOr(int x1, int y1, int x2, int y2)
{
// To store the answer
int ans = 0;
// Loop for each bit
for (int i = 0; i < bitscount; i++)
{
// To store the number of variables
// with ith bit set
int p;
if (x1 == 0 && y1 == 0)
p = prefix_count[i, x2, y2];
else if (x1 == 0)
p = prefix_count[i, x2, y2]
- prefix_count[i, x2, y1 - 1];
else if (y1 == 0)
p = prefix_count[i, x2, y2]
- prefix_count[i, x1 - 1, y2];
else
p = prefix_count[i, x2, y2]
- prefix_count[i, x1 - 1, y2]
- prefix_count[i, x2, y1 - 1]
+ prefix_count[i, x1 - 1, y1 - 1];
// If count of variables with ith bit
// set is greater than 0
if (p != 0)
ans = (ans | (1 << i));
}
return ans;
}
// Driver code
public static void Main (String[] args)
{
int [,]arr = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
findPrefixCount(arr);
int [,]queries = { { 1, 1, 1, 1 }, { 1, 2, 2, 2 } };
int q = queries.GetLength(0);
for (int i = 0; i < q; i++)
Console.WriteLine( rangeOr(queries[i,0],
queries[i,1],
queries[i,2],
queries[i,3]) );
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
<script>
// Javascript implementation of the approach
const bitscount = 32;
const n = 3;
// Array to store bit-wise
// prefix count
let prefix_count = new Array(bitscount);
for (let i = 0; i < bitscount; i++) {
prefix_count[i] = new Array(n);
for (let j = 0; j < n; j++)
prefix_count[i][j] = new Array(n);
}
// Function to find the prefix sum
function findPrefixCount(arr)
{
// Loop for each bit
for (let i = 0; i < bitscount; i++)
{
// Loop to find prefix-count
// for each row
for (let j = 0; j < n; j++) {
prefix_count[i][j][0] =
((arr[j][0] >> i) & 1);
for (let k = 1; k < n; k++)
{
prefix_count[i][j][k] =
((arr[j][k] >> i) & 1);
prefix_count[i][j][k] +=
prefix_count[i][j][k - 1];
}
}
}
// Finding column-wise prefix
// count
for (let i = 0; i < bitscount; i++)
for (let j = 1; j < n; j++)
for (let k = 0; k < n; k++)
prefix_count[i][j][k] +=
prefix_count[i][j - 1][k];
}
// Function to return the result for a query
function rangeOr(x1, y1, x2, y2)
{
// To store the answer
let ans = 0;
// Loop for each bit
for (let i = 0; i < bitscount; i++) {
// To store the number of variables
// with ith bit set
let p;
if (x1 == 0 && y1 == 0)
p = prefix_count[i][x2][y2];
else if (x1 == 0)
p = prefix_count[i][x2][y2]
- prefix_count[i][x2][y1 - 1];
else if (y1 == 0)
p = prefix_count[i][x2][y2]
- prefix_count[i][x1 - 1][y2];
else
p = prefix_count[i][x2][y2]
- prefix_count[i][x1 - 1][y2]
- prefix_count[i][x2][y1 - 1]
+ prefix_count[i][x1 - 1][y1 - 1];
// If count of variables with ith bit
// set is greater than 0
if (p != 0)
ans = (ans | (1 << i));
}
return ans;
}
// Driver code
let arr = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ];
findPrefixCount(arr);
let queries = [ [ 1, 1, 1, 1 ], [ 1, 2, 2, 2 ] ];
let q = queries.length;
for (let i = 0; i < q; i++)
document.write(rangeOr(queries[i][0],
queries[i][1],
queries[i][2],
queries[i][3]) + "<br>");
</script>
Producción:
5 15
La complejidad del tiempo para el cálculo previo es O(n 2 ) y cada consulta se puede responder en O(1)
Espacio Auxiliar: O(n 2 )
Publicación traducida automáticamente
Artículo escrito por DivyanshuShekhar1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA