Dada una array de enteros positivos y q consulta que contiene dos enteros, L & R. La tarea es encontrar el número de bits establecidos para un rango dado.
Requisito previo: Bitwise Hacks
Ejemplos:
Input : Arr[] = { 1, 5, 6, 10, 9, 4 } Query : 2 L & R 1 5 2 4 Output : 9 6 Input : Arr[] = { 1, 10, 5, 2, 8, 11, 15 } Query : 2 L & R 2 4 1 5 Output : 4 9
La solución simple a este problema es ejecutar un bucle de L a R y contar el número de bits establecidos en un rango. Esta solución toma O(nlog(s)) (donde s es el tamaño de bits) para cada consulta.
La solución eficiente se basa en el hecho de que si almacenamos el recuento de todos los bits de números establecidos en una array «BitCounts», entonces respondemos cada consulta en tiempo O (1). Entonces, comience a recorrer los elementos de la array y cuente los bits establecidos para cada elemento y almacene en la array. Ahora, encuentre la suma acumulada de esta array. Esta array ayudará a responder consultas.
BitCount[] that will store the count of set bits in a number. Run a Loop from 0 to 31 "for 32 bits size integer " -> mark elements with i'th bit set Run an inner Loop from 0 to size of Array "Arr" -> Check whether the current bit is set or not -> if it's set then mark it. long temp = arr[j] >> i; if (temp %2 != 0) BitCount[j] += 1
A continuación se muestra la implementación de la idea anterior.
C++
// C++ program to Range query for // Count number of set bits #include <bits/stdc++.h> using namespace std; // 2-D array that will stored the count // of bits set in element of array int BitCount[10000] = { 0 }; // Function store the set bit // count in BitCount Array void fillSetBitsMatrix(int arr[], int n) { // traverse over all bits for (int i = 0; i < 32; i++) { // mark elements with i'th bit set for (int j = 0; j < n; j++) { // Check whether the current bit is // set or not if it's set then mark it. long temp = arr[j] >> i; if (temp % 2 != 0) BitCount[j] += 1; } } // store cumulative sum of bits for (int i = 1; i < n; i++) BitCount[i] += BitCount[i - 1]; } // Function to process queries void Query(int Q[][2], int q) { for (int i = 0; i < q; i++) cout << (BitCount[Q[i][1]] - BitCount[Q[i][0] - 1]) << endl; } // Driver Code int main() { int Arr[] = { 1, 5, 6, 10, 9, 4, 67 }; int n = sizeof(Arr) / sizeof(Arr[0]); fillSetBitsMatrix(Arr, n); int q = 2; int Q[2][2] = { { 1, 5 }, { 2, 6 } }; Query(Q, q); return 0; }
Java
// Java program to Range query for // Count number of set bits import java.io.*; class GFG { // 2-D array that will stored the count // of bits set in element of array static int BitCount[] = new int[10000]; // Function store the set bit // count in BitCount Array static void fillSetBitsMatrix(int arr[], int n) { // traverse over all bits for (int i = 0; i < 32; i++) { // mark elements with i'th bit set for (int j = 0; j < n; j++) { // Check whether the current // bit is set or not if it's // set then mark it. long temp = arr[j] >> i; if (temp % 2 != 0) BitCount[j] += 1; } } // store cumulative sum of bits for (int i = 1; i < n; i++) BitCount[i] += BitCount[i - 1]; } // Function to process queries static void Query(int Q[][], int q) { for (int i = 0; i < q; i++) System.out.println( (BitCount[Q[i][1]] - BitCount[Q[i][0] - 1])); } // Driver Code public static void main (String[] args) { int Arr[] = { 1, 5, 6, 10, 9, 4, 67 }; int n = Arr.length; fillSetBitsMatrix(Arr, n); int q = 2; int Q[][] = { { 1, 5 }, { 2, 6 } }; Query(Q, q); } } // This code is contributed by anuj_67.
Python3
# Python3 program to Range query for # Count number of set bits # 2-D array that will stored the count # of bits set in element of array BitCount = [0] * 10000 # Function store the set bit # count in BitCount Array def fillSetBitsmatrix(arr: list, n: int): global BitCount # traverse over all bits for i in range(32): # mark elements with i'th bit set for j in range(n): # Check whether the current bit is # set or not if it's set then mark it. temp = arr[j] >> i if temp % 2 != 0: BitCount[j] += 1 # store cumulative sum of bits for i in range(1, n): BitCount[i] += BitCount[i - 1] # Function to process queries def Query(Q: list, q: int): for i in range(q): print(BitCount[Q[i][1]] - BitCount[Q[i][0] - 1]) # Driver Code if __name__ == "__main__": Arr = [1, 5, 6, 10, 9, 4, 67] n = len(Arr) fillSetBitsmatrix(Arr, n) q = 2 Q = [(1, 5), (2, 6)] Query(Q, q) # This code is contributed by # sanjeev2552
C#
// C# program to Range query for // Count number of set bits using System; class GFG { // 2-D array that will stored the count // of bits set in element of array static int []BitCount = new int[10000]; // Function store the set bit // count in BitCount Array static void fillSetBitsMatrix(int []arr, int n) { // traverse over all bits for (int i = 0; i < 32; i++) { // mark elements with i'th bit set for (int j = 0; j < n; j++) { // Check whether the current // bit is set or not if it's // set then mark it. long temp = arr[j] >> i; if (temp % 2 != 0) BitCount[j] += 1; } } // store cumulative sum of bits for (int i = 1; i < n; i++) BitCount[i] += BitCount[i - 1]; } // Function to process queries static void Query(int [,]Q, int q) { for (int i = 0; i < q; i++) Console.WriteLine( (BitCount[Q[i,1]] - BitCount[Q[i,0] - 1])); } // Driver Code public static void Main () { int []Arr = { 1, 5, 6, 10, 9, 4, 67 }; int n = Arr.Length; fillSetBitsMatrix(Arr, n); int q = 2; int [,]Q = { { 1, 5 }, { 2, 6 } }; Query(Q, q); } } // This code is contributed by anuj_67.
Javascript
<script> // Javascript program to Range query for // Count number of set bits // 2-D array that will stored the count // of bits set in element of array var BitCount = Array.from({length: 10000}, (_, i) => 0); // Function store the set bit // count in BitCount Array function fillSetBitsMatrix(arr, n) { // traverse over all bits for(i = 0; i < 32; i++) { // mark elements with i'th bit set for(j = 0; j < n; j++) { // Check whether the current // bit is set or not if it's // set then mark it. var temp = arr[j] >> i; if (temp % 2 != 0) BitCount[j] += 1; } } // store cumulative sum of bits for(i = 1; i < n; i++) BitCount[i] += BitCount[i - 1]; } // Function to process queries function Query(Q, q) { for(i = 0; i < q; i++) document.write((BitCount[Q[i][1]] - BitCount[Q[i][0] - 1]) + "<br>"); } // Driver Code var Arr = [ 1, 5, 6, 10, 9, 4, 67 ]; var n = Arr.length; fillSetBitsMatrix(Arr, n); var q = 2; var Q = [ [ 1, 5 ], [ 2, 6 ] ]; Query(Q, q); // This code is contributed by Rajput-Ji </script>
9 10
Complejidad de tiempo: O(1) para cada consulta.
Publicación traducida automáticamente
Artículo escrito por Nishant_Singh y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA