Dada una array binaria arr[] y q consultas de los siguientes tipos:
- k: encuentre el índice del k -ésimo conjunto de bits, es decir , k -ésimo 1 en la array.
- (x, y): actualice arr[x] = y donde y puede ser 0 o 1 .
Ejemplos:
Entrada: arr[] = {1, 0, 1, 0, 0, 1, 1, 1}, q = 2
k = 4
(x, y) = (5, 1)
Salida:
Índice del 4.º conjunto de bits: 6
Array después de la actualización:
1 0 1 0 0 0 1 1
Enfoque: una solución simple es ejecutar un ciclo de 0 a n – 1 y encontrar el k -ésimo 1 . Para actualizar un valor, simplemente haga arr[i] = x . La primera operación toma el tiempo O(n) y la segunda operación toma el tiempo O(1) .
Otra solución es crear otra array y almacenar el índice de cada 1 en el i -ésimo índice de esta array. El índice de Kth 1 ahora se puede calcular en el tiempo O(1) , pero la operación de actualización ahora toma el tiempo O(n) . Esto funciona bien si la cantidad de operaciones de consulta es grande y hay muy pocas actualizaciones.
Pero, ¿y si el número de consultas y actualizaciones es igual? Podemos realizar ambas operaciones en tiempo O (log n) utilizando un árbol de segmentos para realizar ambas operaciones en tiempo O (Logn) .
Representación del árbol de segmentos:
- Los Nodes hoja son los elementos de la array de entrada.
- Cada Node interno representa la suma de la fusión de los Nodes hoja.
Se utiliza una representación de array de árbol para representar árboles de segmento. Para cada Node en el índice i, el hijo izquierdo está en el índice 2*i+1, el hijo derecho en 2*i+2 y el padre está en (i-1)/2.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to build the tree
void buildTree(int* tree, int* a, int s, int e, int idx)
{
// s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e) {
// store each element value at the
// leaf node
tree[idx] = a[s];
return;
}
int mid = (s + e) / 2;
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1);
// Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return;
}
// Function to return the index of the query
int queryTree(int* tree, int s, int e, int k, int idx)
{
// s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return -1;
// leaf node at which kth 1 is stored
if (s == e)
return s;
int mid = (s + e) / 2;
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1, e, k - tree[2 * idx], 2 * idx + 1);
}
// Function to perform the update query
void updateTree(int* tree, int s, int e, int i, int change, int idx)
{
// s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e) {
cout << "error";
return;
}
// Leaf node of the required index i
if (s == e) {
// Replacing the node value with
// the new changed value
tree[idx] = change;
return;
}
int mid = (s + e) / 2;
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i, change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1, e, i, change, 2 * idx + 1);
// Merging both left and right sub-trees
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return;
}
// Function to perform queries
void queries(int* tree, int* a, int q, int p, int k, int change, int n)
{
int s = 0, e = n - 1, idx = 1;
if (q == 1) {
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
cout << "Array after updation:\n";
for (int i = 0; i < n; i++)
cout << a[i] << " ";
cout << "\n";
}
else {
// q = 0, print kth bit
cout << "Index of " << k << "th set bit: "
<< queryTree(tree, s, e, k, idx) << "\n";
}
}
// Driver code
int main()
{
int a[] = { 1, 0, 1, 0, 0, 1, 1, 1 };
int n = sizeof(a) / sizeof(int);
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
int* tree = new int[4 * n + 1];
for (int i = 0; i < 4 * n + 1; ++i) {
tree[i] = 0;
}
// s and e are the starting and ending
// indices respectively
int s = 0, e = n - 1, idx = 1;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
int q = 0, p = 0, change = 0, k = 4;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1, p = 5, change = 0;
queries(tree, a, q, p, k, change, n);
return 0;
}
Java
// Java implementation of the approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to build the tree
static void buildTree(int[] tree, int[] a,
int s, int e, int idx)
{
// s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e)
{
// store each element value at the
// leaf node
tree[idx] = a[s];
return;
}
int mid = (s + e) / 2;
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1);
// Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return;
}
// Function to return the index of the query
static int queryTree(int[] tree, int s,
int e, int k, int idx)
{
// s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return -1;
// leaf node at which kth 1 is stored
if (s == e)
return s;
int mid = (s + e) / 2;
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1, e, k - tree[2 * idx],
2 * idx + 1);
}
// Function to perform the update query
static void updateTree(int[] tree, int s, int e,
int i, int change, int idx)
{
// s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e)
{
System.out.println("Error");
return;
}
// Leaf node of the required index i
if (s == e)
{
// Replacing the node value with
// the new changed value
tree[idx] = change;
return;
}
int mid = (s + e) / 2;
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i, change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1, e, i, change, 2 * idx + 1);
// Merging both left and right sub-trees
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return;
}
// Function to perform queries
static void queries(int[] tree, int[] a, int q, int p,
int k, int change, int n)
{
int s = 0, e = n - 1, idx = 1;
if (q == 1)
{
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
System.out.println("Array after updation:");
for (int i = 0; i < n; i++)
System.out.print(a[i] + " ");
System.out.println();
}
else
{
// q = 0, print kth bit
System.out.printf("Index of %dth set bit: %d\n",
k, queryTree(tree, s, e, k, idx));
}
}
// Driver Code
public static void main(String[] args)
{
int[] a = { 1, 0, 1, 0, 0, 1, 1, 1 };
int n = a.length;
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
int[] tree = new int[4 * n + 1];
for (int i = 0; i < 4 * n + 1; ++i)
{
tree[i] = 0;
}
// s and e are the starting and ending
// indices respectively
int s = 0, e = n - 1, idx = 1;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
int q = 0, p = 0, change = 0, k = 4;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1;
p = 5;
change = 0;
queries(tree, a, q, p, k, change, n);
}
}
// This code is contributed by
// sanjeev2552
Python3
# Python3 implementation of the approach
# Function to build the tree
def buildTree(tree: list, a: list,
s: int, e: int, idx: int):
# s = starting index
# e = ending index
# a = array storing binary string of numbers
# idx = starting index = 1, for recurring the tree
if s == e:
# store each element value at the
# leaf node
tree[idx] = a[s]
return
mid = (s + e) // 2
# Recurring in two sub portions (left, right)
# to build the segment tree
# Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx)
# Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1)
# Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1]
return
# Function to return the index of the query
def queryTree(tree: list, s: int, e: int,
k: int, idx: int) -> int:
# s = starting index
# e = ending index
# k = searching for kth bit
# idx = starting index = 1, for recurring the tree
# k > number of 1's in a binary string
if k > tree[idx]:
return -1
# leaf node at which kth 1 is stored
if s == e:
return s
mid = (s + e) // 2
# If left sub-tree contains more or equal 1's
# than required kth 1
if tree[2 * idx] >= k:
return queryTree(tree, s, mid, k, 2 * idx)
# If left sub-tree contains less 1's than
# required kth 1 then recur in the right sub-tree
else:
return queryTree(tree, mid + 1, e,
k - tree[2 * idx], 2 * idx + 1)
# Function to perform the update query
def updateTree(tree: list, s: int, e: int,
i: int, change: int, idx: int):
# s = starting index
# e = ending index
# i = index at which change is to be done
# change = new changed bit
# idx = starting index = 1, for recurring the tree
# Out of bounds request
if i < s or i > e:
print("error")
return
# Leaf node of the required index i
if s == e:
# Replacing the node value with
# the new changed value
tree[idx] = change
return
mid = (s + e) // 2
# If the index i lies in the left sub-tree
if i >= s and i <= mid:
updateTree(tree, s, mid, i, change, 2 * idx)
# If the index i lies in the right sub-tree
else:
updateTree(tree, mid + 1, e, i, change, 2 * idx + 1)
# Merging both left and right sub-trees
tree[idx] = tree[2 * idx] + tree[2 * idx + 1]
return
# Function to perform queries
def queries(tree: list, a: list, q: int, p: int,
k: int, change: int, n: int):
s = 0
e = n - 1
idx = 1
if q == 1:
# q = 1 update, p = index at which change
# is to be done, change = new bit
a[p] = change
updateTree(tree, s, e, p, change, idx)
print("Array after updation:")
for i in range(n):
print(a[i], end=" ")
print()
else:
# q = 0, print kth bit
print("Index of %dth set bit: %d" % (k,
queryTree(tree, s, e, k, idx)))
# Driver Code
if __name__ == "__main__":
a = [1, 0, 1, 0, 0, 1, 1, 1]
n = len(a)
# Declaring & initializing the tree with
# maximum possible size of the segment tree
# and each value initially as 0
tree = [0] * (4 * n + 1)
# s and e are the starting and ending
# indices respectively
s = 0
e = n - 1
idx = 1
# Build the segment tree
buildTree(tree, a, s, e, idx)
# Find index of kth set bit
q = 0
p = 0
change = 0
k = 4
queries(tree, a, q, p, k, change, n)
# Update query
q = 1
p = 5
change = 0
queries(tree, a, q, p, k, change, n)
# This code is contributed by
# sanjeev2552
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to build the tree
static void buildTree(int[] tree, int[] a,
int s, int e, int idx)
{
// s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e)
{
// store each element value at the
// leaf node
tree[idx] = a[s];
return;
}
int mid = (s + e) / 2;
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1);
// Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return;
}
// Function to return the index of the query
static int queryTree(int[] tree, int s,
int e, int k, int idx)
{
// s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return -1;
// leaf node at which kth 1 is stored
if (s == e)
return s;
int mid = (s + e) / 2;
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1,
e, k - tree[2 * idx],
2 * idx + 1);
}
// Function to perform the update query
static void updateTree(int[] tree, int s, int e,
int i, int change, int idx)
{
// s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e)
{
Console.WriteLine("Error");
return;
}
// Leaf node of the required index i
if (s == e)
{
// Replacing the node value with
// the new changed value
tree[idx] = change;
return;
}
int mid = (s + e) / 2;
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i,
change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1, e, i,
change, 2 * idx + 1);
// Merging both left and right sub-trees
tree[idx] = tree[2 * idx] +
tree[2 * idx + 1];
return;
}
// Function to perform queries
static void queries(int[] tree, int[] a, int q, int p,
int k, int change, int n)
{
int s = 0, e = n - 1, idx = 1;
if (q == 1)
{
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
Console.WriteLine("Array after updation:");
for (int i = 0; i < n; i++)
Console.Write(a[i] + " ");
Console.WriteLine();
}
else
{
// q = 0, print kth bit
Console.Write("Index of {0}th set bit: {1}\n",
k, queryTree(tree, s, e, k, idx));
}
}
// Driver Code
public static void Main(String[] args)
{
int[] a = { 1, 0, 1, 0, 0, 1, 1, 1 };
int n = a.Length;
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
int[] tree = new int[4 * n + 1];
for (int i = 0; i < 4 * n + 1; ++i)
{
tree[i] = 0;
}
// s and e are the starting and ending
// indices respectively
int s = 0, e = n - 1, idx = 1;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
int q = 0, p = 0, change = 0, k = 4;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1;
p = 5;
change = 0;
queries(tree, a, q, p, k, change, n);
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// JavaScript implementation of the approach
// Function to build the tree
function buildTree(tree, a, s, e, idx)
{
// s = starting index
// e = ending index
// a = array storing binary string of numbers
// idx = starting index = 1, for recurring the tree
if (s == e) {
// store each element value at the
// leaf node
tree[idx] = a[s];
return;
}
let mid = Math.floor((s + e) / 2);
// Recurring in two sub portions (left, right)
// to build the segment tree
// Calling for the left sub portion
buildTree(tree, a, s, mid, 2 * idx);
// Calling for the right sub portion
buildTree(tree, a, mid + 1, e, 2 * idx + 1);
// Summing up the number of one's
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return;
}
// Function to return the index of the query
function queryTree(tree, s, e, k, idx)
{
// s = starting index
// e = ending index
// k = searching for kth bit
// idx = starting index = 1, for recurring the tree
// k > number of 1's in a binary string
if (k > tree[idx])
return -1;
// leaf node at which kth 1 is stored
if (s == e)
return s;
let mid = Math.floor((s + e) / 2);
// If left sub-tree contains more or equal 1's
// than required kth 1
if (tree[2 * idx] >= k)
return queryTree(tree, s, mid, k, 2 * idx);
// If left sub-tree contains less 1's than
// required kth 1 then recur in the right sub-tree
else
return queryTree(tree, mid + 1, e, k -
tree[2 * idx], 2 * idx + 1);
}
// Function to perform the update query
function updateTree(tree, s, e, i, change, idx)
{
// s = starting index
// e = ending index
// i = index at which change is to be done
// change = new changed bit
// idx = starting index = 1, for recurring the tree
// Out of bounds request
if (i < s || i > e) {
document.write("error");
return;
}
// Leaf node of the required index i
if (s == e) {
// Replacing the node value with
// the new changed value
tree[idx] = change;
return;
}
let mid = Math.floor((s + e) / 2);
// If the index i lies in the left sub-tree
if (i >= s && i <= mid)
updateTree(tree, s, mid, i, change, 2 * idx);
// If the index i lies in the right sub-tree
else
updateTree(tree, mid + 1, e, i, change, 2 * idx + 1);
// Merging both left and right sub-trees
tree[idx] = tree[2 * idx] + tree[2 * idx + 1];
return;
}
// Function to perform queries
function queries(tree, a, q, p, k, change, n)
{
let s = 0, e = n - 1, idx = 1;
if (q == 1) {
// q = 1 update, p = index at which change
// is to be done, change = new bit
a[p] = change;
updateTree(tree, s, e, p, change, idx);
document.write("Array after updation:<br>");
for (let i = 0; i < n; i++)
document.write(a[i] + " ");
document.write("<br>");
}
else {
// q = 0, print kth bit
document.write("Index of " + k +
"th set bit: " + queryTree(tree, s, e, k, idx) + "<br>");
}
}
// Driver code
let a = [ 1, 0, 1, 0, 0, 1, 1, 1 ];
let n = a.length;
// Declaring & initializing the tree with
// maximum possible size of the segment tree
// and each value initially as 0
let tree = new Array(4 * n + 1);
for (let i = 0; i < 4 * n + 1; ++i) {
tree[i] = 0;
}
// s and e are the starting and ending
// indices respectively
let s = 0, e = n - 1, idx = 1;
// Build the segment tree
buildTree(tree, a, s, e, idx);
// Find index of kth set bit
let q = 0, p = 0, change = 0, k = 4;
queries(tree, a, q, p, k, change, n);
// Update query
q = 1, p = 5, change = 0;
queries(tree, a, q, p, k, change, n);
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
Index of 4th set bit: 6 Array after updation: 1 0 1 0 0 0 1 1
Complejidad de tiempo: O (logN) por consulta y O (NlogN) para construir el árbol de segmentos.
Espacio Auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por NikhilRathor y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA