Dada una array arr[] de N enteros positivos, la tarea es responder Q consultas donde cada consulta consta de un rango [L, R] y debe verificar si el AND bit a bit de los elementos del rango de índice dado es par o extraño.
Ejemplos:
Entrada: arr[] = {1, 1, 2, 3}, Q[][] = {{1, 2}, {0, 1}}
Salida:
Par
Impar
(1 y 2) = 0, que es par.
(1 y 1) = 1 que es impar.
Entrada: arr[] = {2, 1, 5, 7, 6, 8, 9}, Q[][] = {{0, 2}, {1, 2}, {2, 3}, {3, 6}}
Salida:
Par Impar
Impar Par
Acercarse:
- El AND bit a bit en un rango de índice [L, R] será impar si todos los elementos en ese rango de índice son impares; de lo contrario, será par.
- Por lo tanto, verifique si todos los elementos en el rango de índice [L, R] son impares o no.
- Para responder a cada consulta de manera óptima, cree una array y marque las posiciones donde el valor es impar y luego tome la suma del prefijo de esta array.
- Ahora, el recuento de números impares en el rango de índice [L, R] se puede calcular como pre[R] – pre[L – 1] .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000005;
int even[MAXN], odd[MAXN];
// Function to precompute the count
// of even and odd numbers
void precompute(int arr[], int n)
{
for (int i = 0; i < n; i++) {
// If the current element is odd
// then put 1 at odd[i]
if (arr[i] % 2 == 1)
odd[i] = 1;
// If the current element is even
// then put 1 at even[i]
if (arr[i] % 2 == 0)
even[i] = 1;
}
// Taking the prefix sums of these two arrays
// so we can get the count of even and odd
// numbers in a range [L, R] in O(1)
for (int i = 1; i < n; i++) {
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
}
}
// Function that returns true if the bitwise
// AND of the subarray a[L...R] is odd
bool isOdd(int L, int R)
{
// cnt will store the count of odd
// numbers in the range [L, R]
int cnt = odd[R];
if (L > 0)
cnt -= odd[L - 1];
// Check if all the numbers in
// the range are odd or not
if (cnt == R - L + 1)
return true;
return false;
}
// Function to perform the queries
void performQueries(int a[], int n, int q[][2], int m)
{
precompute(a, n);
// Perform queries
for (int i = 0; i < m; i++) {
int L = q[i][0], R = q[i][1];
if (isOdd(L, R))
cout << "Odd\n";
else
cout << "Even\n";
}
}
// Driver code
int main()
{
int a[] = { 2, 1, 5, 7, 6, 8, 9 };
int n = sizeof(a) / sizeof(a[0]);
// Queries
int q[][2] = { { 0, 2 }, { 1, 2 }, { 2, 3 }, { 3, 6 } };
int m = sizeof(q) / sizeof(q[0]);
performQueries(a, n, q, m);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
static final int MAXN = 1000005;
static int even[] = new int[MAXN];
static int odd[] = new int[MAXN];
// Function to precompute the count
// of even and odd numbers
static void precompute(int arr[], int n)
{
for (int i = 0; i < n; i++)
{
// If the current element is odd
// then put 1 at odd[i]
if (arr[i] % 2 == 1)
odd[i] = 1;
// If the current element is even
// then put 1 at even[i]
if (arr[i] % 2 == 0)
even[i] = 1;
}
// Taking the prefix sums of these two arrays
// so we can get the count of even and odd
// numbers in a range [L, R] in O(1)
for (int i = 1; i < n; i++)
{
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
}
}
// Function that returns true if the bitwise
// AND of the subarray a[L...R] is odd
static boolean isOdd(int L, int R)
{
// cnt will store the count of odd
// numbers in the range [L, R]
int cnt = odd[R];
if (L > 0)
cnt -= odd[L - 1];
// Check if all the numbers in
// the range are odd or not
if (cnt == R - L + 1)
return true;
return false;
}
// Function to perform the queries
static void performQueries(int a[], int n,
int q[][], int m)
{
precompute(a, n);
// Perform queries
for (int i = 0; i < m; i++)
{
int L = q[i][0], R = q[i][1];
if (isOdd(L, R))
System.out.println("Odd");
else
System.out.println("Even");
}
}
// Driver code
public static void main(String args[])
{
int []a = { 2, 1, 5, 7, 6, 8, 9 };
int n = a.length;
// Queries
int q[][] = { { 0, 2 }, { 1, 2 }, { 2, 3 }, { 3, 6 } };
int m = q.length;
performQueries(a, n, q, m);
}
}
// This code is contributed by AnkitRai01
Python3
# Python implementation of the approach
MAXN = 1000005;
even = [0] * MAXN;
odd = [0] * MAXN;
# Function to precompute the count
# of even and odd numbers
def precompute(arr, n):
for i in range(n):
# If the current element is odd
# then put 1 at odd[i]
if (arr[i] % 2 == 1):
odd[i] = 1;
# If the current element is even
# then put 1 at even[i]
if (arr[i] % 2 == 0):
even[i] = 1;
# Taking the prefix sums of these two arrays
# so we can get the count of even and odd
# numbers in a range [L, R] in O(1)
for i in range(1, n):
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
# Function that returns True if the bitwise
# AND of the subarray a[L...R] is odd
def isOdd(L, R):
# cnt will store the count of odd
# numbers in the range [L, R]
cnt = odd[R];
if (L > 0):
cnt -= odd[L - 1];
# Check if all the numbers in
# the range are odd or not
if (cnt == R - L + 1):
return True;
return False;
# Function to perform the queries
def performQueries(a, n, q, m):
precompute(a, n);
# Perform queries
for i in range(m):
L = q[i][0];
R = q[i][1];
if (isOdd(L, R)):
print("Odd");
else:
print("Even");
# Driver code
if __name__ == '__main__':
a = [ 2, 1, 5, 7, 6, 8, 9 ];
n = len(a);
# Queries
q = [[ 0, 2 ],[ 1, 2 ],[ 2, 3 ],[ 3, 6 ]];
m = len(q);
performQueries(a, n, q, m);
# This code is contributed by PrinciRaj1992
C#
// C# implementation of the approach
using System;
class GFG
{
static readonly int MAXN = 1000005;
static int []even = new int[MAXN];
static int []odd = new int[MAXN];
// Function to precompute the count
// of even and odd numbers
static void precompute(int []arr, int n)
{
for (int i = 0; i < n; i++)
{
// If the current element is odd
// then put 1 at odd[i]
if (arr[i] % 2 == 1)
odd[i] = 1;
// If the current element is even
// then put 1 at even[i]
if (arr[i] % 2 == 0)
even[i] = 1;
}
// Taking the prefix sums of these two arrays
// so we can get the count of even and odd
// numbers in a range [L, R] in O(1)
for (int i = 1; i < n; i++)
{
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
}
}
// Function that returns true if the bitwise
// AND of the subarray a[L...R] is odd
static bool isOdd(int L, int R)
{
// cnt will store the count of odd
// numbers in the range [L, R]
int cnt = odd[R];
if (L > 0)
cnt -= odd[L - 1];
// Check if all the numbers in
// the range are odd or not
if (cnt == R - L + 1)
return true;
return false;
}
// Function to perform the queries
static void performQueries(int []a, int n,
int [,]q, int m)
{
precompute(a, n);
// Perform queries
for (int i = 0; i < m; i++)
{
int L = q[i, 0], R = q[i, 1];
if (isOdd(L, R))
Console.WriteLine("Odd");
else
Console.WriteLine("Even");
}
}
// Driver code
public static void Main(String []args)
{
int []a = { 2, 1, 5, 7, 6, 8, 9 };
int n = a.Length;
// Queries
int [,]q = { { 0, 2 }, { 1, 2 },
{ 2, 3 }, { 3, 6 } };
int m = q.GetLength(0);
performQueries(a, n, q, m);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript implementation of the approach
var MAXN = 1000005;
var even = Array(MAXN).fill(0);
var odd = Array(MAXN).fill(0);
// Function to precompute the count
// of even and odd numbers
function precompute(arr, n)
{
for (var i = 0; i < n; i++) {
// If the current element is odd
// then put 1 at odd[i]
if (arr[i] % 2 == 1)
odd[i] = 1;
// If the current element is even
// then put 1 at even[i]
if (arr[i] % 2 == 0)
even[i] = 1;
}
// Taking the prefix sums of these two arrays
// so we can get the count of even and odd
// numbers in a range [L, R] in O(1)
for (var i = 1; i < n; i++) {
even[i] = even[i] + even[i - 1];
odd[i] = odd[i] + odd[i - 1];
}
}
// Function that returns true if the bitwise
// AND of the subarray a[L...R] is odd
function isOdd(L, R)
{
// cnt will store the count of odd
// numbers in the range [L, R]
var cnt = odd[R];
if (L > 0)
cnt -= odd[L - 1];
// Check if all the numbers in
// the range are odd or not
if (cnt == R - L + 1)
return true;
return false;
}
// Function to perform the queries
function performQueries(a, n, q, m)
{
precompute(a, n);
// Perform queries
for (var i = 0; i < m; i++)
{
var L = q[i][0], R = q[i][1];
if (isOdd(L, R))
document.write( "Odd"+"<br>");
else
document.write("Even"+"<br>");
}
}
// Driver code
var a = [2, 1, 5, 7, 6, 8, 9];
var n = a.length;
// Queries
var q = [ [ 0, 2 ], [ 1, 2 ], [ 2, 3 ], [ 3, 6 ] ];
var m = q.length;
performQueries(a, n, q, m);
// This code is contributed by itsok.
</script>
Producción:
Even Odd Odd Even
Complejidad temporal: O(Q) donde Q es el número de consultas.