Dado un rango [L, R] , la tarea es encontrar un par (X, Y) tal que L ≤ X < Y ≤ R y X & Y sea el máximo entre todos los pares posibles, luego imprima el AND bit a bit del encontrado par.
Ejemplos:
Entrada: L = 1, R = 9
Salida: 8
En todos los pares posibles, el par (8, 9) da el valor máximo para AND bit a bit.Entrada: L = 523641, R = 985624
Salida: 985622
Enfoque ingenuo: iterar de L a R y verificar el AND bit a bit para cada par posible e imprimir el valor máximo al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum bitwise AND
// possible among all the possible pairs
int maxAND(int L, int R)
{
int maximum = L & R;
for (int i = L; i < R; i++)
for (int j = i + 1; j <= R; j++)
// Maximum among all (i, j) pairs
maximum = max(maximum, (i & j));
return maximum;
}
// Driver code
int main()
{
int L = 1, R = 632;
cout << maxAND(L, R);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the maximum bitwise AND
// possible among all the possible pairs
static int maxAND(int L, int R)
{
int maximum = L & R;
for (int i = L; i < R; i++)
for (int j = i + 1; j <= R; j++)
// Maximum among all (i, j) pairs
maximum = Math.max(maximum, (i & j));
return maximum;
}
// Driver code
public static void main(String[] args)
{
int L = 1, R = 632;
System.out.println(maxAND(L, R));
}
}
// This code contributed by Rajput-Ji
Python3
# Python 3 implementation of the approach # Function to return the maximum bitwise AND # possible among all the possible pairs def maxAND(L, R): maximum = L & R for i in range(L, R, 1): for j in range(i + 1, R + 1, 1): # Maximum among all (i, j) pairs maximum = max(maximum, (i & j)) return maximum # Driver code if __name__ == '__main__': L = 1 R = 632 print(maxAND(L, R)) # This code is contributed by # Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the maximum bitwise AND
// possible among all the possible pairs
static int maxAND(int L, int R)
{
int maximum = L & R;
for (int i = L; i < R; i++)
for (int j = i + 1; j <= R; j++)
// Maximum among all (i, j) pairs
maximum = Math.Max(maximum, (i & j));
return maximum;
}
// Driver code
public static void Main(String[] args)
{
int L = 1, R = 632;
Console.WriteLine(maxAND(L, R));
}
}
// This code has been contributed by 29AjayKumar
PHP
<?php
// PHP implementation of the approach
// Function to return the maximum bitwise AND
// possible among all the possible pairs
function maxAND($L, $R)
{
$maximum = $L & $R;
for ($i = $L; $i < $R; $i++)
for ($j = $i + 1; $j <= $R; $j++)
// Maximum among all (i, j) pairs
$maximum = max($maximum, ($i & $j));
return $maximum;
}
// Driver code
$L = 1; $R = 632;
echo(maxAND($L, $R));
// This code contributed by Code_Mech.
?>
Javascript
<script>
// JavaScript implementation of the approach
// Function to return the maximum bitwise AND
// possible among all the possible pairs
function maxAND(L, R)
{
var maximum = L & R;
for (var i = L; i < R; i++)
for (var j = i + 1; j <= R; j++)
// Maximum among all (i, j) pairs
maximum = Math.max(maximum, (i & j));
return maximum;
}
// Driver code
var L = 1, R = 632;
document.write( maxAND(L, R));
</script>
Producción
630
Complejidad temporal: O(R*R)
Espacio auxiliar: O(1)
Enfoque eficiente: si observamos cuidadosamente en el rango [L, R], el AND máximo es posible entre el AND de (R) y (R – 1) o el AND de (R – 1) y (R – 2) .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum bitwise AND
// possible among all the possible pairs
int maxAND(int L, int R)
{
// If there is only a single value
// in the range [L, R]
if (L == R)
return L;
// If there are only two values
// in the range [L, R]
else if ((R - L) == 1)
return (R & L);
else {
if (((R - 1) & R) > ((R - 2) & (R - 1)))
return ((R - 1) & R);
else
return ((R - 2) & (R - 1));
}
}
// Driver code
int main()
{
int L = 1, R = 632;
cout << maxAND(L, R);
return 0;
}
Java
// Java implementation of the approach
class GfG
{
// Function to return the maximum bitwise AND
// possible among all the possible pairs
static int maxAND(int L, int R)
{
// If there is only a single value
// in the range [L, R]
if (L == R)
return L;
// If there are only two values
// in the range [L, R]
else if ((R - L) == 1)
return (R & L);
else {
if (((R - 1) & R) > ((R - 2) & (R - 1)))
return ((R - 1) & R);
else
return ((R - 2) & (R - 1));
}
}
// Driver code
public static void main(String[] args)
{
int L = 1, R = 632;
System.out.println(maxAND(L, R));
}
}
// This code is contributed by
// Prerna Saini
Python3
# Python3 implementation of the approach # Function to return the maximum bitwise AND # possible among all the possible pairs def maxAND(L, R): # If there is only a single value # in the range [L, R] if (L == R): return L; # If there are only two values # in the range [L, R] elif ((R - L) == 1): return (R & L); else: if (((R - 1) & R) > ((R - 2) & (R - 1))): return ((R - 1) & R); else: return ((R - 2) & (R - 1)); # Driver code L = 1; R = 632; print(maxAND(L, R)); # This code contributed by PrinciRaj1992
C#
// C# implementation of the approach
using System;
class GfG
{
// Function to return the maximum bitwise AND
// possible among all the possible pairs
static int maxAND(int L, int R)
{
// If there is only a single value
// in the range [L, R]
if (L == R)
return L;
// If there are only two values
// in the range [L, R]
else if ((R - L) == 1)
return (R & L);
else
{
if (((R - 1) & R) > ((R - 2) & (R - 1)))
return ((R - 1) & R);
else
return ((R - 2) & (R - 1));
}
}
// Driver code
public static void Main()
{
int L = 1, R = 632;
Console.WriteLine(maxAND(L, R));
}
}
// This code is contributed by Ryuga
PHP
<?php
// PHP implementation of the approach
// Function to return the maximum bitwise AND
// possible among all the possible pairs
function maxAND($L, $R)
{
// If there is only a single value
// in the range [L, R]
if ($L == $R)
return $L;
// If there are only two values
// in the range [L, R]
else if (($R - $L) == 1)
return ($R & $L);
else
{
if ((($R - 1) & $R) > (($R - 2) & ($R - 1)))
return (($R - 1) & $R);
else
return (($R - 2) & ($R - 1));
}
}
// Driver code
$L = 1;
$R = 632;
echo maxAND($L, $R);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript implementation of the approach
// Function to return the maximum bitwise AND
// possible among all the possible pairs
function maxAND(L, R)
{
// If there is only a single value
// in the range [L, R]
if (L == R)
return L;
// If there are only two values
// in the range [L, R]
else if ((R - L) == 1)
return (R & L);
else
{
if (((R - 1) & R) > ((R - 2) & (R - 1)))
return ((R - 1) & R);
else
return ((R - 2) & (R - 1));
}
}
let L = 1, R = 632;
document.write(maxAND(L, R));
</script>
Producción
630
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por Chandan_Agrawal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA