在给定范围[L,R]的情况下,任务是找到不一定是不同的对(X,Y)。找到所选整数的按位与的最大可能值。
例子:
Input: L = 3, R = 7
Output: 7
Explanation:
In all the possible pairs, pair (7, 7) gives the maximum value for bitwise AND.
Input: L = 54, R = 55
Output: 55
Explanation:
In all the possible pairs, pair (55, 55) gives the maximum value for bitwise AND.
幼稚的方法:为了解决上述问题,幼稚的方法是从L迭代到R,并检查每个可能的对的按位与,并在最后打印最大值。
时间复杂度: O(N 2 )
高效方法:
为了优化上述方法,我们必须观察到,这里我们必须对L和R进行整数处理,并且必须从区间[L,R]中选择两个整数,以使它们的按位与最大。 L和R之间的任何两个数字的按位与将始终始终小于或等于R。因此,如果必须从间隔中选择两个整数,则可以将整数选择为R,这是最大化按位与的唯一方法。
下面是上述方法的实现:
C
// C implementation to find the
// Maximum Bitwise AND pair (X, Y)
// from given range such that
// X and Y can be same
#include
// Function to return the
// maximum bitwise AND
int maximumAND(int L, int R)
{
return R;
}
// Driver code
int main()
{
int l = 3;
int r = 7;
printf("%d", maximumAND(l, r));
return 0;
}
C++
// C++ implementation to find the
// Maximum Bitwise AND pair (X, Y)
// from given range such that
// X and Y can be same
#include
using namespace std;
// Function to return the
// maximum bitwise AND
int maximumAND(int L, int R)
{
return R;
}
// Driver code
int main()
{
int l = 3;
int r = 7;
cout << maximumAND(l, r);
return 0;
}
Java
// Java implementation to find the
// Maximum Bitwise AND pair (X, Y)
// from given range such that
// X and Y can be same
class GFG {
// Function to return the
// maximum bitwise AND
static int maximumAND(int L, int R)
{
return R;
}
// Driver code
public static void main(String[] args)
{
int l = 3;
int r = 7;
System.out.print(maximumAND(l, r));
}
}
Python3
# Python3 implementation to find the
# Maximum Bitwise AND pair (X, Y)
# from given range such that
# X and Y can be same
# Function to return the
# maximum bitwise AND
def maximumAND(L, R):
return R
# Driver code
if __name__ == '__main__':
l = 3
r = 7
print(maximumAND(l, r))
C#
// C# implementation to find the
// maximum Bitwise AND pair (X, Y)
// from given range such that
// X and Y can be same
using System;
class GFG{
// Function to return the
// maximum bitwise AND
static int maximumAND(int L, int R)
{
return R;
}
// Driver code
public static void Main(String[] args)
{
int l = 3;
int r = 7;
Console.Write(maximumAND(l, r));
}
}
// This code is contributed by amal kumar choubey
输出:
7
时间复杂度: O(1)