给定一个整数N和无限个不同深度的完整二叉树,任务是选择最小数量的树,以使每棵树中的叶节点总数为N。
例子:
Input: N = 7
Output: 3
Trees with depths 2, 1 and 0 can be picked
with the number of leaf nodes as 4, 2 and 1 respectively.
(4 + 2 + 1) = 7
Input: N = 1
Output: 1
方法:由于完整的二叉树中叶节点的数量始终是2的幂。因此,现在问题被简化为寻找2的幂,当将它们相加时会得到N,以使总和中的项总数最小,这是所需的答案。
由于2的每个幂在其二进制表示中仅包含一个’1’,因此N将包含与总和中的项数相同数量的’1’(假设我们采用最少的项数)。因此,问题进一步减少为找到N中的设置位数,使用本文中使用的方法可以很容易地计算出该位数。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the minimum
// count of trees required
int minTrees(int n)
{
// To store the count of
// set bits in n
int count = 0;
while (n) {
n &= (n - 1);
count++;
}
return count;
}
// Driver code
int main()
{
int n = 7;
cout << minTrees(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the minimum
// count of trees required
static int minTrees(int n)
{
// To store the count of
// set bits in n
int count = 0;
while (n > 0)
{
n &= (n - 1);
count++;
}
return count;
}
// Driver code
public static void main(String[] args)
{
int n = 7;
System.out.print(minTrees(n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach
# Function to return the minimum
# count of trees required
def minTrees(n):
# To store the count of
# set bits in n
count = 0;
while (n):
n &= (n - 1);
count += 1;
return count;
# Driver code
if __name__ == '__main__':
n = 7;
print(minTrees(n));
# This code is contributed by 29AjayKumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the minimum
// count of trees required
static int minTrees(int n)
{
// To store the count of
// set bits in n
int count = 0;
while (n > 0)
{
n &= (n - 1);
count++;
}
return count;
}
// Driver code
public static void Main()
{
int n = 7;
Console.Write(minTrees(n));
}
}
// This code is contributed by AnkitRai01
输出:
3