费拉兰硬币| TCS Mockvita 2020

问题描述

问题解决者找到了一个新的编码岛，并将其命名为Philaland。这些聪明的人被赋予一项任务，通过分发具有不同价值的各种硬币来简化在岛上的物品购买。曼尼什(Manish)提出了一个解决方案，如果我们使硬币的类别从$ 1开始直到Island上商品的最高价格，那么我们可以轻松地购买任何商品。他添加了以下示例来证明自己的观点。

让我们假设一个项目的最高价格是5 $，那么我们可以制作{$ 1，$ 2，$ 3，$ 4，$ 5}的硬币来购买从$ 1到$ 5的任何项目。现在，Manisha是一位敏锐的观察者，他建议我们实际上可以减少所需硬币的数量，并进行以下分配{$ 1，$ 2，$ 3}。据他说，任何物品都可以一次购买，价格从$ 1到$ 5不等。每个人都给他们留下了深刻的印象。

您的任务是帮助Manisha在Philaland中以任意最高价格得出最小面额的货币。

例子：

Input: N = 10
Output: 4
Explanation:
According to Manish {$1, $2, $3, … $10} must be distributed.
But as per Manisha only {$1, $2, $3, $4} coins are enough to purchase any item ranging from $1 to $10. Hence minimum is 4. Likewise denominations could also be {$1, $2, $3, $5}. Hence answer is still 4.

Input: N = 5
Output: 3
Explanation:
According to Manish {$1, $2, $3, $4, $5} must be distributed.
But as per Manisha only {$1, $2, $3} coins are enough to purchase any item ranging from $1 to $5. Hence minimum is 3. Likewise denominations could also be {$1, $2, $4}. Hence answer is still 3.

方法：对该问题的主要观察是，任何数字都可以表示为二的幂。因此，最低面额要求为–

$log_2 N + 1$

例如：

For N = 12, 
If we choose the denominations as {1, 2, 4, 8}
Then every number up to 12 can be represented as –

1 ==> 1
2 ==> 2
3 ==> 2 + 1
4 ==> 4
5 ==> 4 + 1
6 ==> 4 + 2
7 ==> 4 + 2 + 1
8 ==> 8
9 ==> 8 + 1
10 ==> 8 + 2
11 ==> 8 + 2 + 1
12 ==> 8 + 4

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum number of denominations
// required for any number
 
#include 
using namespace std;
 
// Function to find the minimum
// number of denomminations required
int findMinDenomin(int n)
{
    return log2(n) + 1;
}
 
// Driver Code
int main()
{
    int n = 10;
 
    // Function Call
    cout << findMinDenomin(n);
    return 0;
}

Java

// Java implementation to find the
// minimum number of denominations
// required for any number
import java.io.*;
class GFG
{
   
    // Function to find the minimum
    // number of denomminations required
    static int findMinDenomin(int n)
    {
        return ((int)(Math.log(n)/Math.log(2))+1);
    }
   
    // Driver Code
    public static void main (String[] args)
    {
        int n = 10;
       
        // Function Call
        System.out.println(findMinDenomin(n));
    }
}
 
//  This code is contributed by avanitrachhadiya2155

Python3

# Python3 implementation to find the
# minimum number of denominations
# required for any number
from math import log2, floor
 
# Function to find the minimum
# number of denomminations required
def findMinDenomin(n):
 
    return log2(n) + 1
 
# Driver Code
if __name__ == '__main__':
     
    n = 10
 
    # Function call
    print(floor(findMinDenomin(n)))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation to find the
// minimum number of denominations
// required for any number
using System;
class GFG
{
 
  // Function to find the minimum
  // number of denomminations required
  static int findMinDenomin(int n)
  {
    return ((int)(Math.Log(n)/Math.Log(2))+1);
  }
 
  // Driver Code
  static public void Main ()
  {
    int n = 10;
 
    // Function Call
    Console.WriteLine(findMinDenomin(n));
  }
}
 
// This code is contributed by rag2127

输出

4

性能分析：

• 时间复杂度： O(logN)
• 辅助空间： O(1)