要插入 Array 以使用子集总和形成 [1, K] 中的所有值的最小元素数

给定一个由N 个整数组成的排序数组arr[]和一个整数K ，任务是找到要插入数组的最小元素数，使得[1, K]范围内的任何值都可以通过添加修改后的数组的任何子集的元素。

例子：

Input: arr[] = {1, 3}, K = 6
Output: 1
Explanation:
Adding the number 2 to the array modifies the array arr[] to {1, 2, 3}. Now, every sum value  over the range [1, 6] can be formed:

1. Sum = 1: The subset is {1}.
2. Sum = 2: The subset is {2}.
3. Sum = 3: The subset is {3}.
4. Sum = 4: The subset is {1, 3}.
5. Sum = 5: The subset is {2, 3}.
6. Sum = 6: The subset is {1, 2, 3}.

Therefore, the minimum number of elements to be added is 1.

Input: arr[] = {1, 5, 10}, K = 20
Output: 2

方法：给定的问题可以通过使用基于以下观察的贪心方法来解决：

1. 将X视为最大数，这样范围[1, X]中的所有数都可以通过对数组的任何子集求和来形成。
2. 然后可以观察到，通过向数组添加整数Y ，范围修改为[1, X+Y]，因为现在可以形成范围[X, X+Y]中的每个数字。
3. 因此，想法是贪婪地添加最能增加范围的元素，并且不会跳过获得的新范围中的任何数字。可以通过每次将X添加到数组并将X修改为2*X 来完成。

请按照以下步骤解决问题：

• 初始化两个整数变量，例如i和count为0 ，以分别存储数组元素的索引和所需元素的计数。
• 初始化一个变量，比如requiredNum为1 ，以存储可以形成每个数字的数字。
• 迭代直到requiredNum ≤ K，并执行以下操作：
  • 如果i < N且requiredNum >= arr[i]则将requiredNum增加arr[i]并将i增加1 。
  • 否则，将requiredNum增加requiredNum ，并将count增加1 。
• 最后，完成上述步骤后，打印count 中得到的答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the count of minimum
// elements to be inserted to form every
// number in a range
int minElements(int arr[], int N, int K)
{
    // Stores the count of numbers needed
    int count = 0;
 
    // Stores the numbers upto which every
    // numbers can be formed
    long long requiredNum = 1;
 
    // Stores the index of the array arr[]
    int i = 0;
 
    // Iterate until requiredSum is less than
    // or equal to K
    while (requiredNum <= K) {
 
        // If i is less than N and requiredSum
        // is greater than or equal to arr[i]
        if (i < N && requiredNum >= arr[i]) {
 
            // Increment requiredSum
            // by arr[i]
            requiredNum += arr[i];
 
            // Increment i by 1
            i++;
        }
 
        // Otherwise
        else {
 
            // Increment count by 1
            count++;
 
            // Increment requiredSum
            // by requiredSum
            requiredNum += requiredNum;
        }
    }
 
    // Return result
    return count;
}
 
// Driver Code
int main()
{
    // Input
    int arr[] = { 1, 3 };
    int K = 6;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << minElements(arr, N, K) << endl;
 
    return 0;
}

Java

// Java program for the above approach
 
import java.io.*;
 
class GFG {
    // Function to find the count of minimum
    // elements to be inserted to form every
    // number in a range
    public static int minElements(int arr[], int N, int K)
    {
        // Stores the count of numbers needed
        int count = 0;
 
        // Stores the numbers upto which every
        // numbers can be formed
        long requiredNum = 1;
 
        // Stores the index of the array arr[]
        int i = 0;
 
        // Iterate until requiredSum is less than
        // or equal to K
        while (requiredNum <= K) {
 
            // If i is less than N and requiredSum
            // is greater than or equal to arr[i]
            if (i < N && requiredNum >= arr[i]) {
 
                // Increment requiredSum
                // by arr[i]
                requiredNum += arr[i];
 
                // Increment i by 1
                i++;
            }
 
            // Otherwise
            else {
 
                // Increment count by 1
                count++;
 
                // Increment requiredSum
                // by requiredSum
                requiredNum += requiredNum;
            }
        }
 
        // Return result
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Input
        int arr[] = { 1, 3 };
        int K = 6;
        int N = arr.length;
 
        // Function Call
        System.out.println(minElements(arr, N, K));
        // This code is contributed by Potta Lokesh
    }
}

Python3

# Python 3 program for the above approach
 
# Function to find the count of minimum
# elements to be inserted to form every
# number in a range
def minElements(arr, N, K):
   
    # Stores the count of numbers needed
    count = 0
 
    # Stores the numbers upto which every
    # numbers can be formed
    requiredNum = 1
 
    # Stores the index of the array arr[]
    i = 0
 
    # Iterate until requiredSum is less than
    # or equal to K
    while (requiredNum <= K):
       
        # If i is less than N and requiredSum
        # is greater than or equal to arr[i]
        if (i < N and requiredNum >= arr[i]):
 
            # Increment requiredSum
            # by arr[i]
            requiredNum += arr[i]
 
            # Increment i by 1
            i += 1
 
        # Otherwise
        else:
            # Increment count by 1
            count += 1
 
            # Increment requiredSum
            # by requiredSum
            requiredNum += requiredNum
 
    # Return result
    return count
 
# Driver Code
if __name__ == '__main__':
    # Input
    arr = [1, 3]
    K = 6
    N = len(arr)
 
    # Function Call
    print(minElements(arr, N, K))
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
 
class GFG {
    // Function to find the count of minimum
    // elements to be inserted to form every
    // number in a range
    public static int minElements(int[] arr, int N, int K)
    {
        // Stores the count of numbers needed
        int count = 0;
 
        // Stores the numbers upto which every
        // numbers can be formed
        long requiredNum = 1;
 
        // Stores the index of the array arr[]
        int i = 0;
 
        // Iterate until requiredSum is less than
        // or equal to K
        while (requiredNum <= K) {
 
            // If i is less than N and requiredSum
            // is greater than or equal to arr[i]
            if (i < N && requiredNum >= arr[i]) {
 
                // Increment requiredSum
                // by arr[i]
                requiredNum += arr[i];
 
                // Increment i by 1
                i++;
            }
 
            // Otherwise
            else {
 
                // Increment count by 1
                count++;
 
                // Increment requiredSum
                // by requiredSum
                requiredNum += requiredNum;
            }
        }
 
        // Return result
        return count;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        // Input
        int[] arr = { 1, 3 };
        int K = 6;
        int N = arr.Length;
 
        // Function Call
        Console.Write(minElements(arr, N, K));
    }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript program for the above approach
 
 
// Function to find the count of minimum
// elements to be inserted to form every
// number in a range
function minElements(arr, N, K) {
    // Stores the count of numbers needed
    let count = 0;
 
    // Stores the numbers upto which every
    // numbers can be formed
    let requiredNum = 1;
 
    // Stores the index of the array arr[]
    let i = 0;
 
    // Iterate until requiredSum is less than
    // or equal to K
    while (requiredNum <= K) {
 
        // If i is less than N and requiredSum
        // is greater than or equal to arr[i]
        if (i < N && requiredNum >= arr[i]) {
 
            // Increment requiredSum
            // by arr[i]
            requiredNum += arr[i];
 
            // Increment i by 1
            i++;
        }
 
        // Otherwise
        else {
 
            // Increment count by 1
            count++;
 
            // Increment requiredSum
            // by requiredSum
            requiredNum += requiredNum;
        }
    }
 
    // Return result
    return count;
}
 
// Driver Code
 
// Input
let arr = [1, 3];
let K = 6;
let N = arr.length;
 
// Function Call
document.write(minElements(arr, N, K) + "<br>");
 
// This code is contributed by _saurabh_jaiswal.

输出

1

时间复杂度： O(N + log(K))
辅助空间： O(1)

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