计算选择给定范围内的K个数组元素的方法

给出三个正整数，L，R，K和阵列ARR []由N个正整数，任务是计数的方式来选择的数量在从范围内具有数值的给定阵列[L至少K个数组元素， R] 。

例子：

Input: arr[] = {12, 4, 6, 13, 5, 10}, K = 3, L = 4, R = 10 
Output: 5 
Explanation: 
Possible ways to select at least K(= 3) array elements having values in the range [4, 10] are: { (arr[1], arr[2], arr[4]), (arr[1], arr[2], arr[5]), (arr[1], arr[4], arr[5]), (arr[2], arr[4], arr[5]), (arr[1], arr[2], arr[4], arr[5]) } 
Therefore, the required output is 5.

Input: arr[] = {1, 2, 3, 4, 5}, K = 4, L = 1, R = 5 
Output: 6 
 

方法：请按照以下步骤解决问题：

• 初始化一个变量，例如cntWays ，以存储选择至少K个具有值的数组元素的方法计数[L，R] 。
• 初始化一个变量，例如cntNum，以将给定数组的数字计数存储在给定数组中，其值位于给定范围内。
• 最后，打印出所有可能值的总和${cntNum\choose k + i}$使得(K + i)小于或等于cntNum 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to calculate factorial
// of all the numbers up to N
vector calculateFactorial(int N)
{
    vector fact(N + 1);
 
    // Factorial of 0 is 1
    fact[0] = 1;
 
    // Calculate factorial of
    // all the numbers upto N
    for (int i = 1; i <= N; i++) {
 
        // Calculate factorial of i
        fact[i] = fact[i - 1] * i;
    }
 
    return fact;
}
 
// Function to find the count of ways to select
// at least K elements whose values in range [L, R]
int cntWaysSelection(int arr[], int N, int K,
                     int L, int R)
{
 
    // Stores count of ways to select at least
    // K elements whose values in range [L, R]
    int cntWays = 0;
 
    // Stores count of numbers having
    // value lies in the range [L, R]
    int cntNum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Checks if the array elements
        // lie in the given range
        if (arr[i] >= L && arr[i] <= R) {
 
            // Update cntNum
            cntNum++;
        }
    }
 
    // Stores factorial of numbers upto N
    vector fact
        = calculateFactorial(cntNum);
 
    // Calculate total ways to select at least
    // K elements whose values lies in [L, R]
    for (int i = K; i <= cntNum; i++) {
 
        // Update cntWays
        cntWays += fact[cntNum] / (fact[i]
                                   * fact[cntNum - i]);
    }
 
    return cntWays;
}
 
// Driver Code
int main()
{
    int arr[] = { 12, 4, 6, 13, 5, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
    int L = 4;
    int R = 10;
 
    cout << cntWaysSelection(arr, N, K, L, R);
}

Java

// Java program to implement
// the above approach
class GFG{
 
// Function to calculate factorial
// of all the numbers up to N
static int[] calculateFactorial(int N)
{
    int []fact = new int[N + 1];
 
    // Factorial of 0 is 1
    fact[0] = 1;
 
    // Calculate factorial of
    // all the numbers upto N
    for (int i = 1; i <= N; i++) {
 
        // Calculate factorial of i
        fact[i] = fact[i - 1] * i;
    }
 
    return fact;
}
 
// Function to find the count of ways to select
// at least K elements whose values in range [L, R]
static int cntWaysSelection(int arr[], int N, int K,
                     int L, int R)
{
 
    // Stores count of ways to select at least
    // K elements whose values in range [L, R]
    int cntWays = 0;
 
    // Stores count of numbers having
    // value lies in the range [L, R]
    int cntNum = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Checks if the array elements
        // lie in the given range
        if (arr[i] >= L && arr[i] <= R) {
 
            // Update cntNum
            cntNum++;
        }
    }
 
    // Stores factorial of numbers upto N
    int []fact
        = calculateFactorial(cntNum);
 
    // Calculate total ways to select at least
    // K elements whose values lies in [L, R]
    for (int i = K; i <= cntNum; i++) {
 
        // Update cntWays
        cntWays += fact[cntNum] / (fact[i]
                                   * fact[cntNum - i]);
    }
 
    return cntWays;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 12, 4, 6, 13, 5, 10 };
    int N = arr.length;
    int K = 3;
    int L = 4;
    int R = 10;
 
    System.out.print(cntWaysSelection(arr, N, K, L, R));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement the
# above approach
 
# Function to calculate factorial
# of all the numbers up to N
def calculateFactorial(N):
 
    fact = [0] * (N + 1)
 
    # Factorial of 0 is 1
    fact[0] = 1
 
    # Calculate factorial of all
    # the numbers upto N
    for i in range(1, N + 1):
 
        # Calculate factorial of i
        fact[i] = fact[i - 1] * i
         
    return fact
 
# Function to find count of ways to select
# at least K elements whose values in range[L,R]
def cntWaysSelection(arr, N, K, L, R):
     
    # Stores count of ways to select at leas
    # K elements whose values in range[L,R]
    cntWays = 0
 
    # Stores count of numbers having
    # Value lies in the range[L,R]
    cntNum = 0
 
    # Traverse the array
    for i in range(0, N):
         
        # Check if the array elements
        # Lie in the given range
        if (arr[i] >= L and arr[i] <= R):
             
            # Update cntNum
            cntNum += 1
 
    # Stores factorial of numbers upto N
    fact = list(calculateFactorial(cntNum))
 
    # Calculate total ways to select at least
    # K elements whose values Lies in [L,R]
    for i in range(K, cntNum + 1):
         
        # Update cntWays
        cntWays += fact[cntNum] // (fact[i] *
                                    fact[cntNum - i])
                                     
    return cntWays
 
# Driver code
if __name__ == "__main__":
     
    arr = [ 12, 4, 6, 13, 5, 10 ]
    N = len(arr)
    K = 3
    L = 4
    R = 10
     
    print(cntWaysSelection(arr, N, K, L, R))
 
# This code is contributed by Virusbuddah

C#

// C# program to implement
// the above approach
using System;
  
class GFG{
  
// Function to calculate factorial
// of all the numbers up to N
static int[] calculateFactorial(int N)
{
    int[] fact = new int[(N + 1)];
     
    // Factorial of 0 is 1
    fact[0] = 1;
     
    // Calculate factorial of
    // all the numbers upto N
    for(int i = 1; i <= N; i++)
    {
         
        // Calculate factorial of i
        fact[i] = fact[i - 1] * i;
    }
    return fact;
}
  
// Function to find the count of ways to select
// at least K elements whose values in range [L, R]
static int cntWaysSelection(int[] arr, int N, int K,
                            int L, int R)
{
     
    // Stores count of ways to select at least
    // K elements whose values in range [L, R]
    int cntWays = 0;
     
    // Stores count of numbers having
    // value lies in the range [L, R]
    int cntNum = 0;
     
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Checks if the array elements
        // lie in the given range
        if (arr[i] >= L && arr[i] <= R)
        {
             
            // Update cntNum
            cntNum++;
        }
    }
  
    // Stores factorial of numbers upto N
    int[] fact = calculateFactorial(cntNum);
  
    // Calculate total ways to select at least
    // K elements whose values lies in [L, R]
    for(int i = K; i <= cntNum; i++)
    {
         
        // Update cntWays
        cntWays += fact[cntNum] / (fact[i] *
                   fact[cntNum - i]);
    }
    return cntWays;
}
  
// Driver Code
public static void Main()
{
    int[] arr = { 12, 4, 6, 13, 5, 10 };
    int N = arr.Length;
    int K = 3;
    int L = 4;
    int R = 10;
     
    Console.WriteLine(cntWaysSelection(
        arr, N, K, L, R));
}
}
 
// This code is contributed by code_hunt

输出：

5

时间复杂度： O(N)
辅助空间： O(N)