给定范围内的整数计数，仅包含给定的一组数字

给定两个整数L和R以及一个包含单个数字整数的数组arr[] ，任务是从给定的数字数组中找到由数字组成的范围 [L, R) 中的所有整数。

例子：

Input: L = 1, R = 100, arr[] = {2, 3, 5, 7}
Output: 20
Explanation: The number between 1 and 100 total integers which are made up with 2, 3, 5, 7 are:
2, 3, 5, 7, 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, and 77. Total 20.

Input: L = 50, R = 60, arr[] = 5
Output: 1
Explanation: The only number in range 50 and 60 55.

方法：该问题的解决方案基于贪心方法，使用以下思想：

Traverse each integer in range [L, R) and check if it consists of only given set of digits. 

请按照以下步骤解决问题：

• 迭代范围[L, R) 。
• 在集合的帮助下检查数字是否是arr[]中给出的数字的组合。
• 如果它是给定数字的子集，则将count增加 1，否则不增加。

下面是上述方法的实现：

C++

// C++ code to implement the approach
 
#include 
using namespace std;
 
// Function Check number is
// subset of prime digit of not
bool has_val(int x, set& st)
{
    while (x != 0) {
        if (st.find(x % 10) == st.end())
            return false;
        x /= 10;
    }
    return true;
}
 
// Function to find
// non-prime between range
int total_Num(int A, int B, int arr[], int N)
{
    int ans = 0;
    set st;
    for(int i = 0; i < N; i++)
        st.insert(arr[i]);
   
    // Loop to check if number contains
    // only the digits in given set
    for (int k = A; k < B; k++) {
        if (has_val(k, st))
            ans += 1;
    }
    return ans;
}
 
// Driver Code
int main()
{
    int L = 1, R = 100;
    int arr[] = { 2, 3, 5, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
   
    int ans = total_Num(L, R, arr, N);
    cout << ans;
    return 0;
}

Java

// Java code to implement the approach
import java.util.*;
public class GFG {
 
  // Function Check number is
  // subset of prime digit of not
  static boolean has_val(int x, HashSet st)
  {
    while (x != 0) {
      if (st.contains(x % 10) == false)
        return false;
      x /= 10;
    }
    return true;
  }
 
  // Function to find
  // non-prime between range
  static int total_Num(int A, int B, int arr[], int N)
  {
    int ans = 0;
    HashSet st = new HashSet<>();
    for (int i = 0; i < N; i++)
      st.add(arr[i]);
 
    // Loop to check if number contains
    // only the digits in given set
    for (int k = A; k < B; k++) {
      if (has_val(k, st))
        ans += 1;
    }
    return ans;
  }
 
  // Driver Code
  public static void main(String args[])
  {
    int L = 1, R = 100;
    int[] arr = { 2, 3, 5, 7 };
    int N = arr.length;
 
    int ans = total_Num(L, R, arr, N);
    System.out.print(ans);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# python3 code to implement the approach
 
# Function Check number is
# subset of prime digit of not
def has_val(x, st):
 
    while (x != 0):
        if (not x % 10 in st):
            return False
        x //= 10
 
    return True
 
# Function to find
# non-prime between range
def total_Num(A, B, arr, N):
 
    ans = 0
    st = set()
    for i in range(0, N):
        st.add(arr[i])
 
    # Loop to check if number contains
    # only the digits in given set
    for k in range(A, B):
        if (has_val(k, st)):
            ans += 1
 
    return ans
 
# Driver Code
if __name__ == "__main__":
 
    L, R = 1, 100
    arr = [2, 3, 5, 7]
    N = len(arr)
 
    ans = total_Num(L, R, arr, N)
    print(ans)
 
# This code is contributed by rakeshsahni

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
  // Function Check number is
  // subset of prime digit of not
  static bool has_val(int x, HashSet st)
  {
    while (x != 0) {
      if (st.Contains(x % 10) == false)
        return false;
      x /= 10;
    }
    return true;
  }
 
  // Function to find
  // non-prime between range
  static int total_Num(int A, int B, int[] arr, int N)
  {
    int ans = 0;
    HashSet st = new HashSet();
    for (int i = 0; i < N; i++)
      st.Add(arr[i]);
 
    // Loop to check if number contains
    // only the digits in given set
    for (int k = A; k < B; k++) {
      if (has_val(k, st))
        ans += 1;
    }
    return ans;
  }
 
  // Driver Code
  static public void Main (){
 
    int L = 1, R = 100;
    int[] arr = { 2, 3, 5, 7 };
    int N = arr.Length;
 
    int ans = total_Num(L, R, arr, N);
    Console.Write(ans);
  }
}
 
// This code is contributed by hrithikgarg03188.

Javascript

       // JavaScript code for the above approach
 
       // Function Check number is
       // subset of prime digit of not
       function has_val(x, st) {
           while (x != 0) {
               if (!st.has(x % 10))
                   return false;
               x = Math.floor(x / 10);
           }
           return true;
       }
 
       // Function to find
       // non-prime between range
       function total_Num(A, B, arr, N) {
           let ans = 0;
           let st = new Set();
           for (let i = 0; i < N; i++)
               st.add(arr[i]);
 
           // Loop to check if number contains
           // only the digits in given set
           for (let k = A; k < B; k++) {
               if (has_val(k, st))
                   ans += 1;
           }
           return ans;
       }
 
       // Driver Code
       let L = 1, R = 100;
       let arr = [2, 3, 5, 7];
       let N = arr.length;
 
       let ans = total_Num(L, R, arr, N);
       document.write(ans);
 
    // This code is contributed by Potta Lokesh
   

输出

20

时间复杂度： O((RL)*logN)
辅助空间： O(1)