通过用一些 d 中的余数替换将数组转换为 1 到 N 的排列的最小操作

给定一个大小为N的数组arr[] ，任务是找到将数组转换为[1, n]的排列的最小操作数，在每个操作中，元素a[i]可以替换为a[ i] % d其中 d 在执行的每个操作中可能不同。如果不可能打印-1 。

例子：

Input: arr[] = {5, 4, 10, 8, 1}
Output: 2 
Explanation: In first operation choosing d = 7  , 10 can be replaced by 10 % 7  , 
In second operation d = 6, 8 can be replaced by 8 %6 so two operations.

Input : arr[] = {1, 2, 3, 7}
Output: -1

方法：可以使用贪心方法来解决任务。这种方法基于这样一个事实，即当要获得余数r 时， a[i] > 2*r即r位于[0, a[i]-1 /2]范围之间

Let us take an example: 8 for different d

Taking, 8 % 7= 1

           8%6 = 2

           8%5 = 3

          8%4 = 0

          8%3 = 2

          8%2 = 0

          8%1=0

So maximum number that can be obtained is 3 using mod operation, so when we want to obtain a number i in a permutation then the number should a[i] > 2* i+1

请按照以下步骤解决此问题：

• 初始化一个集合
• 遍历数组arr[]并将 arr[] 的所有元素插入到集合中。
• 初始化变量ops = 0
• 现在从 n 迭代到 1
• 检查 s 是否已经 i 是否已将其从集合中删除。
• 否则增加操作并检查集合的最大元素是否 < 2* i +1
• 如果集合中的最大值 < 2* i +1则设置ops = -1并跳出循环。
• 否则将其从集合中删除，因为我们可以使用 mod 操作使其成为。
• 打印操作

`下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to find the minimum operations
// to convert the array into a permutation of [1, n]
void minimum_operations(int arr[], int n)
{
    // Initialize the set
    set s;
 
    // Insert all the elements into the set
    for (int i = 0; i < n; i++) {
        s.insert(arr[i]);
    }
 
    // Initialize ops to count the operations
    int ops = 0;
 
    // Traverse from [n to 1]
    for (int i = n; i >= 1; i--) {
 
        // If we already have i in our
        // array erase it from the set
        if (s.find(i) != s.end()) {
            s.erase(s.find(i));
        }
 
        // count the ops because there is no element
        else {
            ops++;
 
            // Check the largest element of the set
            auto it = s.end();
            it--;
 
            // If it is < 2*i +1 we cant get that i
            // using % operation so there is no way to
            // create a permutation
            if (*it < 2 * i + 1) {
                ops = -1;
                break;
            }
 
            // Erase it if we have processed
            // it to i by % operation
            s.erase(it);
        }
    }
 
    // Print the result
    cout << ops << endl;
}
 
// Driver Code
int main()
{
    // Initialize the value n
    int arr[] = { 5, 4, 10, 8, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    minimum_operations(arr, n);
 
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
 
class GFG{
 
  // Function to find the minimum operations
  // to convert the array into a permutation of [1, n]
  static void minimum_operations(int arr[], int n)
  {
    // Initialize the set
    SortedSet s = new TreeSet();
 
    // Insert all the elements into the set
    for (int i = 0; i < n; i++) {
      s.add(arr[i]);
    }
 
    // Initialize ops to count the operations
    int ops = 0;
 
    // Traverse from [n to 1]
    for (int i = n; i >= 1; i--) {
 
      // If we already have i in our
      // array erase it from the set
      if (s.contains(i)) {
        s.remove(i);
      }
 
      // count the ops because there is no element
      else {
        ops++;
 
        // Check the largest element of the set
        Integer it = s.last();
        it--;
 
        // If it is < 2*i +1 we cant get that i
        // using % operation so there is no way to
        // create a permutation
        if (it < 2 * i + 1) {
          ops = -1;
          break;
        }
 
        // Erase it if we have processed
        // it to i by % operation
        s.remove(it);
      }
    }
 
    // Print the result
    System.out.print(ops +"\n");
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    // Initialize the value n
    int arr[] = { 5, 4, 10, 8, 1 };
    int n = arr.length;
    minimum_operations(arr, n);
 
  }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python 3 code for the above approach
 
# Function to find the minimum operations
# to convert the array into a permutation of [1, n]
def minimum_operations(arr, n):
 
    # Initialize the set
    s = set([])
 
    # Insert all the elements into the set
    for i in range(n):
        s.add(arr[i])
 
    # Initialize ops to count the operations
    ops = 0
 
    # Traverse from [n to 1]
    for i in range(n, 0, -1):
 
        # If we already have i in our
        # array erase it from the set
        if (i in s):
            list(s).remove(i)
 
        # count the ops because there is no element
        else:
            ops += 1
 
            # Check the largest element of the set
            it = len(s)
            it -= 1
 
            # If it is < 2*i +1 we cant get that i
            # using % operation so there is no way to
            # create a permutation
            if (list(s)[it] < 2 * i + 1):
                ops = -1
                break
 
            # Erase it if we have processed
            # it to i by % operation
            list(s).pop(it)
 
    # Print the result
    print(ops)
 
# Driver Code
if __name__ == "__main__":
 
    # Initialize the value n
    arr = [5, 4, 10, 8, 1]
    n = len(arr)
    minimum_operations(arr, n)
 
    # This code is contributed by ukasp.

C#

// C# code for the above approach
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // Function to find the minimum operations
  // to convert the array into a permutation of [1, n]
  static void minimum_operations(int []arr, int n)
  {
 
    // Initialize the set
    SortedSet s = new SortedSet();
 
    // Insert all the elements into the set
    for (int i = 0; i < n; i++) {
      s.Add(arr[i]);
    }
 
    // Initialize ops to count the operations
    int ops = 0;
 
    // Traverse from [n to 1]
    for (int i = n; i >= 1; i--) {
 
      // If we already have i in our
      // array erase it from the set
      if (s.Contains(i)) {
        s.Remove(i);
      }
 
      // count the ops because there is no element
      else {
        ops++;
 
        // Check the largest element of the set
        int it = s.Max;
        it--;
 
        // If it is < 2*i +1 we cant get that i
        // using % operation so there is no way to
        // create a permutation
        if (it < 2 * i + 1) {
          ops = -1;
          break;
        }
 
        // Erase it if we have processed
        // it to i by % operation
        s.Remove(it);
      }
    }
 
    // Print the result
    Console.Write(ops +"\n");
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    // Initialize the value n
    int []arr = { 5, 4, 10, 8, 1 };
    int n = arr.Length;
    minimum_operations(arr, n);
 
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript code for the above approach
 
// Function to find the minimum operations
// to convert the array into a permutation of [1, n]
function minimum_operations(arr, n)
{
    // Initialize the set
    var s = new Set();
     
    // Insert all the elements into the set
    for (var i = 0; i < n; i++) {
        s.add(arr[i]);
    }
     
    // Initialize ops to count the operations
    var ops = 0;
     
    // Traverse from [n to 1]
    for (var i = n; i >= 1; i--) {
        // If we already have i in our
        // array erase it from the set
        if (s.has(i)) {
            s.delete(i);
        }
         
        // count the ops because there is no element
        else {
            ops++;
            // Check the largest element of the set
            var it = Math.max(...Array.from(s.values()));
            it--;
             
            // If it is < 2*i +1 we cant get that i
            // using % operation so there is no way to
            // create a permutation
            if (it < 2 * i + 1) {
                ops = -1;
                break;
            }
             
            // Erase it if we have processed
            // it to i by % operation
            s.delete(it);
        }
    }
    // Print the result
    document.write(ops +"<br>");
}
 
// Driver Code
 
// Initialize the value n
var arr = [ 5, 4, 10, 8, 1 ];
var n = arr.length;
minimum_operations(arr, n);
 
// This code is contributed by Shubham Singh

输出

2

时间复杂度： O(nlogn)
辅助空间： O(n)