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Minimum operations to convert an Array into a Permutation of 1 to N by replacing with remainder from some d

  • Difficulty Level : Hard
  • Last Updated : 12 Jan, 2022

Given an array arr[] of size N, the task is to find the minimum number of operations to convert the array into a permutation of [1, n], in each operation, an element a[i] can be replaced by a[i] % d where d can be different in each operation performed. If it is not possible print -1.

Examples:

Input: arr[] = {5, 4, 10, 8, 1}
Output:
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

 

Approach: The task can be solved using the greedy approach. This approach is based on the fact that when remainder r is to be obtained, then a[i] > 2*r  i.e  r lies between the range [0, a[i]-1 /2]  

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

Follow these steps to solve this problem:

  • Initialize a set s 
  • Traverse through the array arr[] & insert all the elements of arr[] into the set.
  • Initialize a variable ops = 0
  • Now iterate from n to 1
  • Check if s has i already if it has removed it from the set.
  • Else increment ops  and check if the largest element of the set < 2* i +1
  • If the largest of the set is < 2* i +1 then set ops = -1 and break out of the loop.
  • Else erase it from the set because we can make it i using mod operation.
  • Print the ops

`Below is the implementation of the above approach:

C++




// C++ code for the above approach
#include <bits/stdc++.h>
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<int> 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<Integer> s = new TreeSet<Integer>();
 
    // 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<int> s = new SortedSet<int>();
 
    // 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




<script>
// 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
</script>

 
 

Output
2

 

Time Complexity: O(nlogn)
Auxiliary Space: O(n)

 


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