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A permutation where each element indicates either number of elements before or after it

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Given an array of n elements. The task is to check whether a permutation of given array exists, such that each element indicate number of element present before or after it. Print “Yes” if exists, else print “No”.

Examples : 

Input : arr[] = {1, 3, 3, 2}
Output : Yes
{3, 1, 2, 3} is a permutation in which each element 
indicate number of element present before or after it.
There is one more permutation {3, 2, 1, 3}

Input : arr[] = {4, 1, 2, 3, 0}
Output : Yes
There are two permutations {0, 1, 2, 3, 4} or
{4, 3, 2, 1, 0}

The idea is to use hashing. Observe, for each index i in the array, arr[i] can have value i or n – i. We traverse the given array and find the frequency of each element present in the array. Now, for each index i, check availability of value i and n-i and accordingly decrement the frequency. Note that an item with value i can either go to index i or n-i-1. Similarly, an item with value n-i-1 can go to either index i or n-i-1.

Below is the implementation of this approach: 

C++




// C++ program to check if array permutation
// exists such that each element indicates
// either number of elements before or after
// it.
#include <bits/stdc++.h>
using namespace std;
  
// Check if array permutation exist such that
// each element indicate either number of
// elements before or after it.
bool check(int arr[], int n)
{
    map<int, int> freq;
  
    // Finding the frequency of each number.
    for (int i = 0; i < n; i++)
        freq[arr[i]]++;
  
    for (int i = 0; i < n; i++)
    {
        // Try to find number of element before
        // the current index.
        if (freq[i])
            freq[i]--;
  
        // Try to find number of element after
        // the current index.
        else if (freq[n-i-1])
            freq[n-i-1]--;
  
        // If no such number find, return false.
        else
            return false;
    }
  
    return true;
}
  
// Driven Program
int main()
{
    int arr[] = {1, 3, 3, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
  
    check(arr, n)? (cout << "Yes" << endl) :
                   (cout << "No" << endl);
  
    return 0;
}


Java




// Java program to check if array permutation
// exists such that each element indicates
// either number of elements before or after
// it.
import java.io.*;
class GFG 
{
    // Check if array permutation exist such that
    // each element indicate either number of
    // elements before or after it.
    public static boolean check(int arr[])
    {
        int n = arr.length;
        int[] freq = new int[n];
   
        // Finding the frequency of each number.
        for (int i = 0; i < n; i++)
            freq[arr[i]]++;
   
        for (int i = 0; i < n; i++)
        {
            // Try to find number of element before
            // the current index.
            if (freq[i]!= 0)
                 freq[i]--;
   
            // Try to find number of element after
            // the current index.
            else if (freq[n-i-1]!= 0)
                    freq[n-i-1]--;
   
            // If no such number find, return false.
            else
                return false;
        }
   
    return true;
    }
      
    //Driver program
    public static void main (String[] args) 
    {
          
        int arr[] = {1, 3, 3, 2};
        boolean bool = check(arr);
        if(bool)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}


Python3




# Python program to check if array permutation
# exists such that each element indicates
# either number of elements before or after
# it.
  
# Check if array permutation exist such that
# each element indicate either number of
# elements before or after it.
def check(arr):
  
    n = len(arr);
    freq = [0] * n;
  
    # Finding the frequency of each number.
    for i in range(n):
        freq[arr[i]] += 1;
  
    for i in range(n):
          
        # Try to find number of element before
        # the current index.
        if (freq[i] != 0):
            freq[i] -= 1;
  
        # Try to find number of element after
        # the current index.
        else if (freq[n - i - 1] != 0):
                freq[n - i - 1] -= 1;
  
        # If no such number find, return false.
        else:
            return False;
  
    return True;
      
# Driver program
if __name__ == '__main__':
  
    arr = [1, 3, 3, 2];
    bool = check(arr);
    if(bool):
        print("Yes");
    else:
        print("No");
  
# This code is contributed by 29AjayKumar


C#




// C# program to check if array permutation
// exists such that each element indicates
// either number of elements before or after it.
using System;
  
class GFG 
{
    // Check if array permutation exist such that
    // each element indicate either number of
    // elements before or after it.
    public static bool check(int []arr)
    {
        int n = arr.Length;
        int []freq = new int[n];
  
        // Finding the frequency of each number.
        for (int i = 0; i < n; i++)
            freq[arr[i]]++;
  
        for (int i = 0; i < n; i++)
        {
            // Try to find number of element 
            // before the current index.
            if (freq[i]!= 0)
                freq[i]--;
  
            // Try to find number of element
            // after the current index.
            else if (freq[n-i-1] != 0)
                    freq[n-i-1]--;
  
            // If no such number find, return false.
            else
                return false;
        }
  
    return true;
    }
      
    //Driver program
    public static void Main () 
    {
        int []arr = {1, 3, 3, 2};
        bool boo = check(arr);
        if(boo)
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
  
// This code is contributed by nitin mittal.


Javascript




<script>
      // JavaScript program to check if array permutation
      // exists such that each element indicates
      // either number of elements before or after it.
      // Check if array permutation exist such that
      // each element indicate either number of
      // elements before or after it.
      function check(arr) 
      {
        var n = arr.length;
        var freq = new Array(n).fill(0);
  
        // Finding the frequency of each number.
        for (var i = 0; i < n; i++) freq[arr[i]]++;
  
        for (var i = 0; i < n; i++)
        {
          
          // Try to find number of element
          // before the current index.
          if (freq[i] !== 0) freq[i]--;
            
          // Try to find number of element
          // after the current index.
          else if (freq[n - i - 1] !== 0) freq[n - i - 1]--;
            
          // If no such number find, return false.
          else return false;
        }
  
        return true;
      }
  
      // Driver program
      var arr = [1, 3, 3, 2];
      var boo = check(arr);
      if (boo) document.write("Yes");
      else document.write("No");
        
      // This code is contributed by rdtank.
    </script>


Output

Yes

Time Complexity: O(n).

 



Last Updated : 13 Sep, 2023
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