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Check if an array can be formed by merging 2 non-empty permutations
  • Difficulty Level : Medium
  • Last Updated : 09 Dec, 2020

Given an array arr[] of length N, the task is to check if it can be formed by merging two permutations of the same or different lengths. Print YES if such merging is possible. Otherwise, print NO.

Permutations of length 3 are {1, 2, 3}, {2, 3, 1}, {1, 3, 2}, {3, 1, 2}, {3, 2, 1}, {2, 1, 3}. 

Examples: 

Input: arr = [1, 3, 2, 4, 3, 1, 2] 
Output: YES 
Explanation: 
The given array can be formed by merging a permutation of length 4 [1, 3, 2, 4] and permutation of length 3 [3, 1, 2]

Input: arr = [1, 2, 3, 2, 3, 2, 1] 
Output: NO 
 



Approach : 
We can observe that the minimum excludant (MEX) of a permutation of length N is N+1
So, if the length of the first permutation is l, then MEX of the prefix arr [0 …… l-1] is l+1 and the MEX of the suffix a[l …… n] will be N – l + 1
So, we can calculate MEX of prefix and suffixes and if the above condition is satisfied, the answer will be “YES”. Otherwise, the answer will be “NO”.

Below is the implementation of the above approach: 

C++




// C++ program for the
// above approach
#include<bits/stdc++.h>
using namespace std;
void if_merged_permutations(int a[],
                            int n)
{
  int pre_mex[n];
 
  // Calculate the mex of the
  // array a[0...i]
  int freq[n + 1];
   
  memset(freq, 0, sizeof(freq));
   
  for(int i = 0; i < n; i++)
  {
    pre_mex[i] = 1;
  }
 
  // Mex of empty
  // array is 1
  int mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = 0; i < n; i++)
  {
    freq[a[i]]++;
    if(freq[a[i]] > 1)
    {
      // In a permutation
      // each element is
      // present one time,
      // So there is no chance
      // of getting permutations
      // for the prefix of
      // length greater than i
      break;
    }
 
    // The current element
    // is the mex
    if(a[i] == mex)
    {
      // While mex is present
      // in the array
      while(freq[mex] != 0)
      {
        mex++;
      }
    }
    pre_mex[i] = mex;
  }
   
  int suf_mex[n];
   
  for(int i = 0; i < n; i++)
  {
    suf_mex[i] = 1;
  }
   
  // Calculate the mex of the
  // array a[i..n]
  memset(freq, 0, sizeof(freq));
 
  // Mex of empty
  // array is 1
  mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = n - 1; i > -1; i--)
  {
    freq[a[i]]++;
    if(freq[a[i]] > 1)
    {
      // In a permutation each
      // element is present
      // one time, So there is
      // no chance of getting
      // permutations for the
      // suffix of length lesser
      // than i
      continue;
    }
     
    // The current element is
    // the mex
    if(a[i] == mex)
    {
      // While mex is present
      // in the array
      while(freq[mex] != 0)
      {
        mex++;
      }
    }
    suf_mex[i] = mex;
  }
 
  // Now check if there is atleast
  // one index i such that mex of
  // the prefix a[0..i]= i +
  // 2(0 based indexing) and  mex
  // of the suffix a[i + 1..n]= n-i
  for(int i = 0; i < n - 1; i++)
  {
    if(pre_mex[i] == i + 2 &&
       suf_mex[i + 1] == n - i)
    {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
 
// Driver code
int main()
{
    int a[] = {1, 3, 2,
               4, 3, 1, 2};
    int n = sizeof(a)/ sizeof(a[0]);
    if_merged_permutations(a, n);   
}
 
//This code is contributed by avanitrachhadiya2155

Java




// Java program for above approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
static void if_merged_permutations(int a[],
                                   int n)
{
  int[] pre_mex = new int[n];
 
  // Calculate the mex of the
  // array a[0...i]
  int[] freq = new int[n + 1];
   
  for(int i = 0; i < n; i++)
  {
    pre_mex[i] = 1;
  }
 
  // Mex of empty
  // array is 1
  int mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = 0; i < n; i++)
  {
    freq[a[i]]++;
     
    if (freq[a[i]] > 1)
    {
       
      // In a permutation
      // each element is
      // present one time,
      // So there is no chance
      // of getting permutations
      // for the prefix of
      // length greater than i
      break;
    }
 
    // The current element
    // is the mex
    if (a[i] == mex)
    {
       
      // While mex is present
      // in the array
      while (freq[mex] != 0)
      {
        mex++;
      }
    }
    pre_mex[i] = mex;
  }
   
  int[] suf_mex = new int[n];
   
  for(int i = 0; i < n; i++)
  {
    suf_mex[i] = 1;
  }
   
  // Calculate the mex of the
  // array a[i..n]
  Arrays.fill(freq, 0);
 
  // Mex of empty
  // array is 1
  mex = 1;
 
  // Calculating the frequency
  // of array elements
  for(int i = n - 1; i > -1; i--)
  {
    freq[a[i]]++;
     
    if (freq[a[i]] > 1)
    {
       
      // In a permutation each
      // element is present
      // one time, So there is
      // no chance of getting
      // permutations for the
      // suffix of length lesser
      // than i
      continue;
    }
     
    // The current element is
    // the mex
    if (a[i] == mex)
    {
       
      // While mex is present
      // in the array
      while (freq[mex] != 0)
      {
        mex++;
      }
    }
    suf_mex[i] = mex;
  }
 
  // Now check if there is atleast
  // one index i such that mex of
  // the prefix a[0..i]= i +
  // 2(0 based indexing) and  mex
  // of the suffix a[i + 1..n]= n-i
  for(int i = 0; i < n - 1; i++)
  {
    if (pre_mex[i] == i + 2 &&
        suf_mex[i + 1] == n - i)
    {
      System.out.println("YES");
      return;
    }
  }
  System.out.println("NO");
}
   
// Driver code
public static void main(String[] args)
{
  int a[] = { 1, 3, 2, 4, 3, 1, 2 };
  int n = a.length;
   
  if_merged_permutations(a, n);   
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program for above approach
def if_merged_permutations(a, n):
    pre_mex =[1 for i in range(n)]
     
    # Calculate the mex of the
    # array a[0...i]
    freq =[0 for i in range(n + 1)]
     
    # Mex of empty array is 1
    mex = 1
     
    # Calculating the frequency
    # of array elements
    for i in range(n):
        freq[a[i]]+= 1
        if freq[a[i]]>1:
            # In a permutation
            # each element is
            # present one time,
            # So there is no chance
            # of getting permutations
            # for the prefix of
            # length greater than i
            break
         
        # The current element
        # is the mex   
        if a[i]== mex:
       
            # While mex is present
            # in the array
            while freq[mex]!= 0 :
                mex+= 1
        pre_mex[i]= mex
                 
    suf_mex =[1 for i in range(n)]
     
    # Calculate the mex of the
    # array a[i..n]
    freq =[0 for i in range(n + 1)]
   
    # Mex of empty array is 1
    mex = 1
     
    # Calculating the frequency
    # of array elements
    for i in range(n-1, -1, -1):
        freq[a[i]]+= 1
        if freq[a[i]]>1:
 
            # In a permutation each
            # element is present
            # one time, So there is
            # no chance of getting
            # permutations for the
            # suffix of length lesser
            # than i
            break
 
        # The current element is
        # the mex
        if a[i]== mex:
            # While mex is present
            # in the array
            while freq[mex]!= 0 :
                mex+= 1
        suf_mex[i]= mex
 
    # Now check if there is atleast
    # one index i such that mex of
    # the prefix a[0..i]= i +
    # 2(0 based indexing) and  mex
    # of the suffix a[i + 1..n]= n-i
 
    for i in range(n-1):
        if pre_mex[i]== i + 2 and suf_mex[i + 1]== n-i:
            print("YES")
            return
    print("NO")
     
a =[1, 3, 2, 4, 3, 1, 2]   
n = len(a)       
if_merged_permutations(a, n)

C#




// C# program for above approach
using System;
 
class GFG{
     
static void if_merged_permutations(int[] a,
                                   int n)
{
    int[] pre_mex = new int[n];
     
    // Calculate the mex of the
    // array a[0...i]
    int[] freq = new int[n + 1];
     
    for(int i = 0; i < n; i++)
    {
        pre_mex[i] = 1;
    }
     
    // Mex of empty
    // array is 1
    int mex = 1;
     
    // Calculating the frequency
    // of array elements
    for(int i = 0; i < n; i++)
    {
        freq[a[i]]++;
     
        if (freq[a[i]] > 1)
        {
             
            // In a permutation
            // each element is
            // present one time,
            // So there is no chance
            // of getting permutations
            // for the prefix of
            // length greater than i
            break;
        }
     
        // The current element
        // is the mex
        if (a[i] == mex)
        {
             
            // While mex is present
            // in the array
            while (freq[mex] != 0)
            {
                mex++;
            }
        }
        pre_mex[i] = mex;
    }
     
    int[] suf_mex = new int[n];
     
    for(int i = 0; i < n; i++)
    {
        suf_mex[i] = 1;
    }
     
    // Calculate the mex of the
    // array a[i..n]
    Array.Fill(freq, 0);
     
    // Mex of empty
    // array is 1
    mex = 1;
     
    // Calculating the frequency
    // of array elements
    for(int i = n - 1; i > -1; i--)
    {
        freq[a[i]]++;
         
        if (freq[a[i]] > 1)
        {
             
            // In a permutation each
            // element is present
            // one time, So there is
            // no chance of getting
            // permutations for the
            // suffix of length lesser
            // than i
            continue;
        }
         
        // The current element is
        // the mex
        if (a[i] == mex)
        {
             
            // While mex is present
            // in the array
            while (freq[mex] != 0)
            {
                mex++;
            }
        }
        suf_mex[i] = mex;
    }
     
    // Now check if there is atleast
    // one index i such that mex of
    // the prefix a[0..i]= i +
    // 2(0 based indexing) and  mex
    // of the suffix a[i + 1..n]= n-i
    for(int i = 0; i < n - 1; i++)
    {
        if (pre_mex[i] == i + 2 &&
            suf_mex[i + 1] == n - i)
        {
            Console.WriteLine("YES");
            return;
        }
    }
    Console.WriteLine("NO");
}
 
// Driver Code
static void Main()
{
    int[] a = { 1, 3, 2, 4, 3, 1, 2 };
    int n = a.Length;
     
    if_merged_permutations(a, n);
}
}
 
// This code is contributed by divyeshrabadiya07
Output: 
YES

 

Time Complexity: O(N)

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