Print the two possible permutations from a given sequence

Given an array arr containing N positive integers, the task is to check if the given array can be dissociated into two permutations or not and print the permutations if possible. A sequence of M integers is called a permutation if it contains all integers from 1 to M exactly once.
Examples:

Input: arr[] = { 1, 2, 5, 3, 4, 1, 2 }, N = 7
Output: {1 2 5 3 4}, {1 2}

Input: arr[] = {2, 1, 1, 3}, N = 4
Output: Not possible

Approach:

  • First of all, we need to check if the array is the concatenation of two permutations.It is explained in this article.
  • If so, find the largest element of the array, say x.
  • If the elements at indices [0, x-1] and [x, n-1] form two valid permutations, print them.
  • Otherwise, print the elements at indices [0, n -1 – x] and [n – x, n – 1] as the two valid permutations.

Below is the implementation of the above approach:

C++

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// C++ program to print two
// permutations from a given sequence
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if the sequence is
// concatenation of two permutations or not
bool checkPermutation(int arr[], int n)
{
    // Computing the sum of all the
    // elements in the array
    long long sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
  
    // Computing the prefix sum
    // for all the elements in the array
    long long prefix[n + 1] = { 0 };
    prefix[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
  
    // Iterating through the i
    // from lengths 1 to n-1
    for (int i = 0; i < n - 1; i++) {
  
        // Sum of first i+1 elements
        long long lsum = prefix[i];
  
        // Sum of remaining n-i-1 elements
        long long rsum = sum - prefix[i];
  
        // Lengths of the 2 permutations
        long long l_len = i + 1,
                  r_len = n - i - 1;
  
        // Checking if the sums
        // satisfy the formula or not
        if (((2 * lsum)
             == (l_len * (l_len + 1)))
            && ((2 * rsum)
                == (r_len * (r_len + 1))))
            return true;
    }
  
    return false;
}
  
// Function to print the
// two permutations
void printPermutations(int arr[], int n,
                       int l1, int l2)
{
    // Print the first permutation
    for (int i = 0; i < l1; i++) {
        cout << arr[i] << " ";
    }
    cout << endl;
  
    // Print the second permutation
    for (int i = l1; i < n; i++) {
        cout << arr[i] << " ";
    }
}
  
// Function to find the two permutations
// from the given sequence
void findPermutations(int arr[], int n)
{
    // If the sequence is not a
    // concatenation of two permutations
    if (!checkPermutation(arr, n)) {
        cout << "Not Possible";
        return;
    }
  
    int l1 = 0, l2 = 0;
  
    // Find the largest element in the
    // array and set the lengths of the
    // permutations accordingly
    l1 = *max_element(arr, arr + n);
    l2 = n - l1;
  
    set<int> s1, s2;
    for (int i = 0; i < l1; i++)
        s1.insert(arr[i]);
  
    for (int i = l1; i < n; i++)
        s2.insert(arr[i]);
  
    if (s1.size() == l1 && s2.size() == l2)
        printPermutations(arr, n, l1, l2);
    else {
        swap(l1, l2);
        printPermutations(arr, n, l1, l2);
    }
}
  
// Driver code
int main()
{
    int arr[] = { 2, 1, 3, 4, 5,
                  6, 7, 8, 9, 1,
                  10, 2 };
    int n = sizeof(arr) / sizeof(int);
  
    findPermutations(arr, n);
    return 0;
}

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Java

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// Java program to print two
// permutations from a given sequence
import java.util.*;
  
class GFG{
   
// Function to check if the sequence is
// concatenation of two permutations or not
static boolean checkPermutation(int arr[], int n)
{
    // Computing the sum of all the
    // elements in the array
    long sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
   
    // Computing the prefix sum
    // for all the elements in the array
    int []prefix = new int[n + 1];
    prefix[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
   
    // Iterating through the i
    // from lengths 1 to n-1
    for (int i = 0; i < n - 1; i++) {
   
        // Sum of first i+1 elements
        long lsum = prefix[i];
   
        // Sum of remaining n-i-1 elements
        long rsum = sum - prefix[i];
   
        // Lengths of the 2 permutations
        long l_len = i + 1,
                  r_len = n - i - 1;
   
        // Checking if the sums
        // satisfy the formula or not
        if (((2 * lsum)
             == (l_len * (l_len + 1)))
            && ((2 * rsum)
                == (r_len * (r_len + 1))))
            return true;
    }
   
    return false;
}
   
// Function to print the
// two permutations
static void printPermutations(int arr[], int n,
                       int l1, int l2)
{
    // Print the first permutation
    for (int i = 0; i < l1; i++) {
        System.out.print(arr[i]+ " ");
    }
    System.out.println();
   
    // Print the second permutation
    for (int i = l1; i < n; i++) {
        System.out.print(arr[i]+ " ");
    }
}
   
// Function to find the two permutations
// from the given sequence
static void findPermutations(int arr[], int n)
{
    // If the sequence is not a
    // concatenation of two permutations
    if (!checkPermutation(arr, n)) {
        System.out.print("Not Possible");
        return;
    }
   
    int l1 = 0, l2 = 0;
   
    // Find the largest element in the
    // array and set the lengths of the
    // permutations accordingly
    l1 = Arrays.stream(arr).max().getAsInt();
    l2 = n - l1;
   
    HashSet<Integer> s1 = new HashSet<Integer>(),
            s2 = new HashSet<Integer>();
    for (int i = 0; i < l1; i++)
        s1.add(arr[i]);
   
    for (int i = l1; i < n; i++)
        s2.add(arr[i]);
   
    if (s1.size() == l1 && s2.size() == l2)
        printPermutations(arr, n, l1, l2);
    else {
        l1 = l1+l2;
        l2 = l1-l2;
        l1 = l1-l2;
        printPermutations(arr, n, l1, l2);
    }
}
   
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 1, 3, 4, 5,
                  6, 7, 8, 9, 1,
                  10, 2 };
    int n = arr.length;
   
    findPermutations(arr, n);
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 program to print two
# permutations from a given sequence
  
# Function to check if the sequence is
# concatenation of two permutations or not
def checkPermutation(arr, n):
    # Computing the sum of all the
    # elements in the array
    sum = 0
    for i in range(n):
        sum += arr[i]
  
    # Computing the prefix sum
    # for all the elements in the array
    prefix = [0 for i in range(n+1)]
    prefix[0] = arr[0]
    for i in range(1,n):
        prefix[i] = prefix[i - 1] + arr[i]
  
    # Iterating through the i
    # from lengths 1 to n-1
    for i in range(n - 1):
          
        # Sum of first i+1 elements
        lsum = prefix[i]
  
        # Sum of remaining n-i-1 elements
        rsum = sum - prefix[i]
  
        # Lengths of the 2 permutations
        l_len = i + 1
        r_len = n - i - 1
  
        # Checking if the sums
        # satisfy the formula or not
        if (((2 * lsum) == (l_len * (l_len + 1))) and 
                ((2 * rsum) == (r_len * (r_len + 1)))):
            return True
  
    return False
  
# Function to print the
# two permutations
def printPermutations(arr,n,l1,l2):
    # Print the first permutation
    for i in range(l1):
        print(arr[i],end = " ")
  
    print("\n",end = "");
  
    # Print the second permutation
    for i in range(l1, n, 1):
        print(arr[i], end = " ")
  
# Function to find the two permutations
# from the given sequence
def findPermutations(arr,n):
      
    # If the sequence is not a
    # concatenation of two permutations
    if (checkPermutation(arr, n) == False):
        print("Not Possible")
        return
  
    l1 = 0
    l2 = 0
  
    # Find the largest element in the
    # array and set the lengths of the
    # permutations accordingly
    l1 = max(arr)
    l2 = n - l1
  
    s1 = set()
    s2 = set()
    for i in range(l1):
        s1.add(arr[i])
  
    for i in range(l1,n):
        s2.add(arr[i])
  
    if (len(s1) == l1 and len(s2) == l2):
        printPermutations(arr, n, l1, l2)
    else:
        temp = l1
        l1 = l2
        l2 = temp
        printPermutations(arr, n, l1, l2)
  
# Driver code
if __name__ == '__main__':
    arr = [2, 1, 3, 4, 5,6, 7, 8, 9, 1,10, 2]
    n = len(arr)
  
    findPermutations(arr, n)
  
# This code is contributed by Surendra_Gangwar

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C#

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// C# program to print two
// permutations from a given sequence
using System;
using System.Linq;
using System.Collections.Generic;
  
class GFG{
    
// Function to check if the sequence is
// concatenation of two permutations or not
static bool checkPermutation(int []arr, int n)
{
    // Computing the sum of all the
    // elements in the array
    long sum = 0;
    for (int i = 0; i < n; i++)
        sum += arr[i];
    
    // Computing the prefix sum
    // for all the elements in the array
    int []prefix = new int[n + 1];
    prefix[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
    
    // Iterating through the i
    // from lengths 1 to n-1
    for (int i = 0; i < n - 1; i++) {
    
        // Sum of first i+1 elements
        long lsum = prefix[i];
    
        // Sum of remaining n-i-1 elements
        long rsum = sum - prefix[i];
    
        // Lengths of the 2 permutations
        long l_len = i + 1,
                  r_len = n - i - 1;
    
        // Checking if the sums
        // satisfy the formula or not
        if (((2 * lsum)
             == (l_len * (l_len + 1)))
            && ((2 * rsum)
                == (r_len * (r_len + 1))))
            return true;
    }
    
    return false;
}
    
// Function to print the
// two permutations
static void printPermutations(int []arr, int n,
                       int l1, int l2)
{
    // Print the first permutation
    for (int i = 0; i < l1; i++) {
        Console.Write(arr[i]+ " ");
    }
    Console.WriteLine();
    
    // Print the second permutation
    for (int i = l1; i < n; i++) {
        Console.Write(arr[i]+ " ");
    }
}
    
// Function to find the two permutations
// from the given sequence
static void findPermutations(int []arr, int n)
{
    // If the sequence is not a
    // concatenation of two permutations
    if (!checkPermutation(arr, n)) {
        Console.Write("Not Possible");
        return;
    }
    
    int l1 = 0, l2 = 0;
    
    // Find the largest element in the
    // array and set the lengths of the
    // permutations accordingly
    l1 = arr.Max();
    l2 = n - l1;
    
    HashSet<int> s1 = new HashSet<int>(),
            s2 = new HashSet<int>();
    for (int i = 0; i < l1; i++)
        s1.Add(arr[i]);
    
    for (int i = l1; i < n; i++)
        s2.Add(arr[i]);
    
    if (s1.Count == l1 && s2.Count == l2)
        printPermutations(arr, n, l1, l2);
    else {
        l1 = l1+l2;
        l2 = l1-l2;
        l1 = l1-l2;
        printPermutations(arr, n, l1, l2);
    }
}
    
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 1, 3, 4, 5,
                  6, 7, 8, 9, 1,
                  10, 2 };
    int n = arr.Length;
    
    findPermutations(arr, n);
}
}
  
// This code contributed by Rajput-Ji

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Output:

2 1 
3 4 5 6 7 8 9 1 10 2

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