Check if an array can be Arranged in Left or Right Positioned Array

Given an array arr[] of size n>4, the task is to check whether the given array can be arranged in the form of Left or Right positioned array?
Left or Right Positioned Array means each element in the array is equal to the number of elements to its left or number of elements to its right.

Examples :

Input  : arr[] = {1, 3, 3, 2}
Output : "YES"  
This array has one such arrangement {3, 1, 2, 3}. 
In this arrangement, first element '3' indicates 
that three numbers are after it, the 2nd element 
'1' indicates that one number is before it, the 
3rd element '2' indicates that two elements are 
before it.

Input : arr[] = {1, 6, 5, 4, 3, 2, 1}
Output: "NO"
// No such arrangement is possible

Input : arr[] = {2, 0, 1, 3}
Output: "YES"
// Possible arrangement is {0, 1, 2, 3}

Input : arr[] = {2, 1, 5, 2, 1, 5}
Output: "YES"
// Possible arrangement is {5, 1, 2, 2, 1, 5}

A simple solution is to generate all possible arrangements (see this article) and check for the Left or Right Positioned Array condition, if each element in the array satisfies the condition then “YES” else “NO”. Time complexity for this approach is O(n*n! + n), n*n! to generate all arrangements and n for checking the condition using temporary array.

An efficient solution for this problem needs little bit observation and pen-paper work. To satisfy the Left or Right Positioned Array condition all the numbers in the array should either be equal to index, i or (n-1-i) and arr[i] < n. So we create an visited[] array of size n and initialize its element with 0. Then we traverse array and follow given steps :

  • If visited[arr[i]] = 0 then make it 1, which checks for the condition that number of elements on the left side of array arr[0]…arr[i-1] is equal to arr[i].
  • Else make visited[n-arr[i]-1] = 1, which checks for the condition that number of elements on the right side of array arr[i+1]…arr[n-1] is equal to arr[i].
  • Now traverse visited[] array and if all the elements of visited[] array become 1 that means arrangement is possible “YES” else “NO”.

C++

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// C++ program to check if an array can be arranged
// to left or right positioned array.
#include<bits/stdc++.h>
using namespace std;
  
// Function to check Left or Right Positioned
// Array.
// arr[] is array of n elements
// visited[] is boolean array of size n
bool leftRight(int arr[],int n)
{
    // Initially no element is placed at any position
    int visited[n] = {0};
  
    // Traverse each element of array
    for (int i=0; i<n; i++)
    {
        // Element must be smaller than n.
        if (arr[i] < n)
        {
            // Place "arr[i]" at position "i"
            // or at position n-arr[i]-1
            if (visited[arr[i]] == 0)
                visited[arr[i]] = 1;
            else
                visited[n-arr[i]-1] = 1;
        }
    }
  
    // All positions must be occupied
    for (int i=0; i<n; i++)
        if (visited[i] == 0)
            return false;
  
    return true;
}
  
// Driver program to test the case
int main()
{
    int arr[] = {2, 1, 5, 2, 1, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    if (leftRight(arr, n) == true)
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

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Java

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// Java program to check if an array
// can be arranged to left or
// right positioned array.
  
class GFG {
      
    // Function to check Left or 
    // Right Positioned Array.
    // arr[] is array of n elements
    // visited[] is boolean array of size n
    static boolean leftRight(int arr[], int n) {
      
    // Initially no element is 
    // placed at any position
    int visited[] = new int[n];
  
    // Traverse each element of array
    for (int i = 0; i < n; i++) {
          
    // Element must be smaller than n.
    if (arr[i] < n) {
          
        // Place "arr[i]" at position "i"
        // or at position n-arr[i]-1
        if (visited[arr[i]] == 0)
        visited[arr[i]] = 1;
        else
        visited[n - arr[i] - 1] = 1;
    }
    }
  
    // All positions must be occupied
    for (int i = 0; i < n; i++)
    if (visited[i] == 0)
        return false;
  
    return true;
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = {2, 1, 5, 2, 1, 5};
    int n = arr.length;
    if (leftRight(arr, n) == true)
    System.out.print("YES");
    else
    System.out.print("NO");
}
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# Python3 program to check
# if an array can be arranged
# to left or right positioned array.
  
# Function to check Left
# or Right Positioned
# Array.
# arr[] is array of n elements
# visited[] is boolean array of size n
def leftRight(arr,n):
  
    # Initially no element
    # is placed at any position
    visited=[]
    for i in range(n+1):
        visited.append(0)
   
    # Traverse each element of array
    for i in range(n):
      
        # Element must be smaller than n.
        if (arr[i] < n):
          
            # Place "arr[i]" at position "i"
            # or at position n-arr[i]-1
            if (visited[arr[i]] == 0):
                visited[arr[i]] = 1
            else:
                visited[n-arr[i]-1] = 1
   
    # All positions must be occupied
    for i in range(n):
        if (visited[i] == 0):
            return False
   
    return True
      
# Driver code
  
arr = [2, 1, 5, 2, 1, 5]
n = len(arr)
  
if (leftRight(arr, n) == True):
    print("YES")
else:
    print("NO")
  
# This code is contributed
# by Anant Agarwal.

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C#

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// C# program to check if an array
// can be arranged to left or
// right positioned array.
using System; 
public class GFG {
  
        // Function to check Left or 
        // Right Positioned Array.
        // arr[] is array of n elements
        // visited[] is boolean array of size n
        static bool leftRight(int []arr, int n) {
  
        // Initially no element is 
        // placed at any position
        int []visited = new int[n];
  
        // Traverse each element of array
        for (int i = 0; i < n; i++) {
  
        // Element must be smaller than n.
        if (arr[i] < n) {
  
            // Place "arr[i]" at position "i"
            // or at position n-arr[i]-1
            if (visited[arr[i]] == 0)
            visited[arr[i]] = 1;
            else
            visited[n - arr[i] - 1] = 1;
        }
        }
  
        // All positions must be occupied
        for (int i = 0; i < n; i++)
        if (visited[i] == 0)
            return false;
  
        return true;
    }
  
    // Driver code
    public static void Main() 
    {
        int []arr = {2, 1, 5, 2, 1, 5};
        int n = arr.Length;
        if (leftRight(arr, n) == true)
        Console.WriteLine("YES");
        else
        Console.WriteLine("NO");
    }
}
// This code is contributed by PrinciRaj1992

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PHP

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<?php
// PHP program to check if an 
// array can be arranged to 
// left or right positioned array.
  
// Function to check Left or
// Right Positioned Array.
// arr[] is array of n elements
// visited[] is boolean array of size n
function leftRight($arr, $n)
{
    // Initially no element is
    // placed at any position
    $visited[$n] = array(0);
  
    // Traverse each element of array
    for ($i = 0; $i < $n; $i++)
    {
        // Element must be smaller than n.
        if ($arr[$i] < $n)
        {
            // Place "arr[i]" at position "i"
            // or at position n-arr[i]-1
                $visited[$arr[$i]] = 1;
                $visited[$n - $arr[$i] - 1] = 1;
        }
    }
  
    // All positions must be occupied
    for ($i = 0; $i < $n; $i++)
        if ($visited[$i] == 0)
            return false;
  
    return true;
}
  
// Driver Code
$arr = array(2, 1, 5, 2, 1, 5);
$n = sizeof($arr);
if (leftRight($arr, $n) == true)
    echo "YES";
else
    echo "NO";
  
// This code is contributed by ajit
?>

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Output :

"YES"

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

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : jit_t, princiraj1992



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