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

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

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”.

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;

}

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