# Arrange N elements in circular fashion such that all elements are strictly less than sum of adjacent elements

Given an array of N integers, the task is to arrange them in a circular arrangement in such a way that the element is strictly less than the sum of its adjacent elements. In case such an arrangement is not possible, then print -1.
Note that there can be multiple ways of arranging the elements such that the condition is satisfied and the task is to find any such arrangement.

Examples:

Input: arr[] = {1, 4, 4, 3, 2}
Output: 1 3 4 4 2
arr = 1 < (2 + 3)
arr = 4 < (1 + 4)
arr = 4 < (4 + 3)
arr = 3 < (4 + 2)
arr = 2 < (3 + 1)

Input: arr[] = {8, 13, 5}
Output: -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem can be solved using a greedy approach, we first sort the array and then place the smallest element at the beginning, the second smallest at the end, the third smallest at the second position and the fourth smallest at the second last position in another array. Once the arrangement is completed, check if the given condition is satisfied or not.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the arrangement that ` `// satisifes the given condition ` `void` `printArrangement(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// Sort the array initially ` `    ``sort(a, a + n); ` ` `  `    ``// Array that stores the arrangement ` `    ``int` `b[n]; ` ` `  `    ``// Once the array is sorted ` `    ``// Re-fill the array again in the ` `    ``// mentioned way in the approach ` `    ``int` `low = 0, high = n - 1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(i % 2 == 0) ` `            ``b[low++] = a[i]; ` `        ``else` `            ``b[high--] = a[i]; ` `    ``} ` ` `  `    ``// Iterate in the array ` `    ``// and check if the arrangement made ` `    ``// satisfies the given condition or not ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// For the first element ` `        ``// the adjacents will be a and a[n-1] ` `        ``if` `(i == 0) { ` `            ``if` `(b[n - 1] + b <= b[i]) { ` `                ``cout << -1; ` `                ``return``; ` `            ``} ` `        ``} ` ` `  `        ``// For the last element ` `        ``// the adjacents will be a and a[n-2] ` `        ``else` `if` `(i == (n - 1)) { ` `            ``if` `(b[n - 2] + b <= b[i]) { ` `                ``cout << -1; ` `                ``return``; ` `            ``} ` `        ``} ` `        ``else` `{ ` `            ``if` `(b[i - 1] + b[i + 1] <= b[i]) { ` `                ``cout << -1; ` `                ``return``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If we reach this position then ` `    ``// the arrangement is possible ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << b[i] << ``" "``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 4, 4, 3, 2 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``printArrangement(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.Arrays; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to print the arrangement that ` `// satisifes the given condition ` `static` `void` `printArrangement(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// Sort the array initially ` `    ``Arrays.sort(a); ` ` `  `    ``// Array that stores the arrangement ` `    ``int` `b[] = ``new` `int``[n]; ` ` `  `    ``// Once the array is sorted ` `    ``// Re-fill the array again in the ` `    ``// mentioned way in the approach ` `    ``int` `low = ``0``, high = n - ``1``; ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``if` `(i % ``2` `== ``0``) ` `            ``b[low++] = a[i]; ` `        ``else` `            ``b[high--] = a[i]; ` `    ``} ` ` `  `    ``// Iterate in the array ` `    ``// and check if the arrangement made ` `    ``// satisfies the given condition or not ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` ` `  `        ``// For the first element ` `        ``// the adjacents will be a and a[n-1] ` `        ``if` `(i == ``0``) ` `        ``{ ` `            ``if` `(b[n - ``1``] + b[``1``] <= b[i]) ` `            ``{ ` `                ``System.out.print(-``1``); ` `                ``return``; ` `            ``} ` `        ``} ` ` `  `        ``// For the last element ` `        ``// the adjacents will be a and a[n-2] ` `        ``else` `if` `(i == (n - ``1``))  ` `        ``{ ` `            ``if` `(b[n - ``2``] + b[``0``] <= b[i]) ` `            ``{ ` `                ``System.out.print(-``1``); ` `                ``return``; ` `            ``} ` `        ``} ` `        ``else`  `        ``{ ` `            ``if` `(b[i - ``1``] + b[i + ``1``] <= b[i]) ` `            ``{ ` `                ``System.out.print(-``1``); ` `                ``return``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If we reach this position then ` `    ``// the arrangement is possible ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``System.out.print(b[i] + ``" "``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `a[] = { ``1``, ``4``, ``4``, ``3``, ``2` `}; ` `    ``int` `n = a.length; ` ` `  `    ``printArrangement(a, n); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to print the arrangement that ` `# satisifes the given condition ` `def` `printArrangement(a, n): ` ` `  `    ``# Sort the array initially ` `    ``a ``=` `sorted``(a) ` ` `  `    ``# Array that stores the arrangement ` `    ``b ``=` `[``0` `for` `i ``in` `range``(n)] ` ` `  `    ``# Once the array is sorted ` `    ``# Re-fill the array again in the ` `    ``# mentioned way in the approach ` `    ``low ``=` `0` `    ``high ``=` `n ``-` `1` `    ``for` `i ``in` `range``(n): ` `        ``if` `(i ``%` `2` `=``=` `0``): ` `            ``b[low] ``=` `a[i] ` `            ``low ``+``=` `1` `        ``else``: ` `            ``b[high] ``=` `a[i] ` `            ``high ``-``=` `1` ` `  `    ``# Iterate in the array ` `    ``# and check if the arrangement made ` `    ``# satisfies the given condition or not ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# For the first element ` `        ``# the adjacents will be a and a[n-1] ` `        ``if` `(i ``=``=` `0``): ` `            ``if` `(b[n ``-` `1``] ``+` `b[``1``] <``=` `b[i]): ` `                ``print``(``"-1"``) ` `                ``return` `                 `  `        ``# For the last element ` `        ``# the adjacents will be a and a[n-2] ` `        ``elif` `(i ``=``=` `(n ``-` `1``)) : ` `            ``if` `(b[n ``-` `2``] ``+` `b[``0``] <``=` `b[i]): ` `                ``print``(``"-1"``) ` `                ``return` ` `  `        ``else``: ` `            ``if` `(b[i ``-` `1``] ``+` `b[i ``+` `1``] <``=` `b[i]): ` `                ``print``(``"-1"``) ` `                ``return` ` `  `    ``# If we reach this position then ` `    ``# the arrangement is possible ` `    ``for` `i ``in` `range``(n): ` `        ``print``(b[i], end ``=` `" "``) ` ` `  `# Driver code ` `a ``=` `[ ``1``, ``4``, ``4``, ``3``, ``2` `] ` `n ``=` `len``(a) ` ` `  `printArrangement(a, n) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to print the arrangement that ` `// satisifes the given condition ` `static` `void` `printArrangement(``int` `[]a, ``int` `n) ` `{ ` ` `  `    ``// Sort the array initially ` `    ``Array.Sort(a); ` ` `  `    ``// Array that stores the arrangement ` `    ``int` `[]b = ``new` `int``[n]; ` ` `  `    ``// Once the array is sorted ` `    ``// Re-fill the array again in the ` `    ``// mentioned way in the approach ` `    ``int` `low = 0, high = n - 1; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``if` `(i % 2 == 0) ` `            ``b[low++] = a[i]; ` `        ``else` `            ``b[high--] = a[i]; ` `    ``} ` ` `  `    ``// Iterate in the array ` `    ``// and check if the arrangement made ` `    ``// satisfies the given condition or not ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` ` `  `        ``// For the first element ` `        ``// the adjacents will be a and a[n-1] ` `        ``if` `(i == 0) ` `        ``{ ` `            ``if` `(b[n - 1] + b <= b[i]) ` `            ``{ ` `                ``Console.Write(-1); ` `                ``return``; ` `            ``} ` `        ``} ` ` `  `        ``// For the last element ` `        ``// the adjacents will be a and a[n-2] ` `        ``else` `if` `(i == (n - 1))  ` `        ``{ ` `            ``if` `(b[n - 2] + b <= b[i]) ` `            ``{ ` `                ``Console.Write(-1); ` `                ``return``; ` `            ``} ` `        ``} ` `        ``else` `        ``{ ` `            ``if` `(b[i - 1] + b[i + 1] <= b[i]) ` `            ``{ ` `                ``Console.Write(-1); ` `                ``return``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If we reach this position then ` `    ``// the arrangement is possible ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``Console.Write(b[i] + ``" "``); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `[]a = { 1, 4, 4, 3, 2 }; ` `    ``int` `n = a.Length; ` ` `  `    ``printArrangement(a, n); ` `} ` `} ` ` `  `// This code is contributed by anuj_67.. `

Output:

```1 3 4 4 2
```

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

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