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

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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[0] = 1 < (2 + 4)
arr[1] = 4 < (1 + 4)
arr[2] = 4 < (4 + 3)
arr[3] = 3 < (4 + 2)
arr[4] = 2 < (3 + 1)
Input: arr[] = {8, 13, 5}
Output: -1

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 <bits/stdc++.h>
using namespace std;
 
// Function to print the arrangement that
// satisfies 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[1] and a[n-1]
        if (i == 0) {
            if (b[n - 1] + b[1] <= b[i]) {
                cout << -1;
                return;
            }
        }
 
        // For the last element
        // the adjacents will be a[0] and a[n-2]
        else if (i == (n - 1)) {
            if (b[n - 2] + b[0] <= 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[0]);
 
    printArrangement(a, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;
 
class GFG 
{
 
// Function to print the arrangement that
// satisfies 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[1] 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[0] 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
# satisfies 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[1] 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[0] 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
// satisfies 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[1] and a[n-1]
        if (i == 0)
        {
            if (b[n - 1] + b[1] <= b[i])
            {
                Console.Write(-1);
                return;
            }
        }
 
        // For the last element
        // the adjacents will be a[0] and a[n-2]
        else if (i == (n - 1)) 
        {
            if (b[n - 2] + b[0] <= 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..

Javascript

<script>
// javascript implementation of the approach
 
    // Function to print the arrangement that
    // satisfies the given condition
    function printArrangement(a, n)
    {
 
        // Sort the array initially
        a.sort();
 
        // Array that stores the arrangement
        var b = Array(n).fill(0);
 
        // Once the array is sorted
        // Re-fill the array again in the
        // mentioned way in the approach
        var low = 0, high = n - 1;
        for (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 (i = 0; i < n; i++) {
 
            // For the first element
            // the adjacents will be a[1] and a[n-1]
            if (i == 0) {
                if (b[n - 1] + b[1] <= b[i]) {
                    document.write(-1);
                    return;
                }
            }
 
            // For the last element
            // the adjacents will be a[0] and a[n-2]
            else if (i == (n - 1)) {
                if (b[n - 2] + b[0] <= b[i]) {
                    document.write(-1);
                    return;
                }
            } else {
                if (b[i - 1] + b[i + 1] <= b[i]) {
                    document.write(-1);
                    return;
                }
            }
        }
 
        // If we reach this position then
        // the arrangement is possible
        for (i = 0; i < n; i++)
            document.write(b[i] + " ");
    }
 
    // Driver code
        var a = [ 1, 4, 4, 3, 2 ];
        var n = a.length;
 
        printArrangement(a, n);
 
// This code is contributed by todaysgaurav 
</script>

Output:

1 3 4 4 2

Time Complexity: O(N log N) //the inbuilt sort function takes N log N time to complete all operations, hence the overall time taken by the algorithm is N log N
Auxiliary Space: O(n) // an extra array is used to store all the elements and in worst case all elements will be stored hence algorithm takes up linear space

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