Given an array arr[] of N unique integers, the task is to find the cost to arrange them in a circular arrangement in such a way that every element is less than or equal to the the sum of its adjacent elements.
Cost of moving an element from index i in original array to index j in final arrangement is |i – j|
In case such an arrangement is not possible, then print -1.
Examples:
Input: arr[] = {2, 4, 5, 1, 3}
Output: 10
Explanation:
One of the possible arrangment is {1, 2, 3, 4, 5}
For index 1, 1 ≤ 4 + 2, cost = 4 – 1 = 3
For index 2, 2 ≤ 1 + 3, cost = 3 + (2 – 1) = 4
For index 3, 3 ≤ 2 + 4, cost = 4 + (5 – 3) = 6
For index 4, 4 ≤ 3 + 4, cost = 6 + (4 – 2) = 8
For index 5, 5 ≤ 4 + 1, cost = 8 + (5 – 3) = 10Input: arr[] = {1, 10, 100, 1000}
Output: -1
Approach: The problem can be solved using a Greedy Approach. The idea is to store the the original index of the elements in a hashmap and then sort the array. Now check if the given condition is satisfied or not. If found to be true, then calculate the cost by adding up the difference between the current and previous indices. Otherwise, print -1.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to check if given elements // can be arranged such that sum of // its neighbours is strictly greater void Arrange(int arr[], int n) { // Initialize the total cost int cost = 0; // Storing the orginal index of // elements in a hashmap unordered_map<int, int> index; for (int i = 0; i < n; i++) { index[arr[i]] = i; } // Sort the given array sort(arr, arr + n); // Check if a given condition // is satisfies or not for (int i = 0; i < n; i++) { // First number if (i == 0) { if (arr[i] > arr[i + 1] + arr[n - 1]) { cout << "-1"; return; } else { // Add the cost to overall cost cost += abs(index[arr[i]] - i); } } // Last number else if (i == n - 1) { if (arr[i] > arr[i - 1] + arr[0]) { cout << "-1"; return; } else { // Add the cost to // overall cost cost += abs(index[arr[i]] - i); } } else { if (arr[i] > arr[i - 1] + arr[i + 1]) { cout << "-1"; return; } else { // Add the cost to // overall cost cost += abs(index[arr[i]] - i); } } } // Printing the cost cout << cost; return; } // Driver Code int main() { // Given array int arr[] = { 2, 4, 5, 1, 3 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call Arrange(arr, N); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ // Function to check if given elements // can be arranged such that sum of // its neighbors is strictly greater static void Arrange(int arr[], int n) { // Initialize the total cost int cost = 0; // Storing the orginal index of // elements in a hashmap HashMap<Integer, Integer> index = new HashMap<Integer, Integer>(); for(int i = 0; i < n; i++) { index.put(arr[i], i); } // Sort the given array Arrays.sort(arr); // Check if a given condition // is satisfies or not for(int i = 0; i < n; i++) { // First number if (i == 0) { if (arr[i] > arr[i + 1] + arr[n - 1]) { System.out.print("-1"); return; } else { // Add the cost to overall cost cost += Math.abs(index.get(arr[i]) - i); } } // Last number else if (i == n - 1) { if (arr[i] > arr[i - 1] + arr[0]) { System.out.print("-1"); return; } else { // Add the cost to // overall cost cost += Math.abs(index.get(arr[i]) - i); } } else { if (arr[i] > arr[i - 1] + arr[i + 1]) { System.out.print("-1"); return; } else { // Add the cost to // overall cost cost += Math.abs(index.get(arr[i]) - i); } } } // Printing the cost System.out.print(cost); return; } // Driver Code public static void main(String[] args) { // Given array int arr[] = { 2, 4, 5, 1, 3 }; int N = arr.length; // Function call Arrange(arr, N); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to implement # the above approach # Function to check if given elements # can be arranged such that sum of # its neighbours is strictly greater def Arrange(arr, n): # Initialize the total cost cost = 0 # Storing the orginal index of # elements in a hashmap index = {} for i in range(n): index[arr[i]] = i # Sort the given array arr.sort() # Check if a given condition # is satisfies or not for i in range(n): # First number if(i == 0): if(arr[i] > arr[i + 1] + arr[-1]): print("-1") return else: # Add the cost to overall cost cost += abs(index[arr[i]] - i) # Last number elif(i == n - 1): if(arr[i] > arr[i - 1] + arr[0]): print("-1") return else: # Add the cost to # overall cost cost += abs(index[arr[i]] - i) else: if(arr[i] > arr[i - 1] + arr[i + 1]): print("-1") return else: # Add the cost to # overall cost cost += abs(index[arr[i]] - i) # Printing the cost print(cost) return # Driver Code # Given array arr = [ 2, 4, 5, 1, 3 ] N = len(arr) # Function call Arrange(arr, N) # This code is contributed by Shivam Singh |
C#
// C# program to implement // the above approach using System; using System.Collections.Generic; class GFG{ // Function to check if given elements // can be arranged such that sum of // its neighbors is strictly greater static void Arrange(int []arr, int n) { // Initialize the total cost int cost = 0; // Storing the orginal index of // elements in a hashmap Dictionary<int, int> index = new Dictionary<int, int>(); for(int i = 0; i < n; i++) { index.Add(arr[i], i); } // Sort the given array Array.Sort(arr); // Check if a given condition // is satisfies or not for(int i = 0; i < n; i++) { // First number if (i == 0) { if (arr[i] > arr[i + 1] + arr[n - 1]) { Console.Write("-1"); return; } else { // Add the cost to overall cost cost += Math.Abs(index[arr[i]] - i); } } // Last number else if (i == n - 1) { if (arr[i] > arr[i - 1] + arr[0]) { Console.Write("-1"); return; } else { // Add the cost to // overall cost cost += Math.Abs(index[arr[i]] - i); } } else { if (arr[i] > arr[i - 1] + arr[i + 1]) { Console.Write("-1"); return; } else { // Add the cost to // overall cost cost += Math.Abs(index[arr[i]] - i); } } } // Printing the cost Console.Write(cost); return; } // Driver Code public static void Main(String[] args) { // Given array int []arr = { 2, 4, 5, 1, 3 }; int N = arr.Length; // Function call Arrange(arr, N); } } // This code is contributed by shikhasingrajput |
Output:
10
Time Complexity: O(N log N)
Auxiliary Space: O(N)
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