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Rearrange Array to minimize difference of sum of squares of odd and even index elements

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  • Last Updated : 31 Mar, 2022

Given an array arr[] of size N (multiple of 8) where the values in the array will be in the range [a, (a+8*N) -1] (a can be any positive integer), the task is to rearrange the array in a way such that the difference between the sum of squares at odd indices and sum of squares of the elements at even indices is the minimum among all possible rearrangements.

Note: If there are multiple rearrangements return any one of those.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}
Output: 1 2 4 3 7 8 6 5 
Explanation:  The difference is 0 as 1 + 42 + 72 + 62 = 102 = 22 + 32 + 82 + 52

Input: arr[] = { 9, 11, 12, 15, 16, 13, 10, 14}
Output: 9 10 12 11 15 16 14 13
Explanation: The difference is 0 as 92 + 122 + 152 + 142 = 102 + 112 + 162 + 132 = 646

 

Approach: This problem can be solved based on the following mathematical observation:

For any positive integer S, ( S )2 + ( S+3 )2 – 4 = ( S+1 )2 + ( S+2 )2. As N is a multiple of 8 so it can be divided into N/8 groups where the difference of sum of squares of elements at odd and even indices for each group is 0.
For the first four elements keep the sum of squares at odd indices greater and four the next four just the opposite to keep the sum of squares of even indices more.  So this group of 8 elements will have difference 0. As the similar is done for all N/8 groups the overall difference will be 0.

The sequence of each group can be like: S, S+1, S+3, S+2, S+6, S+7, S+5, S+4

Follow the below steps to solve this problem: 

  • Divide the array into groups of size 8.
  • Arrange elements in each group as derived from the observation.
  • Return the rearranged array.

Below is the implementation of the above approach:

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find
// maximum element of the array
int maximum(int arr[], int size)
{
    int ma = INT_MIN;
    for (int i = 0; i < size; i++) {
        ma = max(ma, arr[i]);
    }
    return ma;
}
 
// Function to find
// minimum element of the array
int minimum(int arr[], int size)
{
    int mi = INT_MAX;
    for (int i = 0; i < size; i++) {
        mi = min(mi, arr[i]);
    }
    return mi;
}
 
// Function to print the array
void print_min(int arr[], int size)
{
    int low = minimum(arr, size);
    int high = maximum(arr, size);
 
    // using the fact that
    // s^2 + (s+3)^2 = (s+1)^2 + (s+2)^2 + 4.
    for (int i = 0; i < size; i += 4) {
 
        // Making the difference +4
        // for the odd indices
        if (i % 8 == 0) {
            arr[i] = low;
            arr[i + 2] = low + 3;
            arr[i + 1] = low + 1;
            arr[i + 3] = low + 2;
        }
 
        // Making the difference -4 for
        // odd indices +4 - 4 = 0 (balanced)
        else {
            arr[i] = low + 2;
            arr[i + 2] = low + 1;
            arr[i + 1] = low + 3;
            arr[i + 3] = low;
        }
        low += 4;
    }
 
    // Printing the array
    for (int i = 0; i < size; i++) {
        cout << arr[i] << " ";
    }
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
    int N = sizeof(arr) / (sizeof(int));
 
    // Function call
    print_min(arr, N);
    return 0;
}

Java




// JAVA code to implement the approach
import java.util.*;
class GFG
{
 
  // Function to find
  // maximum element of the array
  public static int maximum(int arr[], int size)
  {
    int ma = Integer.MIN_VALUE;
    for (int i = 0; i < size; i++) {
      ma = Math.max(ma, arr[i]);
    }
    return ma;
  }
 
  // Function to find
  // minimum element of the array
  public static int minimum(int arr[], int size)
  {
    int mi = Integer.MAX_VALUE;
    for (int i = 0; i < size; i++) {
      mi = Math.min(mi, arr[i]);
    }
    return mi;
  }
 
  // Function to print the array
  public static void print_min(int arr[], int size)
  {
    int low = minimum(arr, size);
    int high = maximum(arr, size);
 
    // using the fact that
    // s^2 + (s+3)^2 = (s+1)^2 + (s+2)^2 + 4.
    for (int i = 0; i < size; i += 4) {
 
      // Making the difference +4
      // for the odd indices
      if (i % 8 == 0) {
        arr[i] = low;
        arr[i + 2] = low + 3;
        arr[i + 1] = low + 1;
        arr[i + 3] = low + 2;
      }
 
      // Making the difference -4 for
      // odd indices +4 - 4 = 0 (balanced)
      else {
        arr[i] = low + 2;
        arr[i + 2] = low + 1;
        arr[i + 1] = low + 3;
        arr[i + 3] = low;
      }
      low += 4;
    }
 
    // Printing the array
    for (int i = 0; i < size; i++) {
      System.out.print(arr[i] + " ");
    }
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
    int N = arr.length;
 
    // Function call
    print_min(arr, N);
  }
}
 
// This code is contributed by Taranpreet

Python3




# Python code to implement the approach
INT_MIN = -2147483647 - 1
INT_MAX = 2147483647
 
# Function to find
# maximum element of the array
def maximum(arr, size):
    ma = INT_MIN
    for i in range(size):
        ma = max(ma, arr[i])
     
    return ma
 
# Function to find
# minimum element of the array
def minimum(arr, size):
    mi = INT_MAX
    for i in range(size):
        mi = min(mi, arr[i])
    
    return mi
 
# Function to print the array
def print_min(arr, size):
    low = minimum(arr, size)
    high = maximum(arr, size)
 
    # using the fact that
    # s^2 + (s+3)^2 = (s+1)^2 + (s+2)^2 + 4.
    for i in range(0,size,4):
 
        # Making the difference +4
        # for the odd indices
        if (i % 8 == 0):
            arr[i] = low
            arr[i + 2] = low + 3
            arr[i + 1] = low + 1
            arr[i + 3] = low + 2
 
        # Making the difference -4 for
        # odd indices +4 - 4 = 0 (balanced)
        else:
            arr[i] = low + 2
            arr[i + 2] = low + 1
            arr[i + 1] = low + 3
            arr[i + 3] = low
         
        low += 4
 
    # Printing the array
    for i in range(size):
        print(arr[i],end=" ")
 
 
# Driver code
 
arr = [1, 2, 3, 4, 5, 6, 7, 8]
N = len(arr)
 
# Function call
print_min(arr, N)
 
# This code is contributed by shinjanpatra

C#




// C# code to implement the approach
using System;
class GFG {
 
  // Function to find
  // maximum element of the array
  static int maximum(int[] arr, int size)
  {
    int ma = Int32.MinValue;
    for (int i = 0; i < size; i++) {
      ma = Math.Max(ma, arr[i]);
    }
    return ma;
  }
 
  // Function to find
  // minimum element of the array
  static int minimum(int[] arr, int size)
  {
    int mi = Int32.MaxValue;
    for (int i = 0; i < size; i++) {
      mi = Math.Min(mi, arr[i]);
    }
    return mi;
  }
 
  // Function to print the array
  static void print_min(int[] arr, int size)
  {
    int low = minimum(arr, size);
    int high = maximum(arr, size);
 
    // using the fact that
    // s^2 + (s+3)^2 = (s+1)^2 + (s+2)^2 + 4.
    for (int i = 0; i < size; i += 4) {
 
      // Making the difference +4
      // for the odd indices
      if (i % 8 == 0) {
        arr[i] = low;
        arr[i + 2] = low + 3;
        arr[i + 1] = low + 1;
        arr[i + 3] = low + 2;
      }
 
      // Making the difference -4 for
      // odd indices +4 - 4 = 0 (balanced)
      else {
        arr[i] = low + 2;
        arr[i + 2] = low + 1;
        arr[i + 1] = low + 3;
        arr[i + 3] = low;
      }
      low += 4;
    }
 
    // Printing the array
    for (int i = 0; i < size; i++) {
      Console.Write(arr[i] + " ");
    }
  }
 
  // Driver code
  public static void Main()
  {
    int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8 };
    int N = arr.Length;
 
    // Function call
    print_min(arr, N);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
    // JavaScript code to implement the approach
    const INT_MIN = -2147483647 - 1;
    const INT_MAX = 2147483647;
 
    // Function to find
    // maximum element of the array
    const maximum = (arr, size) => {
        let ma = INT_MIN;
        for (let i = 0; i < size; i++) {
            ma = Math.max(ma, arr[i]);
        }
        return ma;
    }
 
    // Function to find
    // minimum element of the array
    const minimum = (arr, size) => {
        let mi = INT_MAX;
        for (let i = 0; i < size; i++) {
            mi = Math.min(mi, arr[i]);
        }
        return mi;
    }
 
    // Function to print the array
    const print_min = (arr, size) => {
        let low = minimum(arr, size);
        let high = maximum(arr, size);
 
        // using the fact that
        // s^2 + (s+3)^2 = (s+1)^2 + (s+2)^2 + 4.
        for (let i = 0; i < size; i += 4) {
 
            // Making the difference +4
            // for the odd indices
            if (i % 8 == 0) {
                arr[i] = low;
                arr[i + 2] = low + 3;
                arr[i + 1] = low + 1;
                arr[i + 3] = low + 2;
            }
 
            // Making the difference -4 for
            // odd indices +4 - 4 = 0 (balanced)
            else {
                arr[i] = low + 2;
                arr[i + 2] = low + 1;
                arr[i + 1] = low + 3;
                arr[i + 3] = low;
            }
            low += 4;
        }
 
        // Printing the array
        for (let i = 0; i < size; i++) {
            document.write(`${arr[i]} `);
        }
    }
 
    // Driver code
 
    let arr = [1, 2, 3, 4, 5, 6, 7, 8];
    let N = arr.length;
 
    // Function call
    print_min(arr, N);
 
// This code is contributed by rakeshsahni
 
</script>

Output

1 2 4 3 7 8 6 5 

Time Complexity: O(N)
Auxiliary Space: O(1)


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