Given an array a[] with n integers the task is to rearrange the elements of the array in such a way that the differences of the adjacent elements are in descending order.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 6} Output : 6 1 5 2 4 3 Explanation: For first two elements the difference is abs(6-1)=5 For next two elements the difference is abs(1-5)=4 For next two elements the difference is abs(5-2)=3 For next two elements the difference is abs(2-4)=2 For next two elements the difference is abs(4-3)=1 Hence, difference array is 5, 4, 3, 2, 1. Input : arr[] = {7, 10, 2, 4, 5} Output : 10 2 7 4 5 Explanation: For first two elements the difference is abs(10-2)=8 For next two elements the difference is abs(2-7)=5 For next two elements the difference is abs(7-4)=3 For next two elements the difference is abs(4-5)=1 Hence, difference array is 8, 5, 3, 1.
Approach:
To solve the problem mentioned above we have to sort the array and then print the elements in the following order for the even-sized array :
a[n], a[1], a[n-1], a[2] ….. a[(n/2)], [(n/2)+1]
And for odd-sized array, we have to print array elements in the following order:
a[n], a[1], a[n-1], a[2] ….. a[(n/2)+1]
Below is the implementation of the above approach:
// C++ implementation to Rearrange array // such that difference of adjacent // elements is in descending order #include <bits/stdc++.h> using namespace std;
// Function to print array in given order void printArray( int * a, int n)
{ // Sort the array
sort(a, a + n);
int i = 0;
int j = n - 1;
// Check elements till the middle index
while (i <= j) {
// check if length is odd
// print the middle index at last
if (i == j) {
cout << a[i] << " " ;
}
// Print the remaining elements
// in the described order
else {
cout << a[j] << " " ;
cout << a[i] << " " ;
}
i = i + 1;
j = j - 1;
}
cout << endl;
} // Driver code int main()
{ // array declaration
int arr1[] = { 1, 2, 3, 4, 5, 6 };
// size of array
int n1 = sizeof (arr1) / sizeof (arr1[0]);
printArray(arr1, n1);
} |
// Java implementation to Rearrange array // such that difference of adjacent // elements is in descending order import java.util.*;
class GFG {
// Function to print array in given order static void printArray( int []a, int n)
{ // Sort the array
Arrays.sort(a);
int i = 0 ;
int j = n - 1 ;
// Check elements till the
// middle index
while (i <= j)
{
// Check if length is odd print
// the middle index at last
if (i == j)
{
System.out.print(a[i] + " " );
}
// Print the remaining elements
// in the described order
else
{
System.out.print(a[j] + " " );
System.out.print(a[i] + " " );
}
i = i + 1 ;
j = j - 1 ;
}
System.out.println();
} // Driver code public static void main (String[] args)
{ // Array declaration
int arr1[] = { 1 , 2 , 3 , 4 , 5 , 6 };
// Size of array
int n1 = arr1.length;
printArray(arr1, n1);
} } // This code is contributed by AnkitRai01 |
# Python3 implementation to rearrange # array such that difference of adjacent # elements is in descending order # Function to print array in given order def printArray(a, n):
# Sort the array
a.sort();
i = 0 ;
j = n - 1 ;
# Check elements till the middle index
while (i < = j):
# Check if length is odd print
# the middle index at last
if (i = = j):
print (a[i], end = " " );
# Print the remaining elements
# in the described order
else :
print (a[j], end = " " );
print (a[i], end = " " );
i = i + 1 ;
j = j - 1 ;
print ();
# Driver code if __name__ = = "__main__" :
# Array declaration
arr1 = [ 1 , 2 , 3 , 4 , 5 , 6 ];
# Size of array
n1 = len (arr1);
printArray(arr1, n1);
# This code is contributed by AnkitRai01 |
// C# implementation to rearrange array // such that difference of adjacent // elements is in descending order using System;
class GFG {
// Function to print array in given order static void printArray( int []a, int n)
{ // Sort the array
Array.Sort(a);
int i = 0;
int j = n - 1;
// Check elements till the
// middle index
while (i <= j)
{
// Check if length is odd print
// the middle index at last
if (i == j)
{
Console.Write(a[i] + " " );
}
// Print the remaining elements
// in the described order
else
{
Console.Write(a[j] + " " );
Console.Write(a[i] + " " );
}
i = i + 1;
j = j - 1;
}
Console.WriteLine();
} // Driver code public static void Main ( string [] args)
{ // Array declaration
int []arr1 = { 1, 2, 3, 4, 5, 6 };
// Size of array
int n1 = arr1.Length;
printArray(arr1, n1);
} } // This code is contributed by AnkitRai01 |
<script> // JavaScript implementation to Rearrange array
// such that difference of adjacent
// elements is in descending order
// Function to print array in given order
function printArray(a, n)
{
// Sort the array
a.sort( function (a, b) {
return a - b;
});
var i = 0;
var j = n - 1;
// Check elements till the middle index
while (i <= j) {
// check if length is odd
// print the middle index at last
if (i == j) {
document.write(a[i] + " " );
}
// Print the remaining elements
// in the described order
else {
document.write(a[j] + " " );
document.write(a[i] + " " );
}
i = i + 1;
j = j - 1;
}
document.write( "<br>" );
}
// Driver code
// array declaration
var arr1 = [1, 2, 3, 4, 5, 6];
// size of array
var n1 = arr1.length;
printArray(arr1, n1);
</script> |
6 1 5 2 4 3
Time Complexity: O(N * logN) 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(1) since no extra array is used so the space taken by the algorithm is constant