Hello Everyone,
Problem Statement.: Given an array of integers, task is to print the array in the order – smallest number, Largest number, 2nd smallest number, 2nd largest number, 3rd smallest number, 3rd largest number and so on……
Examples:
Input : arr[] = [5, 8, 1, 4, 2, 9, 3, 7, 6]
Output :arr[] = {1, 9, 2, 8, 3, 7, 4, 6, 5}
Input : arr[] = [1, 2, 3, 4]
Output :arr[] = {1, 4, 2, 3}
A simple solution is to first find the smallest element, swap it with first element. Then find largest element, swap it with second element and so on. Time complexity of this solution is O(n2).
An efficient solution is to use sorting
- Sort the elements of array.
- Take two variables say i and j and point them to the first and last index of the array respectively.
- Now run a loop and store the elements in the array one by one by incrementing i and decrementing j.
Let’s take an array with input 5, 8, 1, 4, 2, 9, 3, 7, 6 and sort them so the array become 1, 2, 3, 4, 5, 6, 7, 8, 9. Now take two variables say i and j and point them to the first and last index of the array respectively, run a loop and store value into new array by incrementing i and decrementing j. We get final result as 1 9 2 8 3 7 4 6 5.
// C++ program to print the array in given order
#include <bits/stdc++.h>
using namespace std;
// Function which arrange the array.
void rearrangeArray( int arr[], int n)
{
// Sorting the array elements
sort(arr, arr + n);
int tempArr[n]; // To store modified array
// Adding numbers from sorted array to
// new array accordingly
int ArrIndex = 0;
// Traverse from begin and end simultaneously
for ( int i = 0, j = n-1; i <= n / 2 ||
j > n / 2; i++, j--) {
tempArr[ArrIndex] = arr[i];
ArrIndex++;
tempArr[ArrIndex] = arr[j];
ArrIndex++;
}
// Modifying original array
for ( int i = 0; i < n; i++)
arr[i] = tempArr[i];
}
// Driver Code
int main()
{
int arr[] = { 5, 8, 1, 4, 2, 9, 3, 7, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
rearrangeArray(arr, n);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
Output:
1 9 2 8 3 7 4 6 5
Time Complexity : O(n Log n)
Auxiliary Space : O(n)