Java Program for Sorting all array elements except one
Given an array, a positive integer, sort the array in ascending order such that the element at index K in the unsorted array stays unmoved and all other elements are sorted.
Examples:
Input : arr[] = {10, 4, 11, 7, 6, 20} k = 2; Output : arr[] = {4, 6, 11, 7, 10, 20} Input : arr[] = {30, 20, 10} k = 0 Output : arr[] = {30, 10, 20}
A simple solution is to copy all elements except k-th of a given array to another array. Then sort the other array using a sorting algorithm. Finally, again copy the sorted array to the original array. While copying, skip k-th element.
Java
// Java code for the approach import java.util.Arrays; public class GFG { // Driver code public static void main(String[] args) { int [] arr = { 10 , 4 , 11 , 7 , 6 , 20 }; int k = 2 ; int n = arr.length; // Function Call sortExceptK(arr, k); // Print final array for ( int i = 0 ; i < n; i++) System.out.print(arr[i] + " " ); } // Function to sort an array in ascending order // except for the element at index k public static void sortExceptK( int [] arr, int k) { int [] temp = new int [arr.length - 1 ]; int index = 0 ; // Copy all elements except k-th to temp array for ( int i = 0 ; i < arr.length; i++) { if (i != k) { temp[index++] = arr[i]; } } // Sort the temp array Arrays.sort(temp); // Copy the sorted array back to original array index = 0 ; for ( int i = 0 ; i < arr.length; i++) { if (i != k) { arr[i] = temp[index++]; } } } } |
Output
4 6 11 7 10 20
Time Complexity: O(n*log2n) as sorting takes n*log2n time.
Space Complexity: O(n) as temp array has been created.
Below is an efficient solution.
- Swap k-th element with the last element.
- Sort all elements except the last.
- For every element from (k+1)-th to last, move them one position ahead.1
- Copy k-th element back to position k.
Java
// Java program to sort all elements except // element at index k. import java.util.Arrays; class GFG { static int sortExceptK( int arr[], int k, int n) { // Move k-th element to end int temp = arr[k]; arr[k] = arr[n- 1 ]; arr[n- 1 ] = temp; // Sort all elements except last Arrays.sort(arr, 0 , n- 1 ); // Store last element (originally k-th) int last = arr[n- 1 ]; // Move all elements from k-th to one // position ahead. for ( int i = n- 1 ; i > k; i--) arr[i] = arr[i- 1 ]; // Restore k-th element arr[k] = last; return 0 ; } //Driver code public static void main (String[] args) { int a[] = { 10 , 4 , 11 , 7 , 6 , 20 }; int k = 2 ; int n = a.length; sortExceptK(a, k, n); for ( int i = 0 ; i < n; i++) System.out.print(a[i] + " " ); } } //This code is contributed by Anant Agarwal. |
Output
4 6 11 7 10 20
Time Complexity: O(n*log(n)) where n is the number of elements.
Auxiliary Space: O(1)
Please refer complete article on Sorting all array elements except one for more details!
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