Sort an array with swapping only with a special element is allowed

Last Updated : 15 Jun, 2022

Given an array of length n + 1, containing elements 1 through n and a space, Requires the use of a given swap (index i, index j) function to sort the array, You can only swap the gap and a number, in the end, put the gap at the end.
There will be a number 999 in the array as a gap or space.

Examples:

Input  : arr = {1, 5, 4, 999, 3, 2}
Output : arr = {1, 2, 3, 4, 5, 999}
We need to sort only by moving 

Input  : arr = {1, 5, 4, 3, 2, 8, 7, 999, 6}
Output : arr = {1, 2, 3, 4, 5, 6, 7, 8, 999}

We follow a recursive approach to solve this problem. As we can only swap numbers with the space. First of all, we find the index of space. If the index is the start of the array, then move this space to the second last index by swapping with each number in its right.
If space is neither start of an array nor the last element of the array and the element before it is greater than the element next to space then do the following.

Step 1: Swap space and element next to space
In case of {3, 999, 2} make it {3, 2, 999}
Step 2 : Swap space and greater element
eg-convert {3, 2, 999} to {999, 2, 3}

Otherwise, elements next to the index are sorted and swap it with the previous element. Again call the sort function with the size of an array decreased by 1 and index of space – 1 as we will get one sorted element each time.

C++

// CPP program to sort an array by moving one
// space around.
#include <bits/stdc++.h>
using namespace std;
 
// n is total number of elements.
// index is index of 999 or space.
// k is number of elements yet to be sorted.
void sortRec(int arr[], int index, int k, int n)
{
    // print the sorted array when loop reaches
    // the base case
    if (k == 0) {
        for (int i = 1; i < n; i++)
            cout << arr[i] << " ";
        cout << 999;
        return;
    }
 
    // else if k>0 and space is at 0th index swap
    // each number with space and store index at
    // second last index
    else if (k > 0 && index == 0) {
        index = n - 2;
        for (int i = 1; i <= index; i++) {
            arr[i - 1] = arr[i];
        }
        arr[index] = 999;
    }
 
    // if space is neither start of array nor last
    // element of array and element before it greater
    // than/ the element next to space
    if (index - 1 >= 0 && index + 1 < n && 
       arr[index - 1] > arr[index + 1]) {
 
        // first swap space and element next to space
        // in case of {3, 999, 2} make it {3, 2, 999}
        swap(arr[index], arr[index + 1]);
 
        // than swap space and greater element
        // convert {3, 2, 999} to {999, 2, 3}
        swap(arr[index - 1], arr[index + 1]);
    }
 
    else
        swap(arr[index], arr[index - 1]);
 
    sortRec(arr, index - 1, k - 1, n);
}
 
// Wrapper over sortRec.
void sort(int arr[], int n)
{
    // Find index of space (or 999)
    int index = -1;
    for (int i = 0; i < n; i++) {
        if (arr[i] == 999) {
            index = i;
            break;
        }
    }
 
    // Invalid input
    if (index == -1)
        return;
 
    sortRec(arr, index, n, n);
}
 
 
// driver program
int main()
{
    int arr[] = { 3, 2, 999, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    sort(arr, n);
    return 0;
}

Java

// Java program to sort an array by moving one 
// space around. 
 
class GFG
{
 
// n is total number of elements. 
// index is index of 999 or space. 
// k is number of elements yet to be sorted. 
static void sortRec(int arr[], int index, int k, int n) 
{ 
    // print the sorted array when loop reaches 
    // the base case 
    if (k == 0)
    { 
        for (int i = 1; i < n; i++) 
                        System.out.print(arr[i] + " "); 
                System.out.println(999); 
        return; 
    } 
 
    // else if k>0 and space is at 0th index swap 
    // each number with space and store index at 
    // second last index 
    else if (k > 0 && index == 0)
    { 
        index = n - 2; 
        for (int i = 1; i <= index; i++)
        { 
            arr[i - 1] = arr[i]; 
        } 
        arr[index] = 999; 
    } 
 
    // if space is neither start of array nor last 
    // element of array and element before it greater 
    // than/ the element next to space 
    if (index - 1 >= 0 && index + 1 < n && 
    arr[index - 1] > arr[index + 1])
    { 
 
        // first swap space and element next to space 
        // in case of {3, 999, 2} make it {3, 2, 999} 
        swap(arr,index, index + 1); 
 
        // than swap space and greater element 
        // convert {3, 2, 999} to {999, 2, 3} 
        swap(arr,index - 1, index + 1); 
    } 
 
    else
        swap(arr,index, index - 1); 
 
    sortRec(arr, index - 1, k - 1, n); 
} 
 
static int[] swap(int []arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
 
// Wrapper over sortRec. 
static void sort(int arr[], int n) 
{ 
    // Find index of space (or 999) 
    int index = -1; 
    for (int i = 0; i < n; i++)
    { 
        if (arr[i] == 999)
        { 
            index = i; 
            break; 
        } 
    } 
 
    // Invalid input 
    if (index == -1) 
        return; 
 
    sortRec(arr, index, n, n); 
} 
 
 
// Driver code 
public static void main(String[] args)
{
    int arr[] = { 3, 2, 999, 1 }; 
    int n = arr.length; 
    sort(arr, n); 
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 program to sort an array
# by moving one space around.
  
# n is total number of elements.
# index is index of 999 or space.
# k is number of elements yet to be sorted.
def sortRec(arr, index, k, n):
     
    # Print the sorted array when loop reaches
    # the base case
    if (k == 0):
        for i in range(1,n):
            print(arr[i],end=" ")
        print(999,end="")
     
    # Else if k>0 and space is at 0th index swap
    # each number with space and store index at
    # second last index
    elif(k > 0 and index == 0):
        index = n - 2
         
        for i in range(1,index+1):
            arr[i - 1] = arr[i]
        arr[index] = 999
     
    # If space is neither start of array nor last
    # element of array and element before it greater
    # than/ the element next to space
    if (index - 1 >= 0 and index + 1 < n and arr[index - 1] > arr[index + 1]):
         
        # First swap space and element next to space
        # in case of {3, 999, 2} make it {3, 2, 999}
        arr[index],arr[index+1] = arr[index+1],arr[index]
         
        # Than swap space and greater element
        # convert {3, 2, 999} to {999, 2, 3}
        arr[index-1],arr[index+1] = arr[index+1],arr[index-1]
     
    else:
        if(index-1<0):
            return
        arr[index],arr[index-1] = arr[index-1],arr[index]
         
    sortRec(arr, index - 1, k - 1, n)
 
# Wrapper over sortRec.
def sort(arr, n):
     
    # Find index of space (or 999)
    index = -1
     
    for i in range(n):
        if (arr[i] == 999):
            index = i
            break
    #  Invalid input
    if (index == -1):
        return
    sortRec(arr, index, n, n)
 
# Driver Code
arr = [ 3, 2, 999, 1 ]
n=len(arr)
sort(arr, n)
 
# This code is contributed by rag2127.

C#

// C# program to sort an array by moving one 
// space around. 
using System;
 
class GFG 
{ 
 
// n is total number of elements. 
// index is index of 999 or space. 
// k is number of elements yet to be sorted. 
static void sortRec(int []arr, int index, int k, int n) 
{ 
    // print the sorted array when loop reaches 
    // the base case 
    if (k == 0) 
    { 
        for (int i = 1; i < n; i++) 
                        Console.Write(arr[i] + " "); 
                Console.WriteLine(999); 
        return; 
    } 
 
    // else if k>0 and space is at 0th index swap 
    // each number with space and store index at 
    // second last index 
    else if (k > 0 && index == 0) 
    { 
        index = n - 2; 
        for (int i = 1; i <= index; i++) 
        { 
            arr[i - 1] = arr[i]; 
        } 
        arr[index] = 999; 
    } 
 
    // if space is neither start of array nor last 
    // element of array and element before it greater 
    // than/ the element next to space 
    if (index - 1 >= 0 && index + 1 < n && 
    arr[index - 1] > arr[index + 1]) 
    { 
 
        // first swap space and element next to space 
        // in case of {3, 999, 2} make it {3, 2, 999} 
        swap(arr,index, index + 1); 
 
        // than swap space and greater element 
        // convert {3, 2, 999} to {999, 2, 3} 
        swap(arr,index - 1, index + 1); 
    } 
 
    else
        swap(arr,index, index - 1); 
 
    sortRec(arr, index - 1, k - 1, n); 
} 
 
static int[] swap(int []arr, int i, int j) 
{ 
    int temp = arr[i]; 
    arr[i] = arr[j]; 
    arr[j] = temp; 
    return arr; 
} 
 
// Wrapper over sortRec. 
static void sort(int []arr, int n) 
{ 
    // Find index of space (or 999) 
    int index = -1; 
    for (int i = 0; i < n; i++) 
    { 
        if (arr[i] == 999) 
        { 
            index = i; 
            break; 
        } 
    } 
 
    // Invalid input 
    if (index == -1) 
        return; 
 
    sortRec(arr, index, n, n); 
} 
 
 
// Driver code 
public static void Main(String[] args) 
{ 
    int []arr = { 3, 2, 999, 1 }; 
    int n = arr.Length; 
    sort(arr, n); 
} 
} 
 
// This code has been contributed by 29AjayKumar

Javascript

<script>
 
// JavaScript program to sort an array
// by moving one space around.
 
// n is total number of elements. 
// index is index of 999 or space. 
// k is number of elements yet to be sorted. 
function sortRec(arr, index, k, n) 
{ 
     
    // Print the sorted array when loop reaches 
    // the base case 
    if (k == 0)
    { 
        for(let i = 1; i < n; i++) 
           document.write(arr[i] + " "); 
            
        document.write(999); 
        return; 
    } 
   
    // Else if k>0 and space is at 0th index swap 
    // each number with space and store index at 
    // second last index 
    else if (k > 0 && index == 0)
    { 
        index = n - 2; 
        for(let i = 1; i <= index; i++)
        { 
            arr[i - 1] = arr[i]; 
        } 
        arr[index] = 999; 
    } 
   
    // If space is neither start of array nor last 
    // element of array and element before it greater 
    // than/ the element next to space 
    if (index - 1 >= 0 && index + 1 < n && 
     arr[index - 1] > arr[index + 1])
    { 
   
        // First swap space and element next to space 
        // in case of {3, 999, 2} make it {3, 2, 999} 
        swap(arr, index, index + 1); 
   
        // Than swap space and greater element 
        // convert {3, 2, 999} to {999, 2, 3} 
        swap(arr, index - 1, index + 1); 
    } 
    else
        swap(arr,index, index - 1); 
   
    sortRec(arr, index - 1, k - 1, n); 
} 
   
function swap(arr, i, j)
{
    let temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
   
// Wrapper over sortRec. 
function sort(arr, n) 
{ 
     
    // Find index of space (or 999) 
    let index = -1; 
    for(let i = 0; i < n; i++)
    { 
        if (arr[i] == 999)
        { 
            index = i; 
            break; 
        } 
    } 
   
    // Invalid input 
    if (index == -1) 
        return; 
   
    sortRec(arr, index, n, n); 
} 
 
// Driver Code
let arr = [ 3, 2, 999, 1 ]; 
let n = arr.length; 
 
sort(arr, n); 
 
// This code is contributed by susmitakundugoaldanga
 
</script>

Output:

1 2 3 999

Time Complexity: O(n²)
Auxiliary Space: O(n)

Iterative diagonal traversal of binary tree

niteesh_Kr

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Sort an array with swapping only with a special element is allowed

C++

Java

Python3

C#

Javascript

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