Sort the given Array in Spiral manner starting from the center - GeeksforGeeks

01 Sep, 2021
Given an array arr[] of size N, the task is to sort the array in descending order starting from the mid of the array having the next largest element at the right of the middle element and the next largest at the left of the middle element and so on.
Examples:
Input: arr[] = {4, 9, 3, 5, 7}Output: 3 5 9 7 4Explanation:The largest element in the array is 9 is kept in the middle of the array at index = 2.The next largest element is 7 is placed to the right of the middle element at index = 3.The 3rd largest element is 5 is placed to the left of the middle element at index = 1.The 4th largest element is 4 is placed to the right of the element is 7 at index = 4.The smallest element is 3 is placed to left of the element is 5 at index = 0.
Input: arr[] = {4, 5, 3, 7, 6, 9, 7}Output: 3 5 7 9 7 6 4Explanation:The largest element in the array is 9 is kept in the middle of the array at index = 3.The next largest element is 7 is placed to the right of the middle element at index = 3.The 3rd largest element is 7 is placed to the left of the middle element at index = 2.The 4th largest element is 6 is placed to the right of the element 7 at index = 5.The 5th largest element is 5 is placed to the left of the element 7 at index = 1.The 6th largest element is 4 is placed at the right of the element 6 at index = 6.The smallest element is 3 is placed to the left of the element 5 at index = 0.
Approach: The idea is to use bubble sort and to start sorting the array from the smallest element. First, place the smallest element in the left-most index and then, place the next smallest element in the rightmost index. Below are the steps:
Initialize variables, say left = 0, right = N – 1 and i = 1.
Start at the left position i.e. from left = 0 then perform bubble sort in the remaining part of the array finding the smallest element and placing it at the left-most position until i == right.
Increase the value of left by 1 since its element is already placed, and doesn’t need to be changed.
Start at the right position and perform bubble sort traversing in reverse direction again finding the smallest element and placing it in the position right until i == left.
Decrement the value of right by 1 since its element is already placed, and doesn’t need to be changed.
If N is even, the smallest element should come at the right extreme of the array but in our approach, the smallest element is placed at the left-most extreme. So, for achieving the required pattern, reverse the first half of the array.
Print the modified sorted array.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach #include <bits/stdc++.h>using namespace std; // Function to sort the array// according to the given patternvoid centerSpiralSort(int arr[], int N){    // Initializing the variables    int left = 0;    int right = N - 1;    int i = 1;     while (left < right) {         // Performing bubble sort        for (i = left + 1; i <= right;             i++) {            if (arr[left] > arr[i]) {                 // Swapping if                // arr[left] > arr[i]                swap(arr[left], arr[i]);            }        }         // Increment left by 1        left++;         // Performing bubble sort        for (i = right - 1; i >= left;             i--) {            if (arr[right] > arr[i]) {                 // Swapping if                // arr[right] > arr[i]                swap(arr[right], arr[i]);            }        }         // Decrement right by 1        right--;    }     // If N is an even number    // Reversing the array from 0 to N/2-1    if (N % 2 == 0) {         // Reversing the half array        for (int i = 0; i < N / 2; i++) {            swap(arr[i], arr[N - 1 - i]);        }    }     // Print the elements of the array    for (int i = 0; i < N; i++) {        cout << arr[i] << " ";    }} // Driver Codeint main(){    // Given array    int arr[] = { 4, 9, 3, 5, 7 };     // Size of the array    int N = sizeof arr / sizeof arr[0];     // Function Call    centerSpiralSort(arr, N);     return 0;}
// Java program for the above approachimport java.io.*;class GFG{       // Function to sort the array    // according to the given pattern    static void centerSpiralSort(int arr[], int N)    {               // Initializing the variables        int left = 0;        int right = N - 1;        int i = 1;        int temp;         while (left < right) {             // Performing bubble sort            for (i = left + 1; i <= right; i++) {                if (arr[left] > arr[i]) {                     // Swapping if                    // arr[left] > arr[i]                    temp = arr[left];                    arr[left] = arr[i];                    arr[i] = temp;                }            }             // Increment left by 1            left++;             // Performing bubble sort            for (i = right - 1; i >= left; i--) {                if (arr[right] > arr[i]) {                     // Swapping if                    // arr[right] > arr[i]                    temp = arr[right];                    arr[right] = arr[i];                    arr[i] = temp;                }            }             // Decrement right by 1            right--;        }         // If N is an even number        // Reversing the array from 0 to N/2-1        if (N % 2 == 0) {             // Reversing the half array            for (i = 0; i < N / 2; i++) {                temp = arr[i];                arr[i] = arr[N - 1 - i];                arr[N - 1 - i] = temp;            }        }         // Print the elements of the array        for (i = 0; i < N; i++) {            System.out.print(arr[i] + " ");        }    }     // Driver Code    public static void main(String[] args)    {               // Given array        int arr[] = { 4, 9, 3, 5, 7 };         // Size of the array        int N = arr.length;         // Function Call        centerSpiralSort(arr, N);    }} // This code is contributed by Potta Lokesh
# Python3 program for the above approach # Function to sort the array# according to the given patterndef centerSpiralSort(arr, N):         # Initializing the variables    left = 0    right = N - 1    i = 1     while (left < right):                 # Performing bubble sort        for i in range(left + 1, right + 1, 1):            if (arr[left] > arr[i]):                                 # Swapping if                # arr[left] > arr[i]                temp = arr[left]                arr[left] = arr[i]                arr[i] = temp         # Increment left by 1        left += 1         # Performing bubble sort        i = right - 1                 while(i >= left):            if (arr[right] > arr[i]):                                 # Swapping if                # arr[right] > arr[i]                temp = arr[right]                arr[right] = arr[i]                arr[i] = temp                             i -= 1         # Decrement right by 1        right -= 1     # If N is an even number    # Reversing the array from 0 to N/2-1    if (N % 2 == 0):                 # Reversing the half array        for i in range(N / 2):            temp = arr[i]            arr[i] = arr[N - 1 - i]            arr[N - 1 - i] = temp     # Print the elements of the array    for i in range(N):        print(arr[i], end = " ") # Driver Codeif __name__ == '__main__':         # Given array    arr = [ 4, 9, 3, 5, 7 ]     # Size of the array    N = len(arr)     # Function Call    centerSpiralSort(arr, N)     # This code is contributed by ipg2016107
// C# program for the above approachusing System; class GFG{ // Function to sort the array// according to the given patternstatic void centerSpiralSort(int []arr, int N){         // Initializing the variables    int left = 0;    int right = N - 1;    int i = 1;    int temp;     while (left < right)    {                 // Performing bubble sort        for(i = left + 1; i <= right; i++)        {            if (arr[left] > arr[i])            {                                 // Swapping if                // arr[left] > arr[i]                temp = arr[left];                arr[left] = arr[i];                arr[i] = temp;            }        }         // Increment left by 1        left++;         // Performing bubble sort        for(i = right - 1; i >= left; i--)        {            if (arr[right] > arr[i])            {                                 // Swapping if                // arr[right] > arr[i]                temp = arr[right];                arr[right] = arr[i];                arr[i] = temp;            }        }         // Decrement right by 1        right--;    }     // If N is an even number    // Reversing the array from 0 to N/2-1    if (N % 2 == 0)    {                 // Reversing the half array        for(i = 0; i < N / 2; i++)        {            temp = arr[i];            arr[i] = arr[N - 1 - i];            arr[N - 1 - i] = temp;        }    }     // Print the elements of the array    for(i = 0; i < N; i++)    {        Console.Write(arr[i] + " ");    }} // Driver Codepublic static void Main(String[] args){         // Given array    int []arr = { 4, 9, 3, 5, 7 };     // Size of the array    int N = arr.Length;     // Function Call    centerSpiralSort(arr, N);}} // This code is contributed by shivanisinghss2110
<script> // JavaScript program for the above approach // Function to sort the array// according to the given patternfunction centerSpiralSort(arr, N) {  // Initializing the variables  let left = 0;  let right = N - 1;  let i = 1;   while (left < right) {    // Performing bubble sort    for (i = left + 1; i <= right; i++) {      if (arr[left] > arr[i]) {        // Swapping if        // arr[left] > arr[i]        let temp = arr[left];        arr[left] = arr[i];        arr[i] = temp;      }    }     // Increment left by 1    left++;     // Performing bubble sort    for (i = right - 1; i >= left; i--) {      if (arr[right] > arr[i]) {        // Swapping if        // arr[right] > arr[i]        let temp = arr[right];        arr[right] = arr[i];        arr[i] = temp;      }    }     // Decrement right by 1    right--;  }   // If N is an even number  // Reversing the array from 0 to N/2-1  if (N % 2 == 0) {    // Reversing the half array    for (let i = 0; i < N / 2; i++) {      let temp = arr[N - 1 - i];      arr[N - 1 - i] = arr[i];      arr[i] = temp;    }  }   // Print the elements of the array  for (let i = 0; i < N; i++) {    document.write(arr[i] + " ");  }} // Driver Code // Given arraylet arr = [4, 9, 3, 5, 7]; // Size of the arraylet N = arr.length; // Function CallcenterSpiralSort(arr, N); </script>
3 5 9 7 4 
Time Complexity: O(N2)Auxiliary Space: O(1)
lokeshpotta20
_saurabh_jaiswal
ipg2016107
shivanisinghss2110
rajeev0719singh
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