Sort an array where a subarray of a sorted array is in reverse order

Given an array of N numbers where a subarray is sorted in descending order and rest of the numbers in the array are in ascending order. The task is to sort an array where a subarray of a sorted array is in reversed order.

Examples:

Input: 2 5 65 55 50 70 90
Output: 2 5 50 55 65 70 90
The subarray from 2nd index to 4th index is in reverse order.
So the subarray is reversed, and the sorted array is printed.

Input: 1 7 6 5 4 3 2 8
Output: 1 2 3 4 5 6 7 8

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach will be to sort the array and print the array. Time Complexity of this approach will be O(N log n).

An efficient approach will be to find and store the starting index and ending index of the reversed subarray. Since the subarray is in descending order and the rest of the elements are in ascending order, only reversing the subarray will sort the complete array. Reverse the subarray using two pointer approach.

Below is the implementation of the above approach:

C++

 // C++ program to sort an array where // a subarray of a sorted array // is in reversed order #include using namespace std;    // Function to print the sorted array // by reversing the subarray void printSorted(int a[], int n) {     int front = -1, back = -1;        // find the starting index of the     // reversed subarry     for (int i = 1; i < n; i++) {         if (a[i] < a[i - 1]) {             front = i - 1;             break;         }     }        // find the ending index of the     // reversed subarray     for (int i = n - 2; i >= 0; i--) {         if (a[i] > a[i + 1]) {             back = i + 1;             break;         }     }        // if no reversed subarray is present     if (front == -1 and back == -1) {         for (int i = 0; i < n; i++)             cout << a[i] << " ";         return;     }        // swap the reversed subarray     while (front <= back) {            // swaps the front and back element         // using c++ STL         swap(a[front], a[back]);            // move the pointers one step         // ahead and one step back         front++;         back--;     }     for (int i = 0; i < n; i++)         cout << a[i] << " "; }    // Driver Code int main() {     int a[] = { 1, 7, 6, 5, 4, 3, 2, 8 };     int n = sizeof(a) / sizeof(a);     printSorted(a, n);     return 0; }

Java

 // Java program to sort an array where // a subarray of a sorted array // is in reversed order import java.io.*;    class GFG {    // Function to print the sorted array // by reversing the subarray static void printSorted(int a[], int n) {     int front = -1, back = -1;        // find the starting index of the     // reversed subarry     for (int i = 1; i < n; i++)     {         if (a[i] < a[i - 1])          {             front = i - 1;             break;         }     }        // find the ending index of the     // reversed subarray     for (int i = n - 2; i >= 0; i--)      {         if (a[i] > a[i + 1])          {             back = i + 1;             break;         }     }        // if no reversed subarray is present     if (front == -1 && back == -1)      {         for (int i = 0; i < n; i++)             System.out.println(a[i] + " ");         return;     }        // swap the reversed subarray     while (front <= back)     {            // swaps the front and back element         // using c++ STL         int temp = a[front];         a[front] = a[back];         a[back] = temp;            // move the pointers one step         // ahead and one step back         front++;         back--;     }     for (int i = 0; i < n; i++)         System.out.print(a[i] + " "); }    // Driver Code public static void main (String[] args) {     int a[] = { 1, 7, 6, 5, 4, 3, 2, 8 };     int n = a.length;     printSorted(a, n);; } }    // This code is contributed by anuj_67..

Python3

 # Python 3 program to sort an array where # a subarray of a sorted array is in  # reversed order    # Function to print the sorted array # by reversing the subarray def printSorted(a, n):     front = -1     back = -1        # find the starting index of the     # reversed subarry     for i in range(1, n, 1):         if (a[i] < a[i - 1]):             front = i - 1             break        # find the ending index of the     # reversed subarray     i = n - 2     while(i >= 0):         if (a[i] > a[i + 1]):             back = i + 1             break         i -= 1            # if no reversed subarray is present     if (front == -1 and back == -1):         for i in range(0, n, 1):             print(a[i], end = " ")         return        # swap the reversed subarray     while (front <= back):                    # swaps the front and back element         # using c++ STL         temp = a[front]         a[front] = a[back]         a[back] = temp            # move the pointers one step         # ahead and one step back         front += 1         back -= 1            for i in range(0, n, 1):         print(a[i], end = " ")     # Driver Code if __name__ == '__main__':     a = [1, 7, 6, 5, 4, 3, 2, 8]     n = len(a)     printSorted(a, n)    # This code is contributed by # Sahil_Shelangia

C#

 // C# program to sort an array where  // a subarray of a sorted array  // is in reversed order  using System;    class GFG {    // Function to print the sorted array  // by reversing the subarray  static void printSorted(int []a, int n)  {     int front = -1, back = -1;        // find the starting index of the      // reversed subarry      for (int i = 1; i < n; i++)     {         if (a[i] < a[i - 1])          {             front = i - 1;             break;         }     }        // find the ending index of the      // reversed subarray      for (int i = n - 2; i >= 0; i--)     {         if (a[i] > a[i + 1])         {             back = i + 1;             break;         }     }        // if no reversed subarray is present      if (front == -1 && back == -1)      {         for (int i = 0; i < n; i++)          {             Console.Write(a[i] + " ");         }         return;     }        // swap the reversed subarray      while (front <= back)      {            // swaps the front and back element          // using c++ STL          swap(a, front, back);            // move the pointers one step          // ahead and one step back          front++;         back--;     }     for (int i = 0; i < n; i++)      {         Console.Write(a[i] + " ");     } }    static void swap(int[] a, int front,                            int back)  {     int c = a[front];     a[front] = a[back];     a[back] = c; }    // Driver Code  public static void Main()  {     int []a = {1, 7, 6, 5, 4, 3, 2, 8};     int n = a.Length;     printSorted(a, n); } }    // This code contributed by 29AjayKumar

PHP

 = 0; \$i--)      {         if (\$a[\$i] > \$a[\$i + 1])          {             \$back = \$i + 1;             break;         }     }        // if no reversed subarray is present     if (\$front == -1 && \$back == -1)      {         for (\$i = 0; \$i < \$n; \$i++)             echo \$a[\$i] . " ";         return;     }        // swap the reversed subarray     while (\$front <= \$back)     {            // swaps the front and back element         // using c++ STL         \$temp = \$a[\$front];         \$a[\$front] = \$a[\$back];         \$a[\$back] = \$temp;            // move the pointers one step         // ahead and one step back         \$front++;         \$back--;     }     for (\$i = 0; \$i < \$n; \$i++)         echo \$a[\$i] . " "; }    // Driver Code \$a = array(1, 7, 6, 5, 4, 3, 2, 8); \$n = sizeof(\$a); printSorted(\$a, \$n);    // This code is contributed  // by Akanksha Rai

Output:

1 2 3 4 5 6 7 8

Time Complexity: O(n)

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