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Sort a Rotated Sorted Array

  • Difficulty Level : Easy
  • Last Updated : 07 Jul, 2021

You are given a rotated sorted array and your aim is to restore its original sort in place.
Expected to use O(1) extra space and O(n) time complexity. 

Examples: 

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Input : [3, 4, 1, 2] 
Output : [1, 2, 3, 4]

Input : [2, 3, 4, 1]
Output : [1, 2, 3, 4]

We find the point of rotation. Then we rotate array using reversal algorithm



 1. First, find the split point where the sorting breaks.
 2. Then call the reverse function in three steps.
     - From zero index to split index.
     - From split index to end index.
     - From zero index to end index.

C++




// C++ implementation for restoring original
// sort in rotated sorted array
#include <bits/stdc++.h>
using namespace std;
 
// Function to restore the Original Sort
void restoreSortedArray(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        if (arr[i] > arr[i + 1]) {
 
            // In reverse(), the first parameter
            // is iterator to beginning element
            // and second parameter is iterator
            // to last element plus one.  
            reverse(arr, arr+i+1);
            reverse(arr + i + 1, arr + n);
            reverse(arr, arr + n);
        }
    }
}
 
// Function to print the Array
void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        cout << arr[i] << " ";
}
 
// Driver function
int main()
{
    int arr[] = { 3, 4, 5, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    restoreSortedArray(arr, n);
    printArray(arr, n);
    return 0;
}

Java




// Java implementation for restoring original
// sort in rotated sorted array
class GFG
{
 
// Function to restore the Original Sort
static void restoreSortedArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
    {
        if (arr[i] > arr[i + 1])
        {
 
            // In reverse(), the first parameter
            // is iterator to beginning element
            // and second parameter is iterator
            // to last element plus one.
            reverse(arr,0,i);
            reverse(arr , i + 1, n);
            reverse(arr,0, n);
        }
    }
}
 
static void reverse(int[] arr, int i, int j)
{
    int temp;
    while(i < j)
    {
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        i++;
        j--;
    }
}
 
// Function to print the Array
static void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        System.out.print(arr[i] + " ");
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 4, 5, 1, 2 };
    int n = arr.length;
    restoreSortedArray(arr, n - 1);
    printArray(arr, n);
}
}
 
// This code has been contributed by 29AjayKumar

Python3




# Python3 implementation for restoring original
# sort in rotated sorted array
 
# Function to restore the Original Sort
def restoreSortedArray(arr, n):
    for i in range(n):
        if (arr[i] > arr[i + 1]):
            # In reverse(), the first parameter
            # is iterator to beginning element
            # and second parameter is iterator
            # to last element plus one.
            reverse(arr, 0, i);
            reverse(arr, i + 1, n);
            reverse(arr, 0, n);
 
def reverse(arr, i, j):
    while (i < j):
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        i += 1;
        j -= 1;
 
# Function to print the Array
def printArray(arr, size):
    for i in range(size):
        print(arr[i], end="");
 
# Driver code
if __name__ == '__main__':
    arr = [3, 4, 5, 1, 2];
    n = len(arr);
    restoreSortedArray(arr, n - 1);
    printArray(arr, n);
 
# This code is contributed by 29AjayKumar

C#




// C# implementation for restoring original
// sort in rotated sorted array
using System;
     
class GFG
{
 
// Function to restore the Original Sort
static void restoreSortedArray(int []arr, int n)
{
    for (int i = 0; i < n; i++)
    {
        if (arr[i] > arr[i + 1])
        {
 
            // In reverse(), the first parameter
            // is iterator to beginning element
            // and second parameter is iterator
            // to last element plus one.
            reverse(arr,0,i);
            reverse(arr , i + 1, n);
            reverse(arr,0, n);
        }
    }
}
 
static void reverse(int[] arr, int i, int j)
{
    int temp;
    while(i < j)
    {
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        i++;
        j--;
    }
}
 
// Function to print the Array
static void printArray(int []arr, int size)
{
    for (int i = 0; i < size; i++)
        Console.Write(arr[i] + " ");
}
 
// Driver code
public static void Main(String[] args)
{
    int[] arr = { 3, 4, 5, 1, 2 };
    int n = arr.Length;
    restoreSortedArray(arr, n - 1);
    printArray(arr, n);
}
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
// JavaScript implementation for restoring original
// sort in rotated sorted array
 
// Function to restore the Original Sort
function restoreSortedArray(arr, n)
{
    for (let i = 0; i < n; i++)
    {
        if (arr[i] > arr[i + 1])
        {
   
            // In reverse(), the first parameter
            // is iterator to beginning element
            // and second parameter is iterator
            // to last element plus one.
            reverse(arr,0,i);
            reverse(arr , i + 1, n);
            reverse(arr,0, n);
        }
    }
}
   
function reverse(arr, i, j)
{
    let temp;
    while(i < j)
    {
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        i++;
        j--;
    }
}
   
// Function to print the Array
function printArray(arr, size)
{
    for (let i = 0; i < size; i++)
        document.write(arr[i] + " ");
}
  
// Driver Code
 
    let arr = [ 3, 4, 5, 1, 2 ];
    let n = arr.length;
    restoreSortedArray(arr, n - 1);
    printArray(arr, n)
                 
</script>

Output: 

1 2 3 4 5

We can binary search to find the rotation point as discussed here .
Efficient code approach using binary search: 

  1. First find the index of minimum element (split index) in the array using binary search
  2. Then call the reverse function in three steps.
    • From zero index to split index.
    • From split index to end index.
    • From zero index to end index.

C++




// C++ implementation for restoring original
// sort in rotated sorted array using binary search
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to find start index of array
int findStartIndexOfArray(int arr[], int low,int high)
{
    if (low>high)
    {
        return -1;
    }
 
    if (low == high)
    {
        return low;
    }
 
    int mid = low + (high-low)/2;
    if(arr[mid] > arr[mid+1])
        return mid+1;
 
    if(arr[mid-1] > arr[mid])
        return mid;
 
    if(arr[low] > arr[mid])
        return findStartIndexOfArray(arr, low, mid-1);
    else
        return findStartIndexOfArray(arr, mid+1, high);
}
 
// Function to restore the Original Sort
void restoreSortedArray(int arr[], int n)
{
    // array is already sorted
    if (arr[0] < arr[n-1])
        return;
     
    int start = findStartIndexOfArray(arr, 0, n-1);
    // In reverse(), the first parameter
    // is iterator to beginning element
    // and second parameter is iterator
    // to last element plus one.
    reverse(arr, arr + start);
    reverse(arr + start, arr + n);
    reverse(arr, arr + n);
     
}
 
// Function to print the Array
void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        cout << arr[i] << " ";
}
 
// Driver function
int main()
{
    int arr[] = { 1, 2, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    restoreSortedArray(arr, n);
    printArray(arr, n);
    return 0;
}

Java




// Java implementation for restoring original
// sort in rotated sorted array using binary search
import java.util.*;
 
class GFG
{
     
    // Function to find start index of array
    static int findStartIndexOfArray(int arr[],
                            int low, int high)
    {
        if (low > high)
        {
            return -1;
        }
 
        if (low == high)
        {
            return low;
        }
 
        int mid = low + (high - low) / 2;
        if (arr[mid] > arr[mid + 1])
        {
            return mid + 1;
        }
 
        if (arr[mid - 1] > arr[mid])
        {
            return mid;
        }
 
        if (arr[low] > arr[mid])
        {
            return findStartIndexOfArray(arr, low, mid - 1);
        }
        else
        {
            return findStartIndexOfArray(arr, mid + 1, high);
        }
    }
 
    // Function to restore the Original Sort
    static void restoreSortedArray(int arr[], int n)
    {
        // array is already sorted
        if (arr[0] < arr[n - 1])
        {
            return;
        }
 
        int start = findStartIndexOfArray(arr, 0, n - 1);
         
        // In reverse(), the first parameter
        // is iterator to beginning element
        // and second parameter is iterator
        // to last element plus one.
        Arrays.sort(arr, 0, start);
        Arrays.sort(arr, start, n);
        Arrays.sort(arr);
 
    }
 
    // Function to print the Array
    static void printArray(int arr[], int size)
    {
        for (int i = 0; i < size; i++)
        {
            System.out.print(arr[i] + " ");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 2, 3, 4, 5};
        int n = arr.length;
        restoreSortedArray(arr, n);
        printArray(arr, n);
    }
}
 
// This code contributed by Rajput-Ji

Python3




# Python3 implementation for restoring original
# sort in rotated sorted array using binary search
 
# Function to find start index of array
def findStartIndexOfArray(arr, low, high):
    if (low > high):
        return -1;
     
    if (low == high):
        return low;
     
    mid = low + (high - low) / 2;
    if (arr[mid] > arr[mid + 1]):
        return mid + 1;
     
    if (arr[mid - 1] > arr[mid]):
        return mid;
     
    if (arr[low] > arr[mid]):
        return findStartIndexOfArray(arr, low, mid - 1);
    else:
        return findStartIndexOfArray(arr, mid + 1, high);
 
# Function to restore the Original Sort
def restoreSortedArray(arr, n):
 
    # array is already sorted
    if (arr[0] < arr[n - 1]):
        return;
     
    start = findStartIndexOfArray(arr, 0, n - 1);
 
    # In reverse(), the first parameter
    # is iterator to beginning element
    # and second parameter is iterator
    # to last element plus one.
    reverse(arr, 0, start);
    reverse(arr, start, n);
    reverse(arr);
 
# Function to print the Array
def printArray(arr, size):
    for i in range(size):
        print(arr[i], end="");
     
def reverse(arr, i, j):
    while (i < j):
        temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
        i += 1;
        j -= 1;
 
# Driver code
if __name__ == '__main__':
    arr = [ 1, 2, 3, 4, 5 ];
    n = len(arr);
    restoreSortedArray(arr, n);
    printArray(arr, n);
 
# This code is contributed by PrinciRaj1992

C#




// C# implementation for restoring original
// sort in rotated sorted array using binary search
using System;
 
class GFG
{
     
    // Function to find start index of array
    static int findStartIndexOfArray(int []arr,
                            int low, int high)
    {
        if (low > high)
        {
            return -1;
        }
 
        if (low == high)
        {
            return low;
        }
 
        int mid = low + (high - low) / 2;
        if (arr[mid] > arr[mid + 1])
        {
            return mid + 1;
        }
 
        if (arr[mid - 1] > arr[mid])
        {
            return mid;
        }
 
        if (arr[low] > arr[mid])
        {
            return findStartIndexOfArray(arr, low, mid - 1);
        }
        else
        {
            return findStartIndexOfArray(arr, mid + 1, high);
        }
    }
 
    // Function to restore the Original Sort
    static void restoreSortedArray(int []arr, int n)
    {
        // array is already sorted
        if (arr[0] < arr[n - 1])
        {
            return;
        }
 
        int start = findStartIndexOfArray(arr, 0, n - 1);
         
        // In reverse(), the first parameter
        // is iterator to beginning element
        // and second parameter is iterator
        // to last element plus one.
        Array.Sort(arr, 0, start);
        Array.Sort(arr, start, n);
        Array.Sort(arr);
 
    }
 
    // Function to print the Array
    static void printArray(int []arr, int size)
    {
        for (int i = 0; i < size; i++)
        {
            Console.Write(arr[i] + " ");
        }
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = {1, 2, 3, 4, 5};
        int n = arr.Length;
        restoreSortedArray(arr, n);
        printArray(arr, n);
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript




<script>
 
// Javascript implementation for restoring original
// sort in rotated sorted array using binary search
 
// Function to find start index of array
function findStartIndexOfArray(arr, low, high)
{
    if (low > high)
    {
        return -1;
    }
 
    if (low == high)
    {
        return low;
    }
 
    let mid = low + parseInt((high - low) / 2, 10);
    if (arr[mid] > arr[mid + 1])
    {
        return mid + 1;
    }
 
    if (arr[mid - 1] > arr[mid])
    {
        return mid;
    }
 
    if (arr[low] > arr[mid])
    {
        return findStartIndexOfArray(arr, low,
                                     mid - 1);
    }
    else
    {
        return findStartIndexOfArray(arr, mid + 1,
                                     high);
    }
}
 
// Function to restore the Original Sort
function restoreSortedArray(arr, n)
{
     
    // Array is already sorted
    if (arr[0] < arr[n - 1])
    {
        return;
    }
 
    let start = findStartIndexOfArray(arr, 0, n - 1);
      
    // In reverse(), the first parameter
    // is iterator to beginning element
    // and second parameter is iterator
    // to last element plus one.
    arr.sort();
 
}
 
// Function to print the Array
function printArray(arr, size)
{
    for(let i = 0; i < size; i++)
    {
        document.write(arr[i] + " ");
    }
}
 
// Driver code
let arr = [ 1, 2, 3, 4, 5 ];
let n = arr.length;
 
restoreSortedArray(arr, n);
printArray(arr, n);
 
// This code is contributed by decode2207  
 
</script>

Output: 

1 2 3 4 5

This article is contributed by Kshitiz Gupta.If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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