Sort a Rotated Sorted Array - GeeksforGeeks

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: 
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++
Java
Python3
C#
Javascript
// C++ implementation for restoring original// sort in rotated sorted array#include <bits/stdc++.h>using namespace std; // Function to restore the Original Sortvoid 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 Arrayvoid printArray(int arr[], int size){    for (int i = 0; i < size; i++)        cout << arr[i] << " ";} // Driver functionint main(){    int arr[] = { 3, 4, 5, 1, 2 };    int n = sizeof(arr) / sizeof(arr[0]);    restoreSortedArray(arr, n);    printArray(arr, n);    return 0;}
// Java implementation for restoring original// sort in rotated sorted arrayclass GFG{ // Function to restore the Original Sortstatic 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 Arraystatic void printArray(int arr[], int size){    for (int i = 0; i < size; i++)        System.out.print(arr[i] + " ");} // Driver codepublic 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 implementation for restoring original# sort in rotated sorted array # Function to restore the Original Sortdef 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 Arraydef printArray(arr, size):    for i in range(size):        print(arr[i], end=""); # Driver codeif __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# implementation for restoring original// sort in rotated sorted arrayusing System;     class GFG{ // Function to restore the Original Sortstatic 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 Arraystatic void printArray(int []arr, int size){    for (int i = 0; i < size; i++)        Console.Write(arr[i] + " ");} // Driver codepublic 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
<script>// JavaScript implementation for restoring original// sort in rotated sorted array // Function to restore the Original Sortfunction 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 Arrayfunction 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: 
First find the index of minimum element (split index) in the array using binary searchThen 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.
First find the index of minimum element (split index) in the array using binary search
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.
From zero index to split index.
From split index to end index.
From zero index to end index.
C++
Java
Python3
C#
Javascript
// 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 arrayint 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 Sortvoid 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 Arrayvoid printArray(int arr[], int size){    for (int i = 0; i < size; i++)        cout << arr[i] << " ";} // Driver functionint main(){    int arr[] = { 1, 2, 3, 4, 5};    int n = sizeof(arr) / sizeof(arr[0]);    restoreSortedArray(arr, n);    printArray(arr, n);    return 0;}
// Java implementation for restoring original// sort in rotated sorted array using binary searchimport 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 implementation for restoring original# sort in rotated sorted array using binary search # Function to find start index of arraydef 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 Sortdef 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 Arraydef 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 codeif __name__ == '__main__':    arr = [ 1, 2, 3, 4, 5 ];    n = len(arr);    restoreSortedArray(arr, n);    printArray(arr, n); # This code is contributed by PrinciRaj1992
// C# implementation for restoring original// sort in rotated sorted array using binary searchusing 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 */
<script> // Javascript implementation for restoring original// sort in rotated sorted array using binary search // Function to find start index of arrayfunction 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 Sortfunction 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 Arrayfunction printArray(arr, size){    for(let i = 0; i < size; i++)    {        document.write(arr[i] + " ");    }} // Driver codelet 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
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