Sort a Rotated Sorted Array
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:
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:
- 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.
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
Similar Reads
Rotation Count in a Rotated Sorted array Given an array arr[] having distinct numbers sorted in increasing order and the array has been right rotated (i.e, the last element will be cyclically shifted to the starting position of the array) k number of times, the task is to find the value of k.Examples: Input: arr[] = {15, 18, 2, 3, 6, 12}Ou
12 min read
Circularly Sorted Array (Sorted and Rotated Array) Circularly sorted arrays are arrays that are sorted in ascending or descending order and then rotated by a number of steps. Let us take an example to know more about circularly sorted arrays: Consider an array: arr[] = {23, 34, 45, 12, 17, 19}The elements here, {12, 17, 19, 23, 34, 45} are sorted 'I
7 min read
Minimum in a Sorted and Rotated Array Given a sorted array of distinct elements arr[] of size n that is rotated at some unknown point, the task is to find the minimum element in it. Examples: Input: arr[] = [5, 6, 1, 2, 3, 4]Output: 1Explanation: 1 is the minimum element present in the array.Input: arr[] = [3, 1, 2]Output: 1Explanation:
9 min read
Check if an array is sorted and rotated Given an array arr[] of size n, the task is to return true if it was originally sorted in non-decreasing order and then rotated (including zero rotations). Otherwise, return false. The array may contain duplicates.Examples:Input: arr[] = { 3, 4, 5, 1, 2 }Output: YESExplanation: The above array is so
7 min read
Row wise sorting a 2D array Given a 2D array, sort each row of this array and print the result.Examples: Input : mar[][] = [ [77, 11, 22, 3], [11, 89, 1, 12], [32, 11, 56, 7], [11, 22, 44, 33] ] Output : mat[][] = [ [3, 11, 22, 77], [1, 11, 12, 89], [7, 11, 32, 56], [11, 22, 33, 44] ] Input : mat[][] = [ [8, 6, 4, 5], [3, 5, 2
4 min read