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