Given a sorted array, the task is to remove the duplicate elements from the array.
Examples:
Input: arr[] = {2, 2, 2, 2, 2} Output: arr[] = {2} new size = 1 Input: arr[] = {1, 2, 2, 3, 4, 4, 4, 5, 5} Output: arr[] = {1, 2, 3, 4, 5} new size = 5
Method 1: (Using extra space)
- Create an auxiliary array temp[] to store unique elements.
- Traverse input array and one by one copy unique elements of arr[] to temp[]. Also keep track of count of unique elements. Let this count be j.
- Copy j elements from temp[] to arr[] and return j
C++
// Simple C++ program to remove duplicates #include <iostream> using namespace std;
// Function to remove duplicate // elements This function returns // new size of modified array. int removeDuplicates( int arr[], int n)
{ // Return, if array is empty or
// contains a single element
if (n == 0 || n == 1)
return n;
int temp[n];
// Start traversing elements
int j = 0;
// If current element is not equal
// to next element then store that
// current element
for ( int i = 0; i < n - 1; i++)
if (arr[i] != arr[i + 1])
temp[j++] = arr[i];
// Store the last element as whether
// it is unique or repeated, it hasn't
// stored previously
temp[j++] = arr[n - 1];
// Modify original array
for ( int i = 0; i < j; i++)
arr[i] = temp[i];
return j;
} // Driver code int main()
{ int arr[] = {1, 2, 2, 3, 4,
4, 4, 5, 5};
int n = sizeof (arr) / sizeof (arr[0]);
// RemoveDuplicates() returns
// new size of array.
n = removeDuplicates(arr, n);
// Print updated array
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
Output:
1 2 3 4 5
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 2: (Constant extra space)
Just maintain a separate index for same array as maintained for different array in Method 1.
C++
// C++ program to remove // duplicates in-place #include<iostream> using namespace std;
// Function to remove duplicate // elements. This function returns // new size of modified array. int removeDuplicates( int arr[], int n)
{ if (n==0 || n==1)
return n;
// To store index of next
// unique element
int j = 0;
// Doing same as done in Method 1
// Just maintaining another updated
// index i.e. j
for ( int i = 0; i < n - 1; i++)
if (arr[i] != arr[i + 1])
arr[j++] = arr[i];
arr[j++] = arr[n - 1];
return j;
} // Driver code int main()
{ int arr[] = {1, 2, 2, 3, 4,
4, 4, 5, 5};
int n = sizeof (arr) / sizeof (arr[0]);
// removeDuplicates() returns new
// size of array.
n = removeDuplicates(arr, n);
// Print updated array
for ( int i=0; i<n; i++)
cout << arr[i] << " " ;
return 0;
} |
Output:
1 2 3 4 5
Time Complexity : O(n)
Auxiliary Space : O(1)