C++ Program To Remove Duplicates From Sorted Array
Last Updated :
17 Jan, 2023
Improve
Try it on GfG Practice 
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)
