Delete an element from array (Using two traversals and one traversal)
Given an array and a number ‘x’, write a function to delete ‘x’ from the given array.
Example:
Input: arr[] = {3, 1, 2, 5, 90}, x = 2
Output: arr[] = {3, 1, 5, 90}
A simple solution is to first search ‘x’ in array, then elements that are on right side of x to one position back. The following are the implementation of this simple approach.
C/C++
// C++ program to remove a given element from an array #include<iostream> using namespace std; // This function removes an element x from arr[] and // returns new size after removal (size is reduced only // when x is present in arr[] int deleteElement(int arr[], int n, int x) { // Search x in array int i; for (i=0; i<n; i++) if (arr[i] == x) break; // If x found in array if (i < n) { // reduce size of array and move all // elements on space ahead n = n - 1; for (int j=i; j<n; j++) arr[j] = arr[j+1]; } return n; } /* Driver program to test above function */int main() { int arr[] = {11, 15, 6, 8, 9, 10}; int n = sizeof(arr)/sizeof(arr[0]); int x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); cout << "Modified array is \n"; for (int i=0; i<n; i++) cout << arr[i] << " "; return 0; } |
chevron_right
filter_none
Java
// Java program to remove a given element from an array import java.io.*; class Deletion { // This function removes an element x from arr[] and // returns new size after removal (size is reduced only // when x is present in arr[] static int deleteElement(int arr[], int n, int x) { // Search x in array int i; for (i=0; i<n; i++) if (arr[i] == x) break; // If x found in array if (i < n) { // reduce size of array and move all // elements on space ahead n = n - 1; for (int j=i; j<n; j++) arr[j] = arr[j+1]; } return n; } // Driver program to test above function public static void main(String[] args) { int arr[] = {11, 15, 6, 8, 9, 10}; int n = arr.length; int x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); System.out.println("Modified array is"); for (int i = 0; i < n; i++) System.out.print(arr[i]+" "); } } /*This code is contributed by Devesh Agrawal*/ |
chevron_right
filter_none
Python3
# Python 3 program to remove a given # element from an array # This function removes an element x # from arr[] and returns new size after # removal (size is reduced only when x # is present in arr[] def deleteElement(arr, n, x): # Search x in array for i in range(n): if (arr[i] == x): break # If x found in array if (i < n): # reduce size of array and move # all elements on space ahead n = n - 1; for j in range(i, n, 1): arr[j] = arr[j + 1] return n # Driver Code if __name__ == '__main__': arr = [11, 15, 6, 8, 9, 10] n = len(arr) x = 6 # Delete x from arr[] n = deleteElement(arr, n, x) print("Modified array is") for i in range(n): print(arr[i], end = " ") # This code is contributed by # Shashank_Sharma |
chevron_right
filter_none
C#
// C# program to remove a given element from // an array using System; class GFG { // This function removes an element x // from arr[] and returns new size // after removal (size is reduced only // when x is present in arr[] static int deleteElement(int []arr, int n, int x) { // Search x in array int i; for (i = 0; i < n; i++) if (arr[i] == x) break; // If x found in array if (i < n) { // reduce size of array and // move all elements on // space ahead n = n - 1; for (int j = i; j < n; j++) arr[j] = arr[j+1]; } return n; } // Driver program to test above function public static void Main() { int []arr = {11, 15, 6, 8, 9, 10}; int n = arr.Length; int x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); Console.WriteLine("Modified array is"); for (int i = 0; i < n; i++) Console.Write(arr[i]+" "); } } // This code is contributed by nitin mittal. |
chevron_right
filter_none
Output:
Modified array is 11 15 8 9 10
The above method requires two traversals of array, one for searching and other for moving elements.
Can we delete the element using one traversal?
This is possible under the assumption that the element is always present in array. The idea is to start from right most element and keep moving elements while searching for ‘x’. Below are C++ and Java implementations of this approach. Note that this approach may give unexpected result when ‘x’ is not present in array.
C++
// C++ program to remove a given element from an array #include<iostream> using namespace std; // This function removes an element x from arr[] and // returns new size after removal. // Returned size is n-1 when element is present. // Otherwise 0 is returned to indicate failure. int deleteElement(int arr[], int n, int x) { // If x is last element, nothing to do if (arr[n-1] == x) return (n-1); // Start from rightmost element and keep moving // elements one position ahead. int prev = arr[n-1], i; for (i=n-2; i>=0 && arr[i]!=x; i--) { int curr = arr[i]; arr[i] = prev; prev = curr; } // If element was not found if (i < 0) return 0; // Else move the next element in place of x arr[i] = prev; return (n-1); } /* Driver program to test above function */int main() { int arr[] = {11, 15, 6, 8, 9, 10}; int n = sizeof(arr)/sizeof(arr[0]); int x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); cout << "Modified array is \n"; for (int i=0; i<n; i++) cout << arr[i] << " "; return 0; } |
chevron_right
filter_none
Java
// Java program to remove a given element from an array import java.io.*; class Deletion { // This function removes an element x from arr[] and // returns new size after removal. // Returned size is n-1 when element is present. // Otherwise 0 is returned to indicate failure. static int deleteElement(int arr[], int n, int x) { // If x is last element, nothing to do if (arr[n-1] == x) return (n-1); // Start from rightmost element and keep moving // elements one position ahead. int prev = arr[n-1], i; for (i=n-2; i>=0 && arr[i]!=x; i--) { int curr = arr[i]; arr[i] = prev; prev = curr; } // If element was not found if (i < 0) return 0; // Else move the next element in place of x arr[i] = prev; return (n-1); } // Driver program to test above function public static void main(String[] args) { int arr[] = {11, 15, 6, 8, 9, 10}; int n = arr.length; int x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); System.out.println("Modified array is"); for (int i = 0; i < n; i++) System.out.print(arr[i]+" "); } } /*This code is contributed by Devesh Agrawal*/ |
chevron_right
filter_none
Python3
# python program to remove a given element from an array # This function removes an element x from arr[] and # returns new size after removal. # Returned size is n-1 when element is present. # Otherwise 0 is returned to indicate failure. def deleteElement(arr,n,x): # If x is last element, nothing to do if arr[n-1]==x: return n-1 # Start from rightmost element and keep moving # elements one position ahead. prev = arr[n-1] for i in range(n-2,1,-1): if arr[i]!=x: curr = arr[i] arr[i] = prev prev = curr # If element was not found if i<0: return 0 # Else move the next element in place of x arr[i] = prev return n-1 # Driver code arr = [11,15,6,8,9,10] n = len(arr) x = 6n = deleteElement(arr,n,x) print("Modified array is") for i in range(n): print(arr[i],end=" ") # This code is contributed by Shrikant13 |
chevron_right
filter_none
C#
// C# program to remove a given // element from an array using System; class GFG { // This function removes an // element x from arr[] and // returns new size after // removal. Returned size is // n-1 when element is present. // Otherwise 0 is returned to // indicate failure. static int deleteElement(int []arr, int n, int x) { // If x is last element, // nothing to do if (arr[n - 1] == x) return (n - 1); // Start from rightmost // element and keep moving // elements one position ahead. int prev = arr[n - 1], i; for (i = n - 2; i >= 0 && arr[i] != x; i--) { int curr = arr[i]; arr[i] = prev; prev = curr; } // If element was // not found if (i < 0) return 0; // Else move the next // element in place of x arr[i] = prev; return (n - 1); } // Driver Code public static void Main() { int []arr = {11, 15, 6, 8, 9, 10}; int n = arr.Length; int x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); Console.WriteLine("Modified array is"); for(int i = 0; i < n; i++) Console.Write(arr[i]+" "); } } // This code is contributed by anuj_67. |
chevron_right
filter_none
Output:
Modified array is 11 15 8 9 10
Deleting an element from an array takes O(n) time even if we are given index of the element to be deleted. The time complexity remains O(n) for sorted arrays as well.
In linked list, if we know the pointer to the previous node of the node to be deleted, we can do deletion in O(1) time.
This article is contributed by Himanshu. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Recommended Posts:
- Delete array element in given index range [L - R]
- Closest greater element for every array element from another array
- Replace each element by the difference of the total size of the array and frequency of that element
- Search, insert and delete in a sorted array
- Search, insert and delete in an unsorted array
- Range Query on array whose each element is XOR of index value and previous element
- Minimum delete operations to make all elements of array same
- Find last element after deleting every second element in array of n integers
- Replace every array element by Bitwise Xor of previous and next element
- Sum of (maximum element - minimum element) for all the subsets of an array.
- Minimum number of operations required to delete all elements of the array
- Move all zeroes to end of array | Set-2 (Using single traversal)
- Maximum sum in an array such that every element has exactly one adjacent element to it
- Replace every element of the array by its previous element
- Find an element in array such that sum of left array is equal to sum of right array
