Given an array and a number ‘x’, write a function to delete ‘x’ from the given array. We assume that array maintains two things with it, capacity and size. So when we remove an item, capacity does not change, only size changes.
Example:
Input: arr[] = {3, 1, 2, 5, 90}, x = 2, size = 5, capacity = 5
Output: arr[] = {3, 1, 5, 90, _}, size = 4, capacity = 5
Input: arr[] = {3, 1, 2, _, _}, x = 2, size = 3, capacity = 5
Output: arr[] = {3, 1, _, _, _}, size = 2, capacity = 5
Method 1(First Search, then Remove): We 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.
Implementation:
C
#include <stdio.h>// 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 one 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); printf("Modified array is \n"); for (int i = 0; i < n; i++) { printf("%d ", arr[i]); } return 0;} |
C++
// C++ program to remove a given element from an array#include<bits/stdc++.h>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 arrayint i;for (i=0; i<n; i++) if (arr[i] == x) break;// If x found in arrayif (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;} |
Java
// Java program to remove a given element from an arrayimport 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*/ |
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 Codeif __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 |
C#
// C# program to remove a given element from// an arrayusing 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. |
Javascript
<script>// Javascript 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[]function deleteElement( arr, n, x){ // Search x in array let 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 (let j=i; j<n; j++) arr[j] = arr[j+1]; } return n;} // driver code let arr = [11, 15, 6, 8, 9, 10]; let n = arr.length; let x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); document.write("Modified array is </br>"); for (let i=0; i<n; i++) document.write(arr[i] + " "); </script> |
Output
Modified array is 11 15 8 9 10
Time Complexity : O(n)
Auxiliary Space : O(1)
Method 2 (Move elements while searching): This method assumes 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.
Implementation:
C
#include <stdio.h>// 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); printf("Modified array is \n"); for (int i = 0; i < n; i++) printf("%d ", arr[i]); return 0;} |
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;} |
Java
// Java program to remove a given element from an arrayimport 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*/ |
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 codearr = [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 |
C#
// C# program to remove a given// element from an arrayusing 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. |
Javascript
<script>// Java Script 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. function 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. let prev = arr[n-1], i; for (i=n-2; i>=0 && arr[i]!=x; i--) { let 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 let arr = [11, 15, 6, 8, 9, 10]; let n = arr.length; let x = 6; // Delete x from arr[] n = deleteElement(arr, n, x); document.write("Modified array is<br>"); for (let i = 0; i < n; i++) document.write(arr[i]+" "); // This code is contributed by sravan kumar Gottumukkala</script> |
Output
Modified array is 11 15 8 9 10
Time Complexity : O(n)
Auxiliary Space : O(1)
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.
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