Given an array arr[] of integers, the task is to remove duplicates from the given array.
Examples:
Input: arr[] = {1, 2, 3, 2, 5, 4, 4}
Output: arr[] = {1, 2, 3, 4, 5}
Input: arr[] = {127, 234, 127, 654, 355, 789, 355, 355, 999, 654}
Output: arr[] = {127, 234, 355, 654, 789, 999}
The duplicates in the array can be removed using Binary Search Tree. The idea is to create a Binary Search Tree using the array elements with the condition that the first element is taken as the root(parent) element and when the element “less” than root appears, it is made the left child and the element “greater” than root is made the right child of the root. Since no condition for “equal” exists the duplicates are automatically removed when we form a binary search tree from the array elements.
For the array, arr[] = {1, 2, 3, 2, 5, 4, 4}
BST will be:
Approach:
- Form BST using the array elements
- Display the elements using any Tree Traversal method.
Below is the implementation of the above approach.
C++
// C++ Program of above implementation#include <iostream>using namespace std;// Struct declarationstruct Node { int data; struct Node* left; struct Node* right;};// Node creationstruct Node* newNode(int data){ struct Node* nn = new Node; nn->data = data; nn->left = NULL; nn->right = NULL; return nn;}// Function to insert data in BSTstruct Node* insert(struct Node* root, int data){ if (root == NULL) return newNode(data); else { if (data < root->data) root->left = insert(root->left, data); if (data > root->data) root->right = insert(root->right, data); return root; }}// InOrder function to display value of array// in sorted ordervoid inOrder(struct Node* root){ if (root == NULL) return; else { inOrder(root->left); cout << root->data << " "; inOrder(root->right); }}// Driver codeint main(){ int arr[] = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = sizeof(arr) / sizeof(arr[0]); struct Node* root = NULL; for (int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); return 0;}// This code is contributed by shivanisingh |
C
// C Program of above implementation#include <stdio.h>#include <stdlib.h>// Struct declarationstruct Node { int data; struct Node* left; struct Node* right;};// Node creationstruct Node* newNode(int data){ struct Node* nn = (struct Node*)(malloc(sizeof(struct Node))); nn->data = data; nn->left = NULL; nn->right = NULL; return nn;}// Function to insert data in BSTstruct Node* insert(struct Node* root, int data){ if (root == NULL) return newNode(data); else { if (data < root->data) root->left = insert(root->left, data); if (data > root->data) root->right = insert(root->right, data); return root; }}// InOrder function to display value of array// in sorted ordervoid inOrder(struct Node* root){ if (root == NULL) return; else { inOrder(root->left); printf("%d ", root->data); inOrder(root->right); }}// Driver codeint main(){ int arr[] = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = sizeof(arr) / sizeof(arr[0]); struct Node* root = NULL; for (int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); return 0;} |
Java
// Java implementation of the approachimport java.util.Scanner;// Node declaration class Node { int data; public Node left; public Node right; Node(int data) { this.data = data; left = right = null; }}class GFG { // Function to insert data in BST public static Node insert(Node root, int data) { if (root == null) return new Node(data); if (data < root.data) root.left = insert(root.left, data); if (data > root.data) root.right = insert(root.right, data); return root; } // InOrder function to display value of array // in sorted order public static void inOrder(Node root) { if (root == null) return; inOrder(root.left); System.out.print(root.data+" "); inOrder(root.right); } // Driver Code public static void main(String []args){ int arr[] = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = arr.length; Node root = null; for (int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root,arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root); }}// This code is contributed by anishma |
Python3
# Python3 implementation of the approach # Binary tree node consists of data, a # pointer to the left child and a # pointer to the right child class newNode : def __init__(self,data) : self.data = data; self.left = None; self.right = None; # Function to insert data in BST def insert(root, data) : if (root == None) : return newNode(data); else : if (data < root.data) : root.left = insert(root.left, data); if (data > root.data) : root.right = insert(root.right, data); return root; # InOrder function to display value of array # in sorted order def inOrder(root) : if (root == None) : return; else : inOrder(root.left); print(root.data, end = " "); inOrder(root.right); # Driver code if __name__ == "__main__" : arr = [ 1, 2, 3, 2, 5, 4, 4 ]; # Finding size of array arr[] n = len(arr); root = None; for i in range(n) : # Insert element of arr[] in BST root = insert(root, arr[i]); # Inorder Traversal to print nodes of Tree inOrder(root); # This code is contributed by AnkitRai01 |
C#
// C# program of above implementationusing System;// Node declaration public class Node { public int data; public Node left; public Node right; public Node(int data) { this.data = data; left = right = null; } } class GFG{ // Function to insert data in BST public static Node insert(Node root, int data) { if (root == null) return new Node(data); if (data < root.data) root.left = insert(root.left, data); if (data > root.data) root.right = insert(root.right, data); return root; } // InOrder function to display value of array // in sorted order public static void inOrder(Node root) { if (root == null) return; inOrder(root.left); Console.Write(root.data + " "); inOrder(root.right); } // Driver Code static void Main() { int[] arr = { 1, 2, 3, 2, 5, 4, 4 }; // Finding size of array arr[] int n = arr.Length; Node root = null; for(int i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root);}}// This code is contributed by divyeshrabadiya07 |
Javascript
<script>// JavaScript program of above implementation// Node declaration class Node { constructor(data) { this.data = data; this.left = null; this.right = null; }}// Function to insert data in BST function insert(root, data) { if (root == null) return new Node(data); if (data < root.data) root.left = insert(root.left, data); if (data > root.data) root.right = insert(root.right, data); return root; } // InOrder function to display value of array // in sorted order function inOrder(root) { if (root == null) return; inOrder(root.left); document.write(root.data + " "); inOrder(root.right); } // Driver Code var arr = [1, 2, 3, 2, 5, 4, 4 ]; // Finding size of array arr[] var n = arr.length; var root = null; for(var i = 0; i < n; i++) { // Insert element of arr[] in BST root = insert(root, arr[i]); } // Inorder Traversal to print nodes of Tree inOrder(root);</script> |
Output:
1 2 3 4 5
Time Complexity: 
in the worst case(when the array is sorted) where N is the size of the given array.
Auxiliary Space: O(N).
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
