Given a sorted array arr[] of size N, the task is to print all the unique elements in the array.
An array element is said to be unique if the frequency of that element in the array is 1.
Examples:
Input: arr[ ] = {1, 1, 2, 2, 3, 4, 5, 5}
Output: 3 4
Explanation: Since 1, 2, 5 are occurring more than once in the array, the distinct elements are 3 and 4.Input: arr[ ] = {1, 2, 3, 3, 3, 4, 5, 6}
Output: 1 2 4 5 6
Approach: The simplest approach to solve the problem is to traverse the array arr[] and print only those elements whose frequency is 1. Follow the steps below to solve the problem:
- Iterate over the array arr[] and initialize a variable, say cnt = 0, to count the frequency of the current array element.
- Since the array is already sorted, check if the current element is the same as the previous element. If found to be true, then update cnt += 1.
- Otherwise, if cnt = 1, then print the element. Otherwise, continue.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print all unique// elements present in a sorted arrayvoid RemoveDuplicates(int arr[], int n){ int i = 0; // Traverse the array while (i < n) { int cur = arr[i]; // Stores frequency of // the current element int cnt = 0; // Iterate until end of the // array is reached or current // element is not the same as the // previous element while (i < n and cur == arr[i]) { cnt++; i++; } // If current element is unique if (cnt == 1) { cout << cur << " "; } }}// Driver Codeint main(){ // Given Input int arr[] = { 1, 3, 3, 5, 5, 6, 10 }; int N = 7; // Function Call RemoveDuplicates(arr, N); return 0;} |
Java
// Java Program for the above approachimport java.io.*;class GFG { // Function to print all unique // elements present in a sorted array static void RemoveDuplicates(int arr[], int n) { int i = 0; // Traverse the array while (i < n) { int cur = arr[i]; // Stores frequency of // the current element int cnt = 0; // Iterate until end of the // array is reached or current // element is not the same as the // previous element while (i < n && cur == arr[i]) { cnt++; i++; } // If current element is unique if (cnt == 1) { System.out.print(cur +" "); } } } // Driver Code public static void main (String[] args) { // Given Input int arr[] = { 1, 3, 3, 5, 5, 6, 10 }; int N = 7; // Function Call RemoveDuplicates(arr, N); }}// This code is contributed by Potta Lokesh |
Python3
# Function to print all unique# elements present in a sorted arraydef RemoveDuplicates(arr, n): i = 0 while i < n: cur = arr[i] # Stores frequency of # the current element cnt = 0 # Iterate until end of the # array is reached or current # element is not the same as the # previous element while i < n and cur == arr[i]: cnt += 1 i += 1 if cnt == 1: print(cur, end=" ")# Driver codeif __name__ == "__main__": # Given Input arr = [1, 3, 3, 5, 5, 6, 10] N = 7 # Function Call RemoveDuplicates(arr, N) # This code is contributed by Kushagra Bansal |
C#
// C# Program for the above approachusing System;class GFG { // Function to print all unique // elements present in a sorted array static void RemoveDuplicates(int[] arr, int n) { int i = 0; // Traverse the array while (i < n) { int cur = arr[i]; // Stores frequency of // the current element int cnt = 0; // Iterate until end of the // array is reached or current // element is not the same as the // previous element while (i < n && cur == arr[i]) { cnt++; i++; } // If current element is unique if (cnt == 1) { Console.Write(cur + " "); } } } // Driver Code public static void Main() { // Given Input int[] arr = { 1, 3, 3, 5, 5, 6, 10 }; int N = 7; // Function Call RemoveDuplicates(arr, N); }}// This code is contributed by rishavmahato348. |
Javascript
<script> // JavaScript Program for the above approach // Function to print all unique // elements present in a sorted array function RemoveDuplicates(arr, n) { let i = 0; // Traverse the array while (i < n) { let cur = arr[i]; // Stores frequency of // the current element let cnt = 0; // Iterate until end of the // array is reached or current // element is not the same as the // previous element while (i < n && cur == arr[i]) { cnt++; i++; } // If current element is unique if (cnt == 1) { document.write(cur + " "); } } } // Driver Code // Given Input let arr = [1, 3, 3, 5, 5, 6, 10]; let N = 7; // Function Call RemoveDuplicates(arr, N); // This code is contributed by Potta Lokesh </script> |
Output
1 6 10
Time Complexity: O(N)
Auxiliary Space: O(1)
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