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, Â 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;Â
// Function to print all unique// elements present in a sorted arrayvoid RemoveDuplicates(int arr[], int n){Â
    int i = 0;Â
    // Traverse the array    while (i < n) {Â
        int cur = arr[i];Â
        // Stores frequency of        // the current element        int cnt = 0;Â
        // 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++;        }Â
        // If current element is unique        if (cnt == 1) {Â
            cout << cur << " ";        }    }}Â
// Driver Codeint main(){Â
    // Given Input    int arr[] = { 1, 3, 3, 5, 5, 6, 10 };    int N = 7;Â
    // Function Call    RemoveDuplicates(arr, N);Â
    return 0;} |
Java
// Java Program for the above approachimport java.io.*;Â
class GFG {     // Function to print all unique  // elements present in a sorted array  static void RemoveDuplicates(int arr[], int n)  {Â
    int i = 0;Â
    // Traverse the array    while (i < n) {Â
      int cur = arr[i];Â
      // Stores frequency of      // the current element      int cnt = 0;Â
      // 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++;      }Â
      // If current element is unique      if (cnt == 1) {        System.out.print(cur +" ");      }    }  }Â
  // Driver CodeÂ
  public static void main (String[] args)  {Â
    // Given Input    int arr[] = { 1, 3, 3, 5, 5, 6, 10 };    int N = 7;Â
    // Function Call    RemoveDuplicates(arr, N);  }}Â
// 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 = 0Â
        # 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=" ")Â
# Driver codeif __name__ == "__main__":       # Given Input    arr = [1, 3, 3, 5, 5, 6, 10]    N = 7Â
    # Function Call    RemoveDuplicates(arr, N)     # This code is contributed by Kushagra Bansal |
C#
// C# Program for the above approachusing System;Â
class GFG {Â
    // Function to print all unique    // elements present in a sorted array    static void RemoveDuplicates(int[] arr, int n)    {Â
        int i = 0;Â
        // Traverse the array        while (i < n) {Â
            int cur = arr[i];Â
            // Stores frequency of            // the current element            int cnt = 0;Â
            // 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++;            }Â
            // If current element is unique            if (cnt == 1) {                Console.Write(cur + " ");            }        }    }Â
    // Driver CodeÂ
    public static void Main()    {Â
        // Given Input        int[] arr = { 1, 3, 3, 5, 5, 6, 10 };        int N = 7;Â
        // Function Call        RemoveDuplicates(arr, N);    }}Â
// This code is contributed by rishavmahato348. |
Javascript
<script>Â Â Â Â Â Â Â // JavaScript Program for the above approachÂ
       // Function to print all unique       // elements present in a sorted array       function RemoveDuplicates(arr, n) {Â
           let i = 0;Â
           // Traverse the array           while (i < n) {Â
               let cur = arr[i];Â
               // Stores frequency of               // the current element               let cnt = 0;Â
               // 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++;               }Â
               // If current element is unique               if (cnt == 1) {Â
                   document.write(cur + " ");               }           }       }Â
       // Driver CodeÂ
       // Given Input       let arr = [1, 3, 3, 5, 5, 6, 10];       let N = 7;Â
       // Function Call       RemoveDuplicates(arr, N);Â
   // This code is contributed by Potta LokeshÂ
   </script> |
Output
1 6 10
Time Complexity: O(N)
Auxiliary Space: O(1)
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