C++ Program to Find Non Repeating Elements in an Array

October 11, 2022

Non Repeating Elements in an array in C++

Here, in this section we will discuss the program to print non repeating elements in an array in C++ programming language. We will discuss different methods to print the unique elements of the given input array.

Distinct element of an array in C++
While loop in C

Methods Discussed in this Page are :

  • Method 1 : Using Two loops
  • Method 2 : Using hash Map

Let’s discuss each method one by one,

Method 1 :

In this method we will count the frequency of each elements using two for loops and print those elements which occurs on;y one time in the given input array.

  • To check the status of visited elements create a array of size n.
  • Run a loop from index 0 to n and check if (visited[i]==1) then skip that element.
  • Otherwise create a variable count = 1 to keep the count of frequency.
  • Run a loop from index i+1 to n
  • Check if(arr[i]==arr[j]), then increment the count by 1 and set visited[j]=1.
  • After complete iteration of inner for loop check if(count==1), then print that ith element.

Time and Space Complexity :

  • Time Complexity : O(n2)
  • Space Complexity : O(n)

Counting Distinct Elements in an Array

Finding  Repeating elements in an Array

Removing Duplicate elements from an array

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Finding Maximum scalar product of two vectors in an array 

Method 1 : Code in C++

Run 
#include <bits/stdc++.h>
using namespace std;

// Main function to run the program
int main() 
{ 
    int arr[] = {10, 30, 10, 20, 40, 20, 50, 10}; 
    int n = sizeof(arr)/sizeof(arr[0]); 

    int visited[n], count_dis=0;

    for(int i=0; i<n; i++){

        if(visited[i]!=1){
           int count = 1;
           for(int j=i+1; j<n; j++){
              if(arr[i]==arr[j]){
                 count++;
                 visited[j]=1;
              }
            }
            if(count==1)
             cout<<arr[i]<<" ";
         }
     }
 
    return 0; 
}

Output :

30 50

Method 2 :

In this method we will use hash-map to store the frequency of the elements and print those elements which have frequency equals to 1.

  • Create an unordered_map say mp.
  • Run a loop to iterate over array
  • Set mp[arr[i]]++
  • After, complete iteration, run a loop over map
  • Check if value == 1, then print the key.

Time and Space Complexity :

  • Time Complexity : O(n)
  • Space Complexity : O(n)
Non repeating elements in an array in C++

Method 2 : Code in C++

Run
#include <bits/stdc++.h>
using namespace std;

// Main function to run the program
int main() 
{ 
   int arr[] = {10, 30, 40, 20, 10, 20, 50, 10}; 
   int n = sizeof(arr)/sizeof(arr[0]); 

   unordered_map <int, int>mp;
   int count_dis=0;

   for(int i=0; i<n; i++)
      mp[arr[i]]++;

   for(auto it=mp.begin(); it!=mp.end(); it++){
       if(it->second==1)
          cout<<it->first<<" ";
   }
 
}

Output :

50 40 30

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