Count Non-Repeated Elements

Problem Statement - link #

Hashing is very useful to keep track of the frequency of the elements in a list.

You are given an array of integers. You need to print the count of non-repeated elements in the array.

Your Task: You don’t need to read input or print anything. You only need to complete the function countNonRepeated() that takes array arr[] and its size n as parameters and returns the count of non-repeating elements in the array.

Expected Time Complexity: O(n)
Expected Auxiliary Space: O(n)

Examples #

Example 1:

Input:
10
1 1 2 2 3 3 4 5 6 7

Output: 
4

Explanation: 
4, 5, 6 and 7 are the 
elements with frequency 1 and rest 
elements are repeated so the number 
of non-repeated elements are 4.

Example 2:

Input:
5
10 20 30 40 10

Output: 
3

Explanation: 
20, 30, 40 are the 
elements with the frequency 1 and 
10 is the repeated element to 
number of non-repeated elements 
are 3.

Constraints #

• 1 <= N < 1000
• 0 <= arr[i] <= 10^7

Solutions #

class Solution{
    public:
    //Complete this function
    //Function to return the count of non-repeated elements in the array.
    int countNonRepeated(int arr[], int n)
    {
        //Your code here
        int res = 0;
        unordered_map<int, int> m;
        for(int i=0; i<n; i++) m[arr[i]]++;
        for(auto i: m)
            if(i.second == 1) res++;
        return res;
    }
};