CSES - Distinct Numbers

Author: Andrew Wang

Appears In

Time Complexity: O(Nlog⁡N)\mathcal{O}(N\log{N})

Sort the array of numbers. Loop through the array and increment the answer for every distinct number. Distinct numbers can be found if the current number isn't equal to the previous number in the array.

C++

#include <bits/stdc++.h>
using namespace std;
 
int main() {
	int N; cin >> N;
	int arr[N];
	for (int i = 0; i < N; i++)
		cin >> arr[i];
	sort(arr, arr+N);
	int ans = 1;

Java

import java.io.*;
import java.util.*;
 
public class DistinctNumbers {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {

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