Solution 1 - Frequency Count + Unique Frequencies

Count each value, then check if the multiset of counts has duplicates.

TimeComplexity: O(n)

SpaceComplexity: O(n)

""" 1207.1 - Unique Number of Occurrences - Solution 1 - Count + Set """

from collections import Counter
from typing import List

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        vals = Counter(arr).values()
        return len(vals) == len(set(vals))

package main

func uniqueOccurrences(arr []int) bool {
	cnt := map[int]int{}
	for _, x := range arr { cnt[x]++ }
	seen := map[int]bool{}
	for _, v := range cnt {
		if seen[v] { return false }
		seen[v] = true
	}
	return true
}

func main() {}

use std::collections::{HashMap, HashSet};

fn unique_occurrences(arr: &[i32]) -> bool {
	let mut cnt: HashMap<i32, i32> = HashMap::new();
	for &x in arr { *cnt.entry(x).or_insert(0) += 1; }
	let mut seen: HashSet<i32> = HashSet::new();
	for &v in cnt.values() {
		if !seen.insert(v) { return false; }
	}
	true
}

fn main() {}

1207 - Unique Number of Occurrences

Solution 1 - Frequency Count + Unique Frequencies