README.md

1481. Least Number of Unique Integers after K Removals

Given an array of integers arr and an integer k. Find the least number of unique integers after removing exactly k elements.

Example 1:

Input: arr = [5,5,4], k = 1
Output: 1
Explanation: Remove the single 4, only 5 is left.

Example 2:

Input: arr = [4,3,1,1,3,3,2], k = 3
Output: 2
Explanation: Remove 4, 2 and either one of the two 1s or three 3s. 1 and 3 will be left.

Constraints:

• 1 <= arr.length <= 10^5
• 1 <= arr[i] <= 10^9
• 0 <= k <= arr.length

Solutions (Ruby)

1. Sort

# @param {Integer[]} arr
# @param {Integer} k
# @return {Integer}
def find_least_num_of_unique_ints(arr, k)
  counter = {}
  counter.default = 0
  arr.each { |x| counter[x] += 1 }
  counter = counter.values.sort

  (0..counter.size).each do |i|
    if k == 0
      return counter.size - i
    elsif k < 0
      return counter.size - i + 1
    end

    k -= counter[i]
  end
end

Solutions (Rust)

1. Sort

use std::collections::HashMap;

impl Solution {
    pub fn find_least_num_of_unique_ints(arr: Vec<i32>, mut k: i32) -> i32 {
        let mut counter = HashMap::new();
        for x in arr {
            *counter.entry(x).or_insert(0) += 1;
        }

        let mut counter = counter.values().collect::<Vec<_>>();
        counter.sort_unstable();

        for i in 0..=counter.len() {
            if k == 0 {
                return (counter.len() - i) as i32;
            } else if k < 0 {
                return (counter.len() - i) as i32 + 1;
            }
            k -= counter[i];
        }

        0
    }
}