Top K Frequent Elements

leetcode link

Given an integer array numsand an integer k, return the kmost frequent elements. You may return the answer in any order.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

Constraints:

1 <= nums.length <= 105
-104 <= nums[i] <= 104
k is in the range [1, the number of unique elements in the array].
It is guaranteed that the answer is unique.

how to approach

set the Map to check the element in array.
and in for loop, add the element one by one to check the count.
after that, sort out the array from the map so it can have the order by element count.
return the sliced array for top K frequent elements.

function topKFrequent(nums: number[], k: number): number[] {
    let frequent = new Map()
    
    for(let num of nums){
        let count = frequent.get(num) || 0
        frequent.set(num,count + 1)
    }
    
    let sorted = Array.from(frequent)
                      .sort((a,b)=>b[1]-a[1])
                      .slice(0,k)
                      .map(([num])=>num)
  
    return sorted
};

summary

this leetcode problem is solved by typescript language.

the approach is using Map that has key and value in it, set key as element itself and value as counted number of how many elements was in the array.

and lastly sorting the array from counted number of the elements and slice them by the K input to figure out the top K frequent elements in the array.

[Leetcode] Top K Frequent Elements

Top K Frequent Elements

<span style="font-size: 20px;">how to approach</span>

summary