Top K Frequent Elements

main.md

Question

Approach

1. A loop is run to go through each number.
2. A map is used to check if the number is seen for the first time , if it has not been then the loop skips to the next number.
3. if a number is seen for the first time a second inner loop is run to check which other number in the array is the same as the current number.
4. the map is turned into an array of arrays and the k number frequent elements which are the i[0] values are returned.

Complexity

• Time complexity:O(N^2)

• Space complexity:O(N)

Code

function topKFrequent(nums: number[], k: number): number[] {
    let n = nums.length
    let result: number[] = []
    let seen = new Map()
   
    for (let i = 0; i < n; i++) {
        let j = 0
        let size = 0
        if (seen.has(nums[i])) continue

        while (j < n) {
            if (nums[i] === nums[j]) {
                size++
            }
            j++
        }
        seen.set(nums[i],size)

    }
     let sorted = [...seen.entries()].sort((a, b) => b[1] - a[1])

    return sorted.slice(0, k).map(entry => entry[0])
};