Top K Frequent Elements - Golang

LeetCode Problem: Top K Frequent Elements

Problem statement:

Given an array of integers (numbers), you have to return the k most frequent elements

Example:

Input:

numbers: [1,2,3,4,5,6,1,2,1,2], k: 2

Expected output array:

[1,2]

If you would like to solve the problem on Leetcode, here is the link to the problem: https://leetcode.com/problems/top-k-frequent-elements/

Golang Solution:

import (
    "fmt"
)

func topKFrequent(numbers []int, k int) []int {
    
    arrayWithIndicesAsCount := make([]map[int]int, len(numbers)+1)
    numbersCountMap := make(map[int]int, 0)

    for _, num := range numbers {    

        if count, ok := numbersCountMap[num]; ok {
            mapAtIndex := arrayWithIndicesAsCount[numbersCountMap[num]]
            delete(mapAtIndex, num)    
            arrayWithIndicesAsCount[count] = mapAtIndex
            
            numbersCountMap[num] += 1

            mapAtIndexNew := arrayWithIndicesAsCount[numbersCountMap[num]]
            if mapAtIndexNew == nil {
                mapAtIndexNew = make(map[int]int)
            }
            mapAtIndexNew[num] = numbersCountMap[num]
            arrayWithIndicesAsCount[numbersCountMap[num]] = mapAtIndexNew
        } else {
            numbersCountMap[num] = 1
            arrayWithIndicesAsCountMap := arrayWithIndicesAsCount[numbersCountMap[num]]
            if arrayWithIndicesAsCountMap == nil {
                arrayWithIndicesAsCountMap = make(map[int]int)
            }
            arrayWithIndicesAsCountMap[num] = 1
            arrayWithIndicesAsCount[1] = arrayWithIndicesAsCountMap
        }
    }

    index := len(numbers)
    result := make([]int, 0)
    for index > 0 && k > 0 {
        if arrayWithIndicesAsCount[index] != nil {
            mapAtIndex := arrayWithIndicesAsCount[index]
            for key, _ := range mapAtIndex {
                result = append(result, key)
            }
            k -= len(mapAtIndex)
        }
        index -= 1
    }

    return result
}