feat: top k frequent

pkg/top_k_frequent_elements/main.go

@@ -0,0 +1,89 @@
+package top_k_frequent_elements
+
+/// O(n*log(n))
+
+// func ToMultiset(nums []int) map[int]int {
+// 	result := map[int]int{}
+
+// 	for _, num := range nums {
+// 		result[num]++
+// 	}
+
+// 	return result
+// }
+
+// func ToEntries(multiset map[int]int) [][2]int {
+// 	result := [][2]int{}
+
+// 	for num, amount := range multiset {
+// 		result = append(result, [2]int{num, amount})
+// 	}
+
+// 	return result
+// }
+
+// func ByDescendingAmount(a, b [2]int) int {
+// 	return b[1] - a[1]
+// }
+
+// func GetNumbers(entries [][2]int) []int {
+// 	amounts := []int{}
+
+// 	for _, entry := range entries {
+// 		amounts = append(amounts, entry[0])
+// 	}
+
+// 	return amounts
+// }
+
+// func TopKFrequent(nums []int, k int) []int {
+// 	multiset := ToMultiset(nums)
+// 	entries := ToEntries(multiset)
+
+// 	slices.SortFunc(entries, ByDescendingAmount)
+// 	return GetNumbers(entries[0:k])
+// }
+
+/// O(n)
+
+func ToMultiset(nums []int) map[int]int {
+	result := map[int]int{}
+
+	for _, num := range nums {
+		result[num]++
+	}
+
+	return result
+}
+
+func TopKFrequent(nums []int, k int) []int {
+	multiset := ToMultiset(nums)
+
+	// Bucket Sort.
+	//
+	// A multiset's frequency is always less then or equal to the length of the
+	// original list. So, you can use bucket sort to sort the items in O(n).
+	frequencies := make([][]int, len(nums))
+
+	for value, amount := range multiset {
+		if frequencies[amount-1] == nil {
+			frequencies[amount-1] = []int{}
+		}
+
+		frequencies[amount-1] = append(frequencies[amount-1], value)
+	}
+
+	top_frequencies := []int{}
+
+	for i := len(nums) - 1; i >= 0; i-- {
+		for _, num := range frequencies[i] {
+			top_frequencies = append(top_frequencies, num)
+
+			if len(top_frequencies) == k {
+				return top_frequencies
+			}
+		}
+	}
+
+	return top_frequencies
+}