add: solution03

Author: ysle0

top-k-frequent-elements/ysle0.go

@@ -0,0 +1,127 @@
+package top_k_frequent_elements
+
+//package main
+
+import (
+	"container/heap"
+	"slices"
+)
+
+/*
+ 1. 문제
+    nums 에서 가장 많이 나온 숫자들 k 개를 반환.
+
+ 2. 풀이
+    map 에 빈도를 기록하여 내림차순 정렬한 후 k개 뽑기
+
+3. 분석
+
+  - 시간 복잡도: O(N + M logM) --> O(N logN)
+    빈도맵핑을 위한 nums 순회: O(N)
+    오름차순정렬: O(M logM)
+
+  - 공간 복잡도: O(N)
+    주어진 배열 nums: O(N)
+    빈도맵핑용 map: O(N)
+*/
+type Kvp struct {
+	k int
+	v int
+}
+
+func topKFrequent(nums []int, k int) []int {
+	freq := map[int]int{}
+
+	// 빈도를 기록
+	for _, n := range nums {
+		if _, ok := freq[n]; !ok {
+			freq[n] = 1
+		} else {
+			freq[n]++
+		}
+	}
+
+	// map->array
+	tmp := make([]Kvp, 0, len(freq))
+	for key, v := range freq {
+		tmp = append(tmp, Kvp{key, v})
+	}
+
+	// 내림차순 정렬 (time O(M logM)
+	slices.SortFunc(tmp, func(a, b Kvp) int { return b.v - a.v })
+
+	// []int 로 변환
+	res := make([]int, 0, len(tmp))
+	for _, kvp := range tmp {
+		res = append(res, kvp.k)
+	}
+
+	// k 개 뽑기
+	return res[:k]
+}
+
+func topKElements_HeapBasedApproach(nums []int, k int) []int {
+	freq := map[int]int{}
+	for _, n := range nums {
+		freq[n]++
+	}
+
+	h := &IntHeap{}
+	heap.Init(h)
+
+	for k, v := range freq {
+		heap.Push(h, Kvp{k, v})
+		if h.Len() > k {
+			heap.Pop(h)
+		}
+	}
+
+	res := make([]int, k)
+	for i := k - 1; i >= 0; i-- {
+		res[i] = heap.Pop(h).(Kvp).k
+	}
+
+	return res
+}
+
+type IntHeap []Kvp
+
+func (h *IntHeap) Len() int           { return len(*h) }
+func (h *IntHeap) Less(i, j int) bool { return (*h)[i].v < (*h)[j].v }
+func (h *IntHeap) Swap(i, j int)      { (*h)[i], (*h)[j] = (*h)[j], (*h)[i] }
+func (h *IntHeap) Push(x interface{}) { *h = append(*h, x.(Kvp)) }
+func (h *IntHeap) Pop() interface{} {
+	old := *h
+	n := len(old)
+	x := old[n-1]
+	*h = old[0 : n-1]
+	return x
+}
+
+func topKFrequentElements_BucketSort(nums []int, k int) []int {
+	freq := map[int]int{}
+	for _, n := range nums {
+		freq[n]++
+	}
+
+	buc := make([][]int, len(nums)+1)
+	for k, v := range freq {
+		buc[v] = append(buc[v], k)
+	}
+
+	res := []int{}
+	for i := len(buc) - 1; i >= 0 && len(res) < k; i-- {
+		res = append(res, buc[i]...)
+	}
+
+	return res[:k]
+}
+
+//
+//func main() {
+//	r1 := topKFrequent([]int{1, 1, 1, 2, 2, 3}, 2)
+//	fmt.Println(r1)
+//
+//	r2 := topKFrequent([]int{1}, 1)
+//	fmt.Println(r2)
+//}