From 09731e9c4462bd6cb06a625c8eddabf4077288ad Mon Sep 17 00:00:00 2001
From: nathannaveen <42319948+nathannaveen@users.noreply.github.com>
Date: Thu, 9 May 2024 14:21:33 +0000
Subject: [PATCH] Sync LeetCode submission Runtime - 210 ms (43.21%), Memory -
17.5 MB (74.07%)
---
3363-most-frequent-ids/README.md | 49 ++++++++++++++++++++++++++++++
3363-most-frequent-ids/solution.go | 42 +++++++++++++++++++++++++
2 files changed, 91 insertions(+)
create mode 100644 3363-most-frequent-ids/README.md
create mode 100644 3363-most-frequent-ids/solution.go
diff --git a/3363-most-frequent-ids/README.md b/3363-most-frequent-ids/README.md
new file mode 100644
index 0000000..9841786
--- /dev/null
+++ b/3363-most-frequent-ids/README.md
@@ -0,0 +1,49 @@
+The problem involves tracking the frequency of IDs in a collection that changes over time. You have two integer arrays, nums and freq, of equal length n. Each element in nums represents an ID, and the corresponding element in freq indicates how many times that ID should be added to or removed from the collection at each step.
+
+
+ - Addition of IDs: If
freq[i] is positive, it means freq[i] IDs with the value nums[i] are added to the collection at step i.
+ - Removal of IDs: If
freq[i] is negative, it means -freq[i] IDs with the value nums[i] are removed from the collection at step i.
+
+
+Return an array ans of length n, where ans[i] represents the count of the most frequent ID in the collection after the ith step. If the collection is empty at any step, ans[i] should be 0 for that step.
+
+
+Example 1:
+
+
+
Input: nums = [2,3,2,1], freq = [3,2,-3,1]
+
+
Output: [3,3,2,2]
+
+
Explanation:
+
+
After step 0, we have 3 IDs with the value of 2. So ans[0] = 3.
+After step 1, we have 3 IDs with the value of 2 and 2 IDs with the value of 3. So ans[1] = 3.
+After step 2, we have 2 IDs with the value of 3. So ans[2] = 2.
+After step 3, we have 2 IDs with the value of 3 and 1 ID with the value of 1. So ans[3] = 2.
+
Example 2:
+
+
+
Input: nums = [5,5,3], freq = [2,-2,1]
+
+
Output: [2,0,1]
+
+
Explanation:
+
+
After step 0, we have 2 IDs with the value of 5. So ans[0] = 2.
+After step 1, there are no IDs. So ans[1] = 0.
+After step 2, we have 1 ID with the value of 3. So ans[2] = 1.
+
+Constraints:
+
+
+ 1 <= nums.length == freq.length <= 1051 <= nums[i] <= 105-105 <= freq[i] <= 105freq[i] != 0- The input is generated such that the occurrences of an ID will not be negative in any step.
+
diff --git a/3363-most-frequent-ids/solution.go b/3363-most-frequent-ids/solution.go
new file mode 100644
index 0000000..4b4e1b0
--- /dev/null
+++ b/3363-most-frequent-ids/solution.go
@@ -0,0 +1,42 @@
+func mostFrequentIDs(nums []int, freq []int) []int64 {
+ curValues := map[int] int64 {} // value : frequency
+ h := &KeyHeap{}
+ heap.Init(h)
+ res := make([]int64, len(nums))
+
+ for i := 0; i < len(nums); i++ {
+ curValues[nums[i]] += int64(freq[i])
+ heap.Push(h, key{value: nums[i], freq: curValues[nums[i]]})
+
+ for curValues[(*h)[0].value] != (*h)[0].freq {
+ heap.Pop(h)
+ }
+
+ res[i] = (*h)[0].freq
+ }
+
+ return res
+}
+
+type key struct {
+ value int
+ freq int64
+}
+
+type KeyHeap []key
+
+func (h KeyHeap) Len() int { return len(h) }
+func (h KeyHeap) Less(i, j int) bool { return h[i].freq > h[j].freq }
+func (h KeyHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
+
+func (h *KeyHeap) Push(x any) {
+ *h = append(*h, x.(key))
+}
+
+func (h *KeyHeap) Pop() any {
+ old := *h
+ n := len(old)
+ x := old[n-1]
+ *h = old[0 : n-1]
+ return x
+}