Sync LeetCode submission Runtime - 210 ms (43.21%), Memory - 17.5 MB …

…(74.07%)

3363-most-frequent-ids/README.md

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+<p>The problem involves tracking the frequency of IDs in a collection that changes over time. You have two integer arrays, <code>nums</code> and <code>freq</code>, of equal length <code>n</code>. Each element in <code>nums</code> represents an ID, and the corresponding element in <code>freq</code> indicates how many times that ID should be added to or removed from the collection at each step.</p>
+
+<ul>
+	<li><strong>Addition of IDs:</strong> If <code>freq[i]</code> is positive, it means <code>freq[i]</code> IDs with the value <code>nums[i]</code> are added to the collection at step <code>i</code>.</li>
+	<li><strong>Removal of IDs:</strong> If <code>freq[i]</code> is negative, it means <code>-freq[i]</code> IDs with the value <code>nums[i]</code> are removed from the collection at step <code>i</code>.</li>
+</ul>
+
+<p>Return an array <code>ans</code> of length <code>n</code>, where <code>ans[i]</code> represents the <strong>count</strong> of the <em>most frequent ID</em> in the collection after the <code>i<sup>th</sup></code>&nbsp;step. If the collection is empty at any step, <code>ans[i]</code> should be 0 for that step.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [2,3,2,1], freq = [3,2,-3,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[3,3,2,2]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>After step 0, we have 3 IDs with the value of 2. So <code>ans[0] = 3</code>.<br />
+After step 1, we have 3 IDs with the value of 2 and 2 IDs with the value of 3. So <code>ans[1] = 3</code>.<br />
+After step 2, we have 2 IDs with the value of 3. So <code>ans[2] = 2</code>.<br />
+After step 3, we have 2 IDs with the value of 3 and 1 ID with the value of 1. So <code>ans[3] = 2</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [5,5,3], freq = [2,-2,1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[2,0,1]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>After step 0, we have 2 IDs with the value of 5. So <code>ans[0] = 2</code>.<br />
+After step 1, there are no IDs. So <code>ans[1] = 0</code>.<br />
+After step 2, we have 1 ID with the value of 3. So <code>ans[2] = 1</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length == freq.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>-10<sup>5</sup> &lt;= freq[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>freq[i] != 0</code></li>
+	<li>The input is generated<!-- notionvc: a136b55a-f319-4fa6-9247-11be9f3b1db8 --> such that the occurrences of an ID will not be negative in any step.</li>
+</ul>

3363-most-frequent-ids/solution.go

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+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
+}