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.
freq[i] is positive, it means freq[i] IDs with the value nums[i] are added to the collection at step i.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.
\n
Example 1:
\n\nInput: nums = [2,3,2,1], freq = [3,2,-3,1]
\n\nOutput: [3,3,2,2]
\n\nExplanation:
\n\nAfter step 0, we have 3 IDs with the value of 2. So ans[0] = 3.
\nAfter step 1, we have 3 IDs with the value of 2 and 2 IDs with the value of 3. So ans[1] = 3.
\nAfter step 2, we have 2 IDs with the value of 3. So ans[2] = 2.
\nAfter 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:
\n\nInput: nums = [5,5,3], freq = [2,-2,1]
\n\nOutput: [2,0,1]
\n\nExplanation:
\n\nAfter step 0, we have 2 IDs with the value of 5. So ans[0] = 2.
\nAfter step 1, there are no IDs. So ans[1] = 0.
\nAfter step 2, we have 1 ID with the value of 3. So ans[2] = 1.
\n
Constraints:
\n\n1 <= nums.length == freq.length <= 1051 <= nums[i] <= 105-105 <= freq[i] <= 105freq[i] != 0你需要在一个集合里动态记录 ID 的出现频率。给你两个长度都为 n 的整数数组 nums 和 freq ，nums 中每一个元素表示一个 ID ，对应的 freq 中的元素表示这个 ID 在集合中此次操作后需要增加或者减少的数目。
freq[i] 是正数，那么 freq[i] 个 ID 为 nums[i] 的元素在第 i 步操作后会添加到集合中。freq[i] 是负数，那么 -freq[i] 个 ID 为 nums[i] 的元素在第 i 步操作后会从集合中删除。请你返回一个长度为 n 的数组 ans ，其中 ans[i] 表示第 i 步操作后出现频率最高的 ID 数目 ，如果在某次操作后集合为空，那么 ans[i] 为 0 。
\n\n
示例 1：
\n\n输入：nums = [2,3,2,1], freq = [3,2,-3,1]
\n\n输出：[3,3,2,2]
\n\n解释：
\n\n第 0 步操作后，有 3 个 ID 为 2 的元素，所以 ans[0] = 3 。
\n第 1 步操作后，有 3 个 ID 为 2 的元素和 2 个 ID 为 3 的元素，所以 ans[1] = 3 。
\n第 2 步操作后，有 2 个 ID 为 3 的元素，所以 ans[2] = 2 。
\n第 3 步操作后，有 2 个 ID 为 3 的元素和 1 个 ID 为 1 的元素，所以 ans[3] = 2 。
示例 2：
\n\n输入：nums = [5,5,3], freq = [2,-2,1]
\n\n输出：[2,0,1]
\n\n解释：
\n\n第 0 步操作后，有 2 个 ID 为 5 的元素，所以 ans[0] = 2 。
\n第 1 步操作后，集合中没有任何元素，所以 ans[1] = 0 。
\n第 2 步操作后，有 1 个 ID 为 3 的元素，所以 ans[2] = 1 。
\n\n
提示：
\n\n1 <= nums.length == freq.length <= 1051 <= nums[i] <= 105-105 <= freq[i] <= 105freq[i] != 0i find the occurrences of nums[i].",
"Change the occurrences of nums[i] in the ordered set."
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