- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Hash Table, Ordered Set, Heap (Priority Queue), leetcode-3092, Binary Heap, O(nlogn), O(n), Heap, BST, Sorted List
Problem
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 meansfreq[i]
IDs with the valuenums[i]
are added to the collection at stepi
. - Removal of IDs: If
freq[i]
is negative, it means-freq[i]
IDs with the valuenums[i]
are removed from the collection at stepi
.
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 <= 105
1 <= nums[i] <= 105
-105 <= freq[i] <= 105
freq[i] != 0
- The input is generated such that the occurrences of an ID will not be negative in any step.