Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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.
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] != 0Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Segment Tree
[2,3,2,1] [3,2,-3,1]
[5,5,3] [2,-2,1]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3092: Most Frequent IDs
class Solution {
public long[] mostFrequentIDs(int[] nums, int[] freq) {
Map<Integer, Long> cnt = new HashMap<>();
Map<Long, Integer> lazy = new HashMap<>();
int n = nums.length;
long[] ans = new long[n];
PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
for (int i = 0; i < n; ++i) {
int x = nums[i], f = freq[i];
lazy.merge(cnt.getOrDefault(x, 0L), 1, Integer::sum);
cnt.merge(x, (long) f, Long::sum);
pq.add(cnt.get(x));
while (!pq.isEmpty() && lazy.getOrDefault(pq.peek(), 0) > 0) {
lazy.merge(pq.poll(), -1, Integer::sum);
}
ans[i] = pq.isEmpty() ? 0 : pq.peek();
}
return ans;
}
}
// Accepted solution for LeetCode #3092: Most Frequent IDs
func mostFrequentIDs(nums []int, freq []int) []int64 {
n := len(nums)
cnt := map[int]int{}
lazy := map[int]int{}
ans := make([]int64, n)
pq := hp{}
heap.Init(&pq)
for i, x := range nums {
f := freq[i]
lazy[cnt[x]]++
cnt[x] += f
heap.Push(&pq, cnt[x])
for pq.Len() > 0 && lazy[pq.IntSlice[0]] > 0 {
lazy[pq.IntSlice[0]]--
heap.Pop(&pq)
}
if pq.Len() > 0 {
ans[i] = int64(pq.IntSlice[0])
}
}
return ans
}
type hp struct{ sort.IntSlice }
func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
# Accepted solution for LeetCode #3092: Most Frequent IDs
class Solution:
def mostFrequentIDs(self, nums: List[int], freq: List[int]) -> List[int]:
cnt = Counter()
lazy = Counter()
ans = []
pq = []
for x, f in zip(nums, freq):
lazy[cnt[x]] += 1
cnt[x] += f
heappush(pq, -cnt[x])
while pq and lazy[-pq[0]] > 0:
lazy[-pq[0]] -= 1
heappop(pq)
ans.append(0 if not pq else -pq[0])
return ans
// Accepted solution for LeetCode #3092: Most Frequent IDs
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #3092: Most Frequent IDs
// class Solution {
// public long[] mostFrequentIDs(int[] nums, int[] freq) {
// Map<Integer, Long> cnt = new HashMap<>();
// Map<Long, Integer> lazy = new HashMap<>();
// int n = nums.length;
// long[] ans = new long[n];
// PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
// for (int i = 0; i < n; ++i) {
// int x = nums[i], f = freq[i];
// lazy.merge(cnt.getOrDefault(x, 0L), 1, Integer::sum);
// cnt.merge(x, (long) f, Long::sum);
// pq.add(cnt.get(x));
// while (!pq.isEmpty() && lazy.getOrDefault(pq.peek(), 0) > 0) {
// lazy.merge(pq.poll(), -1, Integer::sum);
// }
// ans[i] = pq.isEmpty() ? 0 : pq.peek();
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #3092: Most Frequent IDs
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3092: Most Frequent IDs
// class Solution {
// public long[] mostFrequentIDs(int[] nums, int[] freq) {
// Map<Integer, Long> cnt = new HashMap<>();
// Map<Long, Integer> lazy = new HashMap<>();
// int n = nums.length;
// long[] ans = new long[n];
// PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
// for (int i = 0; i < n; ++i) {
// int x = nums[i], f = freq[i];
// lazy.merge(cnt.getOrDefault(x, 0L), 1, Integer::sum);
// cnt.merge(x, (long) f, Long::sum);
// pq.add(cnt.get(x));
// while (!pq.isEmpty() && lazy.getOrDefault(pq.peek(), 0) > 0) {
// lazy.merge(pq.poll(), -1, Integer::sum);
// }
// ans[i] = pq.isEmpty() ? 0 : pq.peek();
// }
// return ans;
// }
// }
Use this to step through a reusable interview workflow for this problem.
For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.
Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.