Maximum Frequency of an Element After Performing Operations II

Maximum Frequency of an Element After Performing Operations II

1. What is this problem about?

This is the harder version of the previous problem, usually with much larger constraints on the array size and the values of k. You still want to maximize the frequency of an element after performing at most num_operations by changing values within a range [nums[i] - k, nums[i] + k].

2. Why is this asked in interviews?

This Hard-level question from Amazon and Google tests advanced data structure knowledge or highly optimized sliding window techniques. It requires the candidate to handle a large number of 'events' (the start and end of the reachable ranges for each number) and potentially use a Difference Array or Coordinate Compression to manage the overlaps efficiently.

3. Algorithmic pattern used

The primary pattern is the Difference Array (or Sweep Line) technique combined with Coordinate Compression.

1. For each nums[i], the range of values that can be changed into nums[i] is [nums[i] - k, nums[i] + k].
2. Alternatively, for any target V, the elements that can become V are those in [V - k, V + k].
3. We can use a sweep-line approach to find the maximum number of overlapping ranges. For each nums[i], we also need to know its original frequency to correctly apply the num_operations limit.

4. Example explanation

Nums: [1, 10, 10, 10, 20], k = 5, ops = 2.

• For target 10:
  • 1 cannot reach 10 (1+5=6).
  • 10 is 10 (Freq = 3).
  • 20 cannot reach 10 (20-5=15).
  • Count in range = 3. Ops used = 0. Final = 3.
• For target 15:
  • 10 can reach 15 (10+5=15).
  • 20 can reach 15 (20-5=15).
  • Range [10, 20] includes 4 elements (10, 10, 10, 20).
  • We have 2 ops. Target 15 original frequency = 0.
  • Result = min(4, 0 + 2) = 2. Max frequency is 3.

5. Common mistakes candidates make

Failing to use coordinate compression or a sweep-line algorithm will lead to TLE when the values are large. Another mistake is incorrectly handling the 'original frequency' when the target value is not one of the values already in the array.

6. Interview preparation tip

When you see a large range of possible values but only a few 'points' that matter (the array elements), Coordinate Compression is your best friend. It allows you to map large, sparse values into a small, dense range of indices.