Problem Description

You are given an integer array nums and two integers k and numOperations. You need to perform an operation numOperations times on nums, where in each operation you:

• Select an index i that was not selected in any previous operations.
• Add an integer in the range [-k, k] to nums[i].

The task is to return the maximum possible frequency of any element in nums after performing the operations.

Intuition

The problem revolves around maximizing the frequency of an integer in the array nums through controlled modifications. Given the constraints, the idea is to adjust the values in nums strategically by incrementing or decrementing them within the allowable range [−k, k] to form duplicates and increase the frequency of a specific value.

1. Identify Potential Targets: Consider each unique number in nums as a potential target value whose frequency could be maximized.

2. Difference Array Technique: Use the difference array technique where adjustments in the frequency count for potential values are computed. For each element x in nums, calculate the impact of the operation range [x − k, x + k] by incrementing the position x - k and decrementing at x + k + 1.

3. Accumulating Changes: By sorting these impact points and accumulating the frequency changes, determine the maximum frequency achievable for any element. Adjust the count considering the operations allowed (numOperations), to find the highest frequency you can reach.

This approach efficiently evaluates how operations can be distributed to increase the frequency of one of the elements to the maximum possible degree, adhering to the constraints of the problem.

Learn more about Binary Search, Prefix Sum, Sorting and Sliding Window patterns.

Solution Approach

The solution employs a combination of a frequency counter and a difference array to optimize the operations across the nums array. The implementation process is as follows:

1. Initialize Data Structures:

   • Use defaultdict from the collections module to maintain counts of each number in nums (cnt).
   • Another defaultdict is used to track the frequency changes caused by the range operations (d).

2. Populate the Difference Array:

   • Iterate over each number x in nums.
   • Increase the count for x in the frequency counter cnt to record its original occurrences.
   • Adjust the difference array d to incorporate the effect of adding numbers in the range [x−k, x+k]. Specifically:
     • Increment d[x − k] to indicate that from this point, we will start adding potential frequency.
     • Decrement d[x + k + 1] to indicate that beyond this point, the addition effect ceases.

3. Calculate Maximum Frequency:

   • Sort the difference array d by keys and iterate through it.
   • Use a variable s to accumulate ongoing frequency changes as derived from d.
   • For each point in this sorted list, update the running total s and compare to find the maximum frequency any number can achieve, taking into account the number of operations allowed (numOperations).
   • ans keeps track of the maximum frequency encountered.

4. Return Result:

   • The algorithm concludes by returning the highest frequency (ans) that can be achieved after utilizing all possible operations efficiently.

This method effectively exploits the concept of treating range updates through a difference array pattern to maximize the occurrence of any given element, while smartly managing operations within their constraints.

Example Walkthrough

Let's take a small example to illustrate the solution approach:

Suppose we have nums = [1, 2, 2, 3], k = 1, and numOperations = 2. Our goal is to maximize the frequency of any integer in nums by adding a number in the range [-1, 1] to any two indices.

1. Initialize Data Structures:

   • Start with a frequency counter cnt that tells us how many times each number currently appears:

     cnt = {1: 1, 2: 2, 3: 1}

   • Initialize a difference array d to record range effects:

     d = {}

2. Populate the Difference Array:

   • Traverse each element x in nums:

     • x = 1:
       Increment d[0] (start of range for 1-1 to 1+1), decrement d[2] (end of range).

       d = {0: 1, 2: -1}

     • x = 2:
       Increment d[1] (start of range for 2-1 to 2+1), decrement d[3] (end of range).

       d = {0: 1, 1: 1, 2: -1, 3: -1}

     • x = 3:
       Increment d[2] (start of range for 3-1 to 3+1), decrement d[4] (end of range).

       d = {0: 1, 1: 1, 2: 0, 3: -1, 4: -1}

3. Calculate Maximum Frequency:

   • Sort d by keys and accumulate changes. Adjust with original counts from cnt:

     • s = 0, max_count = 0
     • At position 0: s = 1, check if 3 (count of 1+2 operations) > current max_count
     • At position 1: s = 2, possible frequency = 2 + (2) = 4
     • At position 2: s = 2
     • At position 3: s = 1, not better than 4

   • The highest computed frequency is 4 when expanding around 2.

4. Return Result:

   • The maximum achievable frequency after operations is 4. This can be done by:
     • Changing two 3s to 2s in the allowed operation range.

The solution strategically utilizes the difference array to effectively simulate range modifications, efficiently influencing frequency maximization across the array with given constraints.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(N log N), where N is the number of elements in the nums list. This comes primarily from the sorting operation on the dictionary d items, which contains values derived from the list nums.

The space complexity is O(N) due to the use of the cnt and d dictionaries that store counts and modified counts for elements in nums.

Learn more about how to find time and space complexity quickly.