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 objective is to maximize the frequency of some integer in nums using at most numOperations, each selecting a unique index and modifying the element at that index by adding an integer between -k and k. The approach involves considering the potential of each element within the array to increase its frequency, while ensuring the number of operations is not exceeded.

The solution leverages a counting mechanism to track the frequency of each number, and a differential array to efficiently compute the impact of adding values within the permissible range. By calculating the maximum frequency those numbers can reach when increased optimally with available operations, we can determine the highest possible frequency in the modified array.

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

Solution Approach

The solution utilizes the concept of a frequency counter along with a differential array to determine the influence of operations on each element in nums. Here’s a breakdown of the approach:

1. Initialize Data Structures:

   • Use a defaultdict called cnt to store the initial frequency count of each number in nums.
   • Another defaultdict called d is used to maintain a differential array which helps in tracking how additions affect element frequency over its valid range [-k, k].

2. Populate Frequency and Differential Arrays:

   • Traverse each number x in nums.
   • Increase the count in the cnt dictionary for the number x to keep track of its initial frequency.
   • Use d to set up ranges influenced by additions:
     • d[x] remains unchanged and set as 0 initially.
     • Increment d[x - k] because adding any number greater than or equal to -k at position indexed by x can start affecting this frequency.
     • Decrement the position d[x + k + 1] to signify the end of the influence of the addition once it exceeds the range [x-k, x+k].

3. Calculate Maximum Frequency:

   • Initialize ans to 0 to store the result of the maximum frequency.
   • Traverse through the sorted keys of the differential array d, using a cumulative sum approach:
     • Add the current value of t (which represents the change) to s, representing the current frequency influenced by operations.
     • Compute the maximum of ans and min(s, cnt[x] + numOperations). This takes into account that the frequency can't exceed the sum of its initial count and the number of operations allowed.

4. Return the Result:

   • Upon completion, ans holds the maximum achievable frequency of any number in nums after performing the specified operations.

This approach efficiently determines the optimal way to increase element frequencies while adhering to the constraint of allowed operations.

Example Walkthrough

Let's walk through an example to illustrate the solution approach using the given problem description.

Example

Consider the array nums = [1, 2, 2, 3] with k = 1 and numOperations = 2.

Step-by-Step Solution

1. Initialize Data Structures:

   • Use a defaultdict named cnt to keep track of the initial frequencies:

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

   • Set up another defaultdict named d to keep track of the differential array:

     d = {}

2. Populate Frequency and Differential Arrays:

   • Traverse each x in nums:
     • For x = 1, increment the positions:

       d[0] += 1  ->  d = {0: 1}
       d[2] -= 1  ->  d = {0: 1, 2: -1}

     • For x = 2, increment and decrement the positions:

       d[1] += 1  ->  d = {0: 1, 2: -1, 1: 1}
       d[3] -= 1  ->  d = {0: 1, 2: -1, 1: 1, 3: -1}

     • For x = 3, update positions similarly:

       d[2] += 1  ->  d = {0: 1, 2: 0, 1: 1, 3: -1}
       d[4] -= 1  ->  d = {0: 1, 2: 0, 1: 1, 3: -1, 4: -1}

3. Calculate Maximum Frequency:

   • Initialize ans to 0 for storing the maximum frequency.
   • Traverse sorted keys in d and compute cumulative sums:

     s = 0  (cumulative influence)

     • For key 0:

       t = 1 -> s = 1
       max frequency = max(0, min(1, 1 + 2)) = 1

     • For key 1:

       t = 1 -> s = 2
       max frequency = max(1, min(2, 2 + 2)) = 2

     • For key 2:

       t = 0 -> s = 2  (no change)
       max frequency = max(2, min(2, 2 + 2)) = 2

     • For key 3:

       t = -1 -> s = 1
       max frequency = max(2, min(1, 1 + 2)) = 2

     • Finally, s influences no further, adjusting s with subsequent keys.

4. Return the Result:

   • The maximum achievable frequency is 2, which can be a frequent number in the modified array after performing up to numOperations times.

This example highlights using differential arrays and frequency counts to determine the most effective way to increase element frequencies within an integer array using specified constraints.

Solution Implementation

Time and Space Complexity

The maxFrequency function consists of several key parts that determine its computational complexity:

1. Initialization and Dictionary Population:

   • Populating the dictionaries cnt and d involves iterating over the nums list, which takes O(n) time where n is the length of nums.

2. Sorting the Dictionary Items:

   • Sorting the items of dictionary d takes O(m log m) time, where m is the number of unique keys in d.

3. Processing the Sorted Items:

   • Iterating through the sorted items of d takes O(m) time.

Overall, the time complexity becomes O(n + m log m).

Space Complexity Analysis

• The dictionaries cnt and d each store up to n entries in the worst case. Therefore, the space complexity is O(n).

Thus, the overall space complexity is O(n) because of the usage of the dictionaries.

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