Maximum Frequency After Subarray Operation

To solve this problem, focus on scanning the array and using a hash map to track frequencies while simulating subarray addition. Fix each element as a candidate to convert to k and compute how many elements can match it within one operation. This approach ensures we capture the maximum frequency after one subarray modification without unnecessary brute-force enumeration.

Problem Statement

Given an integer array nums of length n and an integer k, perform a single operation on any contiguous subarray: add an integer value to every element in that subarray. The goal is to determine the highest frequency of the number k in nums after performing this operation exactly once.

You must return the maximum possible frequency of k after the operation. Constraints include 1 <= n <= 105, 1 <= nums[i] <= 50, and 1 <= k <= 50. Example: for nums = [1,2,3,4,5,6] and k = 1, applying the optimal subarray addition yields a frequency of 2 for k.

Examples

Example 1

Input: nums = [1,2,3,4,5,6], k = 1

Output: 2

After adding -5 to nums[2..5] , 1 has a frequency of 2 in [1, 2, -2, -1, 0, 1] .

Example 2

Input: nums = [10,2,3,4,5,5,4,3,2,2], k = 10

Output: 4

After adding 8 to nums[1..9] , 10 has a frequency of 4 in [10, 10, 11, 12, 13, 13, 12, 11, 10, 10] .

Constraints

• 1 <= n == nums.length <= 105
• 1 <= nums[i] <= 50
• 1 <= k <= 50

Solution Approach

Use Array Scanning with Sliding Window

Scan the array while maintaining a running count of differences needed to convert subarray elements to k. Expand and shrink a sliding window to include as many elements as possible within the operation limit.

Track Frequencies Using Hash Map

Maintain a hash map of element counts to quickly evaluate which element, when targeted for conversion to k, maximizes frequency. Update counts dynamically as the window changes to reflect current potential frequency.

Greedy Selection of Candidate Elements

For each element considered as a candidate to become k, calculate the number of elements that can reach k using one subarray addition. Choose the candidate yielding the highest resulting frequency, ensuring optimal use of the operation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n log n) if sorting is needed per candidate element, or O(n) with an optimized sliding window approach. Space complexity is O(n) for maintaining frequency counts and auxiliary data structures.

What Interviewers Usually Probe

• Notice if you attempt full enumeration of all subarrays, it will exceed time limits.
• Check if candidate selection for conversion to k is handled efficiently using hash maps.
• Be aware of off-by-one errors in sliding window boundaries when computing frequency.

Common Pitfalls or Variants

Common pitfalls

• Trying to brute-force all subarrays instead of using array scanning plus hash lookup.
• Failing to update the frequency hash map correctly when the window changes.
• Overlooking elements already equal to k in calculating maximum frequency.

Follow-up variants

• Allow multiple subarray operations and track the resulting frequency of k.
• Change k to a target range instead of a single number and maximize elements in that range.
• Use a decrement operation instead of addition and find the maximum achievable frequency.

FAQ

What is the primary pattern used in Maximum Frequency After Subarray Operation?

The problem relies on array scanning plus hash lookup, often combined with a sliding window to track subarray effects efficiently.

Can multiple subarray operations be applied in this problem?

No, the problem restricts you to exactly one subarray addition operation, so the strategy focuses on maximizing frequency in that single step.

How do I handle elements already equal to k?

Include elements equal to k in your frequency count initially; they contribute to the maximum without needing conversion.

What is an efficient approach to simulate the operation?

Use a sliding window with prefix sums or running differences while updating a hash map to track potential conversions to k.

Why is brute-force subarray enumeration not recommended?

Because enumerating all subarrays is O(n^2) and exceeds time limits for large n; optimized scanning with hash maps avoids this inefficiency.