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Maximum Frequency After Subarray Operation

Difficulty: Medium

Problem Description

You are given an array nums of length n. You are also given an integer k. You perform the following operation on nums once: Select a subarray nums[i..j] where 0 <= i <= j <= n - 1 and select an integer x to add x to all the elements in nums[i..j]. Find the maximum frequency of the value k after the operation.

Key Insights

  • The goal is to maximize the frequency of the integer k in the modified array.
  • The operation allows modifying a contiguous subarray, meaning we can increase or decrease certain values to match k.
  • The difference between k and each element in nums can help determine how to adjust the values to maximize occurrences of k.
  • A frequency count can be maintained using a hashmap or array to track how many times each number appears after operations.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1) (since the range of values is limited to 1-50)

Solution

To solve the problem, we can use a sliding window approach combined with a frequency count. First, we will calculate the difference between k and each element in nums. Then, we will iterate through the array to maintain a window where we can keep track of how many adjustments we need to make in order to make those elements equal to k. The size of the window will expand until we exceed the number of allowable adjustments, at which point we will contract the window from the left. The maximum size of the window during this process will give us the maximum frequency of k.

Code Solutions

def maxFrequency(nums, k):
    # To store the frequency of elements
    count = [0] * 51
    max_freq = 0
    left = 0
    
    for right in range(len(nums)):
        # Increase the count of current number
        count[nums[right]] += 1
        
        # Calculate needed adjustments to turn nums[left..right] into k
        while (right - left + 1) - count[k] > 0:
            # Shrink the window from the left
            count[nums[left]] -= 1
            left += 1
            
        # Update the maximum frequency found
        max_freq = max(max_freq, right - left + 1)
    
    return max_freq
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