Maximum Frequency After Subarray Operation

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.
• Select an integer x and add x to all the elements in nums[i..j].

Find the maximum frequency of the value k after the operation.

For example: Input: nums = [1,2,3,4,5,6], k = 1 Output: 2 Explanation: After adding -5 to nums[2..5], 1 has a frequency of 2 in [1, 2, -2, -1, 0, 1].

Input: nums = [10,2,3,4,5,5,4,3,2,2], k = 10 Output: 4 Explanation: After adding 8 to nums[1..9], 10 has a frequency of 4 in [10, 10, 11, 12, 13, 13, 12, 11, 10, 10].

def max_frequency(nums, k):
    n = len(nums)
    max_freq = 0

    for i in range(n):
        for j in range(i, n):
            for x in range(-50, 51):  # Iterate through possible values of x
                temp_nums = nums[:]
                for idx in range(i, j + 1):
                    temp_nums[idx] += x

                freq = temp_nums.count(k)
                max_freq = max(max_freq, freq)

    return max_freq


# Example usage
nums1 = [1, 2, 3, 4, 5, 6]
k1 = 1
print(f"Input: nums = {nums1}, k = {k1}\nOutput: {max_frequency(nums1, k1)}")  # Output: 2

nums2 = [10, 2, 3, 4, 5, 5, 4, 3, 2, 2]
k2 = 10
print(f"\nInput: nums = {nums2}, k = {k2}\nOutput: {max_frequency(nums2, k2)}")  # Output: 4

Naive Approach:

The most straightforward approach involves iterating through all possible subarrays, trying different values of x to add to the subarray, and then counting the frequency of k. We keep track of the maximum frequency found.

Optimal Approach:

To optimize this problem, we can use dynamic programming or memoization to store results of subarray modifications to avoid redundant calculations. However, given the constraints (especially the small range of possible values for array elements and k), the optimization might not significantly improve performance for smaller input sizes. The key is to efficiently count the occurrences of k after adding x to the subarray. A more efficient approach might involve using a prefix sum-like technique to track counts or modifications, but for the given constraints, the improvements might be marginal.

Big(O) Run-Time Analysis:

Naive Approach:

• The outer loops iterate through all possible subarrays, which is O(n^2).
• The inner loop iterates through possible values of x which is O(101) as the range of x is fixed (-50 to 50).
• Counting frequency of k takes O(n).
• Overall complexity is O(n^2 * 101 * n) which simplifies to O(n^3).

Big(O) Space Usage Analysis:

• The space complexity is O(n) due to the creation of temp_nums in each iteration.

Edge Cases:

1. Empty array: The code should handle the edge case where the input array nums is empty. In the provided code, if nums is empty, the loops will not execute and it will return 0, which is appropriate.
2. k not in array: If the target value k is not present in the array, the algorithm should still function correctly and return the maximum frequency achievable after the operation. The provided code correctly handles this.
3. Large array: Although the constraint mentions the length of nums is at most 10^5, the algorithm's time complexity O(n^3) may still be too slow and might lead to a timeout for larger test cases. Optimization could involve different approaches like prefix sums.
4. Large x values: If the range of x values to iterate through is very large, the algorithm may become slow. The problem constraints limit the values in nums and k to between 1 and 50, which implicitly restricts the range of potentially relevant x values. The code as written iterates through a fixed range of x from -50 to 50. More intelligently, you can derive the lower and upper bound for x based on values in nums and k to reduce unnecessary loop iterations.