Maximum Frequency After Subarray Operation Interview Question for Amazon

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:

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

After adding -5 to nums[2..5], 1 has a frequency of 2 in [1, 2, -2, -1, 0, 1]. Thus the output should be 2.

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

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

Constraints:

1 <= n == nums.length <= 10^5
1 <= nums[i] <= 50
1 <= k <= 50

Naive Solution

The naive solution iterates through all possible subarrays and all possible values of x. For each subarray, it adds x to all elements in the subarray and calculates the frequency of k. The maximum frequency is then returned.

Code

def max_frequency_naive(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 l in range(i, j + 1):
                    temp_nums[l] += x
                
                freq = temp_nums.count(k)
                max_freq = max(max_freq, freq)
    return max_freq

Time Complexity

O(n^3 * range_of_x). We iterate through all possible subarrays (O(n^2)), then for each subarray we iterate through all possible values of x. Then, we iterate through the array to get the frequency of k which is O(n). Thus, the time complexity is O(n^2 * range_of_x * n) == O(n^3 * range_of_x).

Space Complexity

O(n), due to the creation of temp_nums.

Optimal Solution

A more optimal solution involves iterating through all possible subarrays and calculating the x needed to transform an element within that subarray into k. Then, by using another loop, we can determine the overall frequency after applying x on the subarray. This avoids needing to test many x values.

Code

def max_frequency_optimal(nums, k):
    n = len(nums)
    max_freq = 0
    for i in range(n):
        for j in range(i, n):
            # Iterate through all subarrays
            temp_nums = nums[:]
            
            # Iterate through the elements of the subarray
            for l in range(i, j + 1):
                x = k - nums[l] # find the x that would transform nums[l] to k 
                temp_nums = nums[:]

                # Update the subarray with x
                for m in range(i, j + 1):
                  temp_nums[m] += x

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

Time Complexity

O(n^3). We iterate through all possible subarrays (O(n^2)). Then, for each subarray, we iterate through it again (O(n)), calculate the frequency and update the max. So O(n^2 * n) = O(n^3).

Space Complexity

O(n) to store temp_nums.

Edge Cases

Empty array: The problem states that n >= 1, so we don't need to worry about an empty array.
k not in array: The algorithm will still work. The maximum frequency will be the initial frequency of k in nums before any operations.
Large value of x: In the naive solution, the code iterates through a range of x from -50 to 50, as nums[i] and k are within the range of 1 to 50. The optimal solution does not require an explicit x.