Problem URL: Most frequent number following key in an array

We will use a hashmap to store the frequency of each number. Then we will traverse the array and check if the current number is equal to the key. If it is, then we will increment the frequency of the next number. We will return the number with the maximum frequency.

class Solution:
    def mostFrequentNumber(self, nums: List[int], key: int) -> int:
        freq = {}
        for i in range(len(nums) - 1):
            if nums[i] == key:
                freq[nums[i + 1]] = freq.get(nums[i + 1], 0) + 1
        return max(freq, key=freq.get)

Time complexity: O(n)
Space complexity: O(n)