most-frequent-number-following-key-in-an-array_2190.py

# You are given a 0-indexed integer array nums.
# You are also given an integer key, which is present in nums.

# For every unique integer target in nums, count the number of times target
# immediately follows an occurrence of key in nums. In other words, count the number of indices i such that:

# 0 <= i <= nums.length - 2,
# nums[i] == key and,
# nums[i + 1] == target.
# Return the target with the maximum count. The test cases will be generated such that the target with maximum count is unique.

# Example 1:

# Input: nums = [1,100,200,1,100], key = 1
# Output: 100
# Explanation: For target = 100, there are 2 occurrences at indices 1 and 4 which follow an occurrence of key.
# No other integers follow an occurrence of key, so we return 100.
# Example 2:

# Input: nums = [2,2,2,2,3], key = 2
# Output: 2
# Explanation: For target = 2, there are 3 occurrences at indices 1, 2, and 3 which follow an occurrence of key.
# For target = 3, there is only one occurrence at index 4 which follows an occurrence of key.
# target = 2 has the maximum number of occurrences following an occurrence of key, so we return 2.

# ---------------------------------------Runtime 90 ms Beats 78.86% Memory 16.5 MB Beats 69.67%---------------------------------------

from typing import List

# My solution
# Time Complexity O(n)


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