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 = 1Output:100Explanation: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 = 2Output:2Explanation: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.

**Constraints:**

`2 <= nums.length <= 1000`

`1 <= nums[i] <= 1000`

- The test cases will be generated such that the answer is unique.

**Solution: Hashtable**

Time complexity: O(n)

Space complexity: O(n)

## C++

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// Author: Huahua class Solution { public: int mostFrequent(vector<int>& nums, int key) { const int n = nums.size(); unordered_map<int, int> m; int count = 0; int ans = 0; for (int i = 1; i < n; ++i) { if (nums[i - 1] == key && ++m[nums[i]] > count) { count = m[nums[i]]; ans = nums[i]; } } return ans; } }; |