Formatted question description: https://leetcode.ca/all/2190.html

# 2190. Most Frequent Number Following Key In an Array (Easy)

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.

**Related Topics**:

Array, Hash Table, Counting

**Similar Questions**:

## Solution 1.

```
// OJ: https://leetcode.com/problems/most-frequent-number-following-key-in-an-array/
// Time: O(N)
// Space: O(R) where `R` is the range of numbers in `A`. We can also make it the count of unique numbers in `A` if we use `unordered_map`.
class Solution {
public:
int mostFrequent(vector<int>& A, int key) {
int N = A.size(), cnt[1001] = {}, mx = 0, ans = 0;
for (int i = 1; i < N; ++i) {
if (A[i - 1] != key) continue;
cnt[A[i]]++;
}
for (int i = 1; i <= 1000; ++i) {
if (cnt[i] > mx) mx = cnt[i], ans = i;
}
return ans;
}
};
```