## DEV Community

Abhishek Chaudhary

Posted on

# Most Frequent Number Following Key In an Array

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.

Constraints:

• `2 <= nums.length <= 1000`
• `1 <= nums[i] <= 1000`
• The test cases will be generated such that the answer is unique.

SOLUTION:

``````class Solution:
def mostFrequent(self, nums: List[int], key: int) -> int:
n = len(nums)
ctr = {}
for i in range(n):
if nums[i] == key:
if i < n - 1:
ctr[nums[i + 1]] = ctr.get(nums[i + 1], 0) + 1
return max((v, k) for k, v in ctr.items())[1]
``````