Leetcode #2190: Most Frequent Number Following Key In an Array

In this guide, we solve Leetcode #2190 Most Frequent Number Following Key In an Array in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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

Quick Facts

Difficulty: Easy
Premium: No
Tags: Array, Hash Table, Counting

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.
Iterate through the input, querying/updating the map.
Return the first valid result or the final computed value.

Example

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.

Python Solution

class Solution:
    def mostFrequent(self, nums: List[int], key: int) -> int:
        cnt = Counter()
        ans = mx = 0
        for a, b in pairwise(nums):
            if a == key:
                cnt[b] += 1
                if mx < cnt[b]:
                    mx = cnt[b]
                    ans = b
        return ans

Complexity

The time complexity is $O(n)$ , and the space complexity is $O(M)$ . The space complexity is $O(M)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

Initialize a hash map for seen items or counts.

Iterate through the input, querying/updating the map.

Return the first valid result or the final computed value.

Python Solution

class Solution:
    def mostFrequent(self, nums: List[int], key: int) -> int:
        cnt = Counter()
        ans = mx = 0
        for a, b in pairwise(nums):
            if a == key:
                cnt[b] += 1
                if mx < cnt[b]:
                    mx = cnt[b]
                    ans = b
        return ans

Leetcode #2190: Most Frequent Number Following Key In an Array

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2190: Most Frequent Number Following Key In an Array

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview