How do you solve Most Frequent Number Following Key In an Array on LeetCode?

Most Frequent Number Following Key In an Array (LeetCode #2190) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient counting of occurrences.

What is the time complexity of Most Frequent Number Following Key In an Array?

The optimal solution for Most Frequent Number Following Key In an Array runs in O(n) time and uses O(n) space. The complexity is O(n) because we traverse the array only once and use a hash map to store counts, which is efficient for lookups and insertions.

What is the brute force solution for Most Frequent Number Following Key In an Array?

The brute force approach for Most Frequent Number Following Key In an Array is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every occurrence of the key in the array and then count how many times each number follows it. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Most Frequent Number Following Key In an Array?

Most Frequent Number Following Key In an Array uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Most Frequent Number Following Key In an Array?

Think about edge cases, such as when the key is at the start or end of the array. Communicate your thought process clearly as you write code. Always consider the time and space complexity of your solution.

What are common mistakes when solving Most Frequent Number Following Key In an Array?

Not checking the bounds when accessing the next element. Forgetting to initialize the count map for new targets.

#2190

Most Frequent Number Following Key In an Array

Easy

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

We can achieve better performance by using a single pass through the array and a hash map to count occurrences, which reduces the time complexity significantly.

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Algorithm

3 steps

1Step 1: Initialize a hash map to count occurrences of each number following the key.
2Step 2: Loop through the array only once, checking for the key and updating counts for the next number.
3Step 3: After the loop, find the number with the maximum count in the hash map.

solution.py9 lines

1# Full working Python code
2
3def most_frequent(nums, key):
4    count_map = {}
5    for i in range(len(nums) - 1):
6        if nums[i] == key:
7            next_num = nums[i + 1]
8            count_map[next_num] = count_map.get(next_num, 0) + 1
9    return max(count_map, key=count_map.get)

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Complexity note: The complexity is O(n) because we traverse the array only once and use a hash map to store counts, which is efficient for lookups and insertions.

1Using a hash map allows for efficient counting of occurrences.
2A single pass through the array is sufficient to gather all necessary data.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.