algoadvance

You are given an integer array nums. Find the most frequent even element.

1. If there is a tie, return the smallest one.
2. If there is no such element, return -1.

Example:

• Input: nums = [0,1,2,2,4,4,1]
• Output: 2

Clarifying Questions

1. Input Constraints:
   • What is the length range of the array nums?
     • The length of nums can be from 1 to (10^5).
   • What are the value ranges for elements in nums?
     • Elements in nums can be from 0 to (10^5).
2. Output Specifics:
   • What should be the output if there are no even numbers in the array?
     • If there are no even numbers in the array, return -1.

Strategy

1. Filter Out Even Numbers:
   • Loop through the array nums and identify all even numbers.
2. Count Frequencies:
   • Use a dictionary to maintain a count of all even numbers.
3. Determine Most Frequent and Smallest:
   • Traverse the dictionary to find the most frequent even number. In case of a tie in frequency, choose the smaller number.
4. Edge Cases:
   • If no even numbers are found, return -1.

Code

Here is the Python code implementation based on our strategy:

def most_frequent_even(nums):
    # Dictionary to store the frequency of even elements
    freq = {}
    
    # Iterate through nums and count frequencies of even numbers
    for num in nums:
        if num % 2 == 0:
            if num in freq:
                freq[num] += 1
            else:
                freq[num] = 1
    
    # No even numbers case
    if not freq:
        return -1
    
    # Determine the most frequent even number, 
    # If there's a tie, return the smallest even number.
    most_frequent = -1
    max_freq = 0
    
    for num, count in freq.items():
        if count > max_freq or (count == max_freq and num < most_frequent):
            max_freq = count
            most_frequent = num
    
    return most_frequent
    
# Example usage:
nums = [0,1,2,2,4,4,1]
print(most_frequent_even(nums))  # Output: 2

Time Complexity

• Filtering and Counting: O(n), where n is the length of the array nums.
• Finding Most Frequent: O(k), where k is the number of unique even numbers (since we are simply iterating through the dictionary entries).

Overall time complexity is O(n), which is efficient given the constraints.

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