LeetCode 2404: Most Frequent Even Element
LeetCode 2404 Solution Explanation
Explanation
To solve this problem, we can iterate through the input array and count the frequency of each even number. We maintain a map to store the counts of each even number. After counting the frequencies, we iterate through the map to find the most frequent even number. If there is a tie, we return the smallest even number. If there are no even numbers in the array, we return -1.
- Time complexity: O(n) where n is the number of elements in the input array.
- Space complexity: O(n) to store the counts of each even number in the map.
LeetCode 2404 Solutions in Java, C++, Python
import java.util.HashMap;
import java.util.Map;
class Solution {
public int mostFrequentEvenElement(int[] nums) {
Map<Integer, Integer> countMap = new HashMap<>();
int maxFreq = 0;
int result = -1;
for (int num : nums) {
if (num % 2 == 0) {
countMap.put(num, countMap.getOrDefault(num, 0) + 1);
maxFreq = Math.max(maxFreq, countMap.get(num));
}
}
for (int num : countMap.keySet()) {
if (num % 2 == 0 && countMap.get(num) == maxFreq) {
if (result == -1 || num < result) {
result = num;
}
}
}
return result;
}
}Interactive Code Editor for LeetCode 2404
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