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2404. Most Frequent Even Element

Description

Given an integer array nums, return the most frequent even element.

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

 

Example 1:

Input: nums = [0,1,2,2,4,4,1]
Output: 2
Explanation:
The even elements are 0, 2, and 4. Of these, 2 and 4 appear the most.
We return the smallest one, which is 2.

Example 2:

Input: nums = [4,4,4,9,2,4]
Output: 4
Explanation: 4 is the even element appears the most.

Example 3:

Input: nums = [29,47,21,41,13,37,25,7]
Output: -1
Explanation: There is no even element.

 

Constraints:

  • 1 <= nums.length <= 2000
  • 0 <= nums[i] <= 105

Solutions

Solution 1: Hash Table

We use a hash table $cnt$ to count the occurrence of all even elements, and then find the even element with the highest occurrence and the smallest value.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

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class Solution:
    def mostFrequentEven(self, nums: List[int]) -> int:
        cnt = Counter(x for x in nums if x % 2 == 0)
        ans, mx = -1, 0
        for x, v in cnt.items():
            if v > mx or (v == mx and ans > x):
                ans, mx = x, v
        return ans
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class Solution {
    public int mostFrequentEven(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : nums) {
            if (x % 2 == 0) {
                cnt.merge(x, 1, Integer::sum);
            }
        }
        int ans = -1, mx = 0;
        for (var e : cnt.entrySet()) {
            int x = e.getKey(), v = e.getValue();
            if (mx < v || (mx == v && ans > x)) {
                ans = x;
                mx = v;
            }
        }
        return ans;
    }
}
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class Solution {
public:
    int mostFrequentEven(vector<int>& nums) {
        unordered_map<int, int> cnt;
        for (int x : nums) {
            if (x % 2 == 0) {
                ++cnt[x];
            }
        }
        int ans = -1, mx = 0;
        for (auto& [x, v] : cnt) {
            if (mx < v || (mx == v && ans > x)) {
                ans = x;
                mx = v;
            }
        }
        return ans;
    }
};
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func mostFrequentEven(nums []int) int {
    cnt := map[int]int{}
    for _, x := range nums {
        if x%2 == 0 {
            cnt[x]++
        }
    }
    ans, mx := -1, 0
    for x, v := range cnt {
        if mx < v || (mx == v && x < ans) {
            ans, mx = x, v
        }
    }
    return ans
}
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function mostFrequentEven(nums: number[]): number {
    const cnt: Map<number, number> = new Map();
    for (const x of nums) {
        if (x % 2 === 0) {
            cnt.set(x, (cnt.get(x) ?? 0) + 1);
        }
    }
    let ans = -1;
    let mx = 0;
    for (const [x, v] of cnt) {
        if (mx < v || (mx === v && ans > x)) {
            ans = x;
            mx = v;
        }
    }
    return ans;
}
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