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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
class Solution {
public:
int mostFrequentEven(vector<int>& nums) {
map<int, int> mp;
for(auto n: nums) mp[n]++;
int ans = -1, mx = -1;
for(auto m: mp){
if(m.first%2 == 0 && m.second > mx){
mx = m.second;
ans = m.first;
}
}
return ans;
}
};
class Solution {
public int mostFrequentEven(int[] nums) {
Map<Integer,Integer> map=new HashMap<>();
int max=-1;
int res=Integer.MAX_VALUE;
for(int i:nums){
if(i%2 == 0){ //Only even element
map.put(i,map.getOrDefault(i,0)+1);
if(map.get(i)>max){ //Check if greater than Max Val
max=Math.max(max,map.get(i));
res=i;
}
else if(map.get(i)==max && res>i){ //Check if equals to Max Val and element is less than current res
res=i;
}
}
}
return res==Integer.MAX_VALUE? -1: res;
}
}
class Solution(object):
def mostFrequentEven(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums = [n for n in nums if n % 2 == 0]
nums.sort()
return max(nums,key=nums.count) if len(nums) > 0 else -1
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