240308 LeetCode 문제 풀이

3005. Count Elements With Maximum Frequency

https://leetcode.com/problems/count-elements-with-maximum-frequency/description/?envType=daily-question&envId=2024-03-08

 

1) 문제 설명

You are given an array nums consisting of positive integers.

Return the total frequencies of elements in nums such that those elements all have the maximum frequency.

The frequency of an element is the number of occurrences of that element in the array.

 

Example 1:

Input: nums = [1,2,2,3,1,4]
Output: 4
Explanation: The elements 1 and 2 have a frequency of 2 which is the maximum frequency in the array.
So the number of elements in the array with maximum frequency is 4.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 5
Explanation: All elements of the array have a frequency of 1 which is the maximum.
So the number of elements in the array with maximum frequency is 5.

 

2) 제한 사항

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

 

3) 도전 과제

X

 

4) 풀이

배열을 한번만 순회하고 싶었기에 Map과 새로운 배열을 이용하여 풀었다.

 

Map으로 각 값의 빈도수를 기록하며 최대 빈도수를 같이 기록하였다. 새 배열에는 빈도수의 갯수를 기록하였다.

 

5) 소스 코드 및 결과

import java.util.HashMap;

class Solution {
    public int maxFrequencyElements(int[] nums) {
        HashMap<Integer, Integer> map = new HashMap<>();

        int[] array = new int[nums.length];
        int max = 0;

        for(int i : nums){
            if(map.get(i)==null){
                map.put(i,1);
                array[0]++;
                max = Math.max(max,1);
                continue;
            };

            Integer count = map.get(i);
            map.put(i, count+1);
            max = Math.max(max, count+1);
            array[count]++;
        }

        return max * array[max-1];
    }
}

6) 다른 사람의 풀이

class Solution {
    public int maxFrequencyElements(int[] nums) {
        int[] freq = new int[101];

        for(int i = 0; i < nums.length; i++){
            freq[nums[i]]++;
        }

        int temp = 0;
        int c = 0;
        for(int i = 1; i < freq.length; i++){
            if(freq[i] == temp){
                c += freq[i];
            }
            else if(freq[i] > temp){
                temp = freq[i];
                c = freq[i];
            }
        }
        return c;
    }
}