README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3000-3099/3005.Count%20Elements%20With%20Maximum%20Frequency/README_EN.md	1216	Weekly Contest 380 Q1	Array Hash Table Counting

3005. Count Elements With Maximum Frequency

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Description

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.

 

Constraints:

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

Solutions

Solution 1: Counting

We can use a hash table or array $cnt$ to record the occurrence of each element.

Then we traverse $cnt$ to find the element with the most occurrences, and let its occurrence be $mx$. We sum up the occurrences of elements that appear $mx$ times, which is the answer.

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

Python3

class Solution:
    def maxFrequencyElements(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        mx = max(cnt.values())
        return sum(x for x in cnt.values() if x == mx)

Java

class Solution {
    public int maxFrequencyElements(int[] nums) {
        int[] cnt = new int[101];
        for (int x : nums) {
            ++cnt[x];
        }
        int ans = 0, mx = -1;
        for (int x : cnt) {
            if (mx < x) {
                mx = x;
                ans = x;
            } else if (mx == x) {
                ans += x;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxFrequencyElements(vector<int>& nums) {
        int cnt[101]{};
        for (int x : nums) {
            ++cnt[x];
        }
        int ans = 0, mx = -1;
        for (int x : cnt) {
            if (mx < x) {
                mx = x;
                ans = x;
            } else if (mx == x) {
                ans += x;
            }
        }
        return ans;
    }
};

Go

func maxFrequencyElements(nums []int) (ans int) {
	cnt := [101]int{}
	for _, x := range nums {
		cnt[x]++
	}
	mx := -1
	for _, x := range cnt {
		if mx < x {
			mx, ans = x, x
		} else if mx == x {
			ans += x
		}
	}
	return
}

TypeScript

function maxFrequencyElements(nums: number[]): number {
    const cnt: number[] = Array(101).fill(0);
    for (const x of nums) {
        ++cnt[x];
    }
    let [ans, mx] = [0, -1];
    for (const x of cnt) {
        if (mx < x) {
            mx = x;
            ans = x;
        } else if (mx === x) {
            ans += x;
        }
    }
    return ans;
}