1399. Count Largest Group

Q:

Given an integer n. Each number from 1 to n is grouped according to the sum of its digits. 

Return how many groups have the largest size.

 

Example 1:

Input: n = 13
Output: 4
Explanation: There are 9 groups in total, they are grouped according sum of its digits of numbers from 1 to 13:
[1,10], [2,11], [3,12], [4,13], [5], [6], [7], [8], [9]. There are 4 groups with largest size.

Example 2:

Input: n = 2
Output: 2
Explanation: There are 2 groups [1], [2] of size 1.

Example 3:

Input: n = 15
Output: 6

Example 4:

Input: n = 24
Output: 5

 

Constraints:

• 1 <= n <= 10^4

A:

class Solution {
public:
    int countLargestGroup(int n) {
        unordered_map<int, vector<int>> M;
        int largestSize = 0;
        for(int i = 1; i<=n;++i)
            largestSize = max(largestSize, helper(M,i));
        // get how many groups with largestSize
        auto iter = M.begin();
        int res = 0;
        while(iter!= M.end()){
            if(iter->second.size()==largestSize){
                res++;
            }
            iter++;
        }        
        return res;
    }
private:
    int helper(unordered_map<int, vector<int>> & M, int i){
        string str = to_string(i);
        int sum = 0;
        for(char ch : str){
            sum += ch -'0';
        }
        if(M.find(sum)==M.end()){
            vector<int> tmp;
            M[sum] = tmp;
        }
        M[sum].push_back(i);
        return M[sum].size();
    }
};

---------------------修改后，我们发现，不需要存储 所有的i.值， 只需要记录有多少个就行

class Solution {
public:
    int countLargestGroup(int n) {
        unordered_map<int, int> M;
        int largestSize = 0;
        for(int i = 1; i<=n;++i)
            largestSize = max(largestSize, helper(M,i));
        // get how many groups with largestSize
        auto iter = M.begin();
        int res = 0;
        while(iter!= M.end()){
            if(iter->second==largestSize){
                res++;
            }
            iter++;
        }        
        return res;
    }
private:
    int helper(unordered_map<int, int> & M, int i){
        string str = to_string(i);
        int sum = 0;
        for(char ch : str){
            sum += ch -'0';
        }
        M[sum]++;
        return M[sum];
    }
};