1742. Maximum Number of Balls in a Box

You are working in a ball factory where you have n balls numbered from lowLimit up to highLimit inclusive (i.e., n == highLimit - lowLimit + 1), and an infinite number of boxes numbered from 1 to infinity.

Your job at this factory is to put each ball in the box with a number equal to the sum of digits of the ball's number. For example, the ball number 321 will be put in the box number 3 + 2 + 1 = 6 and the ball number 10 will be put in the box number 1 + 0 = 1.

Given two integers lowLimit and highLimit, return the number of balls in the box with the most balls.

 

Example 1:

Input: lowLimit = 1, highLimit = 10
Output: 2
Explanation:
Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
Ball Count:  2 1 1 1 1 1 1 1 1 0  0  ...
Box 1 has the most number of balls with 2 balls.
Example 2:

Input: lowLimit = 5, highLimit = 15
Output: 2
Explanation:
Box Number:  1 2 3 4 5 6 7 8 9 10 11 ...
Ball Count:  1 1 1 1 2 2 1 1 1 0  0  ...
Boxes 5 and 6 have the most number of balls with 2 balls in each.
Example 3:

Input: lowLimit = 19, highLimit = 28
Output: 2
Explanation:
Box Number:  1 2 3 4 5 6 7 8 9 10 11 12 ...
Ball Count:  0 1 1 1 1 1 1 1 1 2  0  0  ...
Box 10 has the most number of balls with 2 balls.
 

Constraints:

1 <= lowLimit <= highLimit <= 105

按照题意模拟即可，有一个可以优化的地方是可以使用一个长度为46的数组模拟map，因为最大的和为99999,46即可存下。

class MaximumNumberOfBallsInABox : public Solution {
public:
    void Exec() {

    }
    int countBalls(int lowLimit, int highLimit) {
        int res = 0;
        vector<int> counter(46, 0);
        for (int i = lowLimit; i <= highLimit; i++) {
            int sum = 0, num = i;
            while (num != 0) {
                sum += num % 10;
                num /= 10;
            }
            counter[sum]++;
            res = std::max(res, counter[sum]);
        }
        return res;
    }
};