1742. Maximum Number of Balls in a Box

/*
 * @lc app=leetcode id=1742 lang=cpp
 *
 * [1742] Maximum Number of Balls in a Box
 *
 * https://leetcode.com/problems/maximum-number-of-balls-in-a-box/description/
 *
 * algorithms
 * Easy (73.70%)
 * Likes:    636
 * Dislikes: 167
 * Total Accepted:    75.7K
 * Total Submissions: 102K
 * Testcase Example:  '1\n10'
 *
 * 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 <= 10^5
 *
 *
 */

// @lc code=start
class Solution {
public:
    int countBalls(int lowLimit, int highLimit) {
        unordered_map<int, int> m;
        int res = 0;
        for (int i = lowLimit; i <= highLimit; i++) {
            m[digitsSum(i)]++;
        }
        for (const auto& [k, v] : m) {
            res = max(res, v);
        }
        return res;
    }
    int digitsSum(int n) {
        int sum = 0;
        while (n) {
            sum += n % 10;
            n /= 10;
        }
        return sum;
    }
};
// @lc code=end