1742. Maximum Number of Balls in a Box
Description
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
Solutions
Solution 1: Array + Simulation
Observing the data range of the problem, the maximum number of the ball does not exceed $10^5$, so the maximum value of the sum of each digit of the number is less than $50$. Therefore, we can directly create an array $cnt$ with a length of $50$ to count the number of each digit sum of each number.
The answer is the maximum value in the array $cnt$.
The time complexity is $O(n \times \log_{10}m)$. Here, $n = highLimit - lowLimit + 1$, and $m = highLimit$.
1 2 3 4 5 6 7 8 9 10 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|