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
Solution: Hashtable and base-10
Max sum will be 9+9+9+9+9 = 45
Time complexity: O((hi-lo) * log(hi))
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int countBalls(int lowLimit, int highLimit) { vector<int> balls(46); int ans = 0; for (int i = lowLimit; i <= highLimit; ++i) { int n = i; int box = 0; while (n) { box += n % 10; n /= 10; } ans = max(ans, ++balls[box]); } return ans; } }; |
Python3
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# Author: Huahua class Solution: def countBalls(self, lowLimit: int, highLimit: int) -> int: balls = defaultdict(int) ans = 0 for x in range(lowLimit, highLimit + 1): s = sum(int(d) for d in str(x)) balls[s] += 1 ans = max(ans, balls[s]) return ans |
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