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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 $m = highLimit$.
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class Solution { public int countBalls(int lowLimit, int highLimit) { int[] cnt = new int[50]; for (int i = lowLimit; i <= highLimit; ++i) { int y = 0; for (int x = i; x > 0; x /= 10) { y += x % 10; } ++cnt[y]; } return Arrays.stream(cnt).max().getAsInt(); } } -
class Solution { public: int countBalls(int lowLimit, int highLimit) { int cnt[50] = {0}; int ans = 0; for (int i = lowLimit; i <= highLimit; ++i) { int y = 0; for (int x = i; x; x /= 10) { y += x % 10; } ans = max(ans, ++cnt[y]); } return ans; } }; -
class Solution: def countBalls(self, lowLimit: int, highLimit: int) -> int: cnt = [0] * 50 for x in range(lowLimit, highLimit + 1): y = 0 while x: y += x % 10 x //= 10 cnt[y] += 1 return max(cnt) -
func countBalls(lowLimit int, highLimit int) (ans int) { cnt := [50]int{} for i := lowLimit; i <= highLimit; i++ { y := 0 for x := i; x > 0; x /= 10 { y += x % 10 } cnt[y]++ if ans < cnt[y] { ans = cnt[y] } } return } -
function countBalls(lowLimit: number, highLimit: number): number { const cnt: number[] = Array(50).fill(0); for (let i = lowLimit; i <= highLimit; ++i) { let y = 0; for (let x = i; x; x = Math.floor(x / 10)) { y += x % 10; } ++cnt[y]; } return Math.max(...cnt); }