Question

1742. Maximum Number of Balls in a Box

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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 $m = highLimit$.

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)

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;
  }
};

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);
}