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Formatted question description: https://leetcode.ca/all/1742.html
1742. Maximum Number of Balls in a Box
Level
Easy
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 <= 10^5
Solution
Since highLimit <= 10^5
, the sum of digits of each number is at most 46. Create an array sums
where sums[i]
is the number of balls that have sum of digits equal to i
. Loop over all numbers from lowLimit
to highLimit
. For each number, calculate the sum of digits and update the array. Finally, loop over the array and obtain the maximum count.
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class Solution { public int countBalls(int lowLimit, int highLimit) { int[] sums = new int[50]; for (int i = lowLimit; i <= highLimit; i++) { int sum = getSumOfDigits(i); sums[sum]++; } int maxCount = 0; for (int i = 0; i < 50; i++) maxCount = Math.max(maxCount, sums[i]); return maxCount; } public int getSumOfDigits(int num) { int sum = 0; while (num > 0) { sum += num % 10; num /= 10; } return sum; } } ############ 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(); } }
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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) ############ # 1742. Maximum Number of Balls in a Box # https://leetcode.com/problems/maximum-number-of-balls-in-a-box class Solution: def countBalls(self, lowLimit: int, highLimit: int) -> int: mp = collections.defaultdict(int) for i in range(lowLimit, highLimit+1): value = 0 while i >= 1: value += i%10 i //= 10 mp[value] += 1 return max(mp.values())
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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; } };
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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 }
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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); }