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
class Solution:
def countBalls(self, lowLimit: int, highLimit: int) -> int:
counter = [0] * 60
for i in range(lowLimit, highLimit + 1):
s = 0
while i:
s += i % 10
i //= 10
counter[s] += 1
return max(counter)
class Solution {
public int countBalls(int lowLimit, int highLimit) {
int[] counter = new int[60];
int ans = 0;
for (int i = lowLimit; i <= highLimit; ++i) {
int s = 0;
int j = i;
while (j > 0) {
s += (j % 10);
j /= 10;
}
++counter[s];
ans = Math.max(ans, counter[s]);
}
return ans;
}
}
class Solution {
public:
int countBalls(int lowLimit, int highLimit) {
vector<int> counter(60);
int ans = 0;
for (int i = lowLimit; i <= highLimit; ++i) {
int s = 0, j = i;
while (j) {
s += (j % 10);
j /= 10;
}
++counter[s];
ans = max(ans, counter[s]);
}
return ans;
}
};
func countBalls(lowLimit int, highLimit int) int {
counter := make([]int, 60)
ans := 0
for i := lowLimit; i <= highLimit; i++ {
s, j := 0, i
for j > 0 {
s += (j % 10)
j /= 10
}
counter[s]++
if counter[s] > ans {
ans = counter[s]
}
}
return ans
}