LeetCode-in-Java
1742. Maximum Number of Balls in a Box
Easy
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
public class Solution {
public int countBalls(int lowLimit, int highLimit) {
int maxValue;
int[] countArray = new int[46];
int currentSum = getDigitSum(lowLimit);
countArray[currentSum]++;
maxValue = 1;
for (int i = lowLimit + 1; i <= highLimit; i++) {
if (i % 10 == 0) {
currentSum = getDigitSum(i);
} else {
currentSum++;
}
countArray[currentSum]++;
if (countArray[currentSum] > maxValue) {
maxValue = countArray[currentSum];
}
}
return maxValue;
}
private int getDigitSum(int num) {
int currentSum = 0;
while (num > 0) {
currentSum += num % 10;
num = num / 10;
}
return currentSum;
}
}