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
public class Solution {
public int CountBalls(int lowLimit, int highLimit) {
var map = new Dictionary<int,int>();
int maxCount = 0;
int count = 0;
for(int i=lowLimit; i<=highLimit;i++){
var boxNum = GetBoxNum(i);
if(map.TryGetValue(boxNum, out count)){
count++;
map[boxNum] = count;
}
else{
count = 1;
map.Add(boxNum, 1);
}
if(count > maxCount){
maxCount = count;
}
}
return maxCount;
}
private int GetBoxNum(int num){
if(num<10){
return num;
}
int sum = 0;
while(num>0){
sum+=num%10;
num/=10;
}
return sum;
}
}Time Complexity: O(n)
Space Complexity: O(n)
