Mutating counts without cleanup
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.
Build confidence with an intuition-first walkthrough focused on hash map fundamentals.
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 <= 105Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Hash Map · Math
1 10
5 15
19 28
find-the-number-of-distinct-colors-among-the-balls)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1742: Maximum Number of Balls in a Box
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();
}
}
// Accepted solution for LeetCode #1742: Maximum Number of Balls in a Box
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
}
# Accepted solution for LeetCode #1742: Maximum Number of Balls in a Box
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)
// Accepted solution for LeetCode #1742: Maximum Number of Balls in a Box
impl Solution {
pub fn count_balls(low_limit: i32, high_limit: i32) -> i32 {
let mut cnt = vec![0; 50];
for x in low_limit..=high_limit {
let mut y = 0;
let mut n = x;
while n > 0 {
y += n % 10;
n /= 10;
}
cnt[y as usize] += 1;
}
*cnt.iter().max().unwrap()
}
}
// Accepted solution for LeetCode #1742: Maximum Number of Balls in a Box
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);
}
Use this to step through a reusable interview workflow for this problem.
For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.
One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.