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3870. Count Commas in Range

SourceWeekly Contest 493 Q1DifficultyEasyRating1149

Description

You are given an integer n.

Return the total number of commas used when writing all integers from [1, n] (inclusive) in standard number formatting.

In standard formatting:

  • A comma is inserted after every three digits from the right.
  • Numbers with fewer than 4 digits contain no commas.

 

Example 1:

Input: n = 1002

Output: 3

Explanation:

The numbers "1,000", "1,001", and "1,002" each contain one comma, giving a total of 3.

Example 2:

Input: n = 998

Output: 0

Explanation:

All numbers from 1 to 998 have fewer than four digits. Therefore, no commas are used.

 

Constraints:

  • 1 <= n <= 105

Solutions

Solution 1: Brain Teaser

Thinking

Count thousands-separator commas used when writing \([1,n]\). \(n \le 10^5\), so numbers have at most six digits and at most one comma each.

\(1\) through \(999\) have none; each integer from \(1000\) to \(n\) has exactly one.

The answer is \(\max(0,n-999)\).

Constant time, no enumeration.

Numbers from 1 to 999 contain no commas, so when \(n\) is less than or equal to 999, the answer is 0.

Since the range of \(n\) is \([1, 10^5]\), when \(n\) is greater than or equal to 1000, each number contains exactly one comma, so the answer is \(n - 999\).

Therefore, the answer is \(\max(0, n - 999)\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

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class Solution:
    def countCommas(self, n: int) -> int:
        return max(0, n - 999)

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class Solution {
    public int countCommas(int n) {
        return Math.max(0, n - 999);
    }
}

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class Solution {
public:
    int countCommas(int n) {
        return max(0, n - 999);
    }
};

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func countCommas(n int) int {
    return max(0, n-999)
}

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function countCommas(n: number): number {
    return Math.max(0, n - 999);
}

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