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1399. Count Largest Group

Description

You are given an integer n.

Each number from 1 to n is grouped according to the sum of its digits.

Return the number of groups that have the largest size.

 

Example 1:

Input: n = 13
Output: 4
Explanation: There are 9 groups in total, they are grouped according sum of its digits of numbers from 1 to 13:
[1,10], [2,11], [3,12], [4,13], [5], [6], [7], [8], [9].
There are 4 groups with largest size.

Example 2:

Input: n = 2
Output: 2
Explanation: There are 2 groups [1], [2] of size 1.

 

Constraints:

  • 1 <= n <= 104

Solutions

Solution 1: Hash Table or Array

We note that the number does not exceed $10^4$, so the sum of the digits also does not exceed $9 \times 4 = 36$. Therefore, we can use a hash table or an array of length $40$, denoted as $cnt$, to count the number of each sum of digits, and use a variable $mx$ to represent the maximum count of the sum of digits.

We enumerate each number in $[1,..n]$, calculate its sum of digits $s$, then increment $cnt[s]$ by $1$. If $mx < cnt[s]$, we update $mx = cnt[s]$ and set $ans$ to $1$. If $mx = cnt[s]$, we increment $ans$ by $1$.

Finally, we return $ans$.

The time complexity is $O(n \times \log M)$, and the space complexity is $O(\log M)$. Where $n$ is the given number, and $M$ is the range of $n$.

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class Solution:
    def countLargestGroup(self, n: int) -> int:
        cnt = Counter()
        ans = mx = 0
        for i in range(1, n + 1):
            s = 0
            while i:
                s += i % 10
                i //= 10
            cnt[s] += 1
            if mx < cnt[s]:
                mx = cnt[s]
                ans = 1
            elif mx == cnt[s]:
                ans += 1
        return ans

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class Solution {
    public int countLargestGroup(int n) {
        int[] cnt = new int[40];
        int ans = 0, mx = 0;
        for (int i = 1; i <= n; ++i) {
            int s = 0;
            for (int x = i; x > 0; x /= 10) {
                s += x % 10;
            }
            ++cnt[s];
            if (mx < cnt[s]) {
                mx = cnt[s];
                ans = 1;
            } else if (mx == cnt[s]) {
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countLargestGroup(int n) {
        int cnt[40]{};
        int ans = 0, mx = 0;
        for (int i = 1; i <= n; ++i) {
            int s = 0;
            for (int x = i; x; x /= 10) {
                s += x % 10;
            }
            ++cnt[s];
            if (mx < cnt[s]) {
                mx = cnt[s];
                ans = 1;
            } else if (mx == cnt[s]) {
                ++ans;
            }
        }
        return ans;
    }
};

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func countLargestGroup(n int) (ans int) {
    cnt := [40]int{}
    mx := 0
    for i := 1; i <= n; i++ {
        s := 0
        for x := i; x > 0; x /= 10 {
            s += x % 10
        }
        cnt[s]++
        if mx < cnt[s] {
            mx = cnt[s]
            ans = 1
        } else if mx == cnt[s] {
            ans++
        }
    }
    return
}

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function countLargestGroup(n: number): number {
    const cnt: number[] = new Array(40).fill(0);
    let mx = 0;
    let ans = 0;
    for (let i = 1; i <= n; ++i) {
        let s = 0;
        for (let x = i; x; x = Math.floor(x / 10)) {
            s += x % 10;
        }
        ++cnt[s];
        if (mx < cnt[s]) {
            mx = cnt[s];
            ans = 1;
        } else if (mx === cnt[s]) {
            ++ans;
        }
    }
    return ans;
}

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