Skip to content

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 \(n\).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
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;
}

Comments