1399. Count Largest Group

Description

You are given an integer n.

We need to group the numbers from 1 to n according to the sum of its digits. For example, the numbers 14 and 5 belong to the same group, whereas 13 and 3 belong to different groups.

Return the number of groups that have the largest size, i.e. the maximum number of elements.

 

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 \(O(n \times \log n)\), and the space complexity is \(O(\log n)\), where \(n\) is the given number.

Python3JavaC++GoTypeScriptRust

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

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;
    }
}

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;
    }
};

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
}

function countLargestGroup(n: number): number {
    const cnt: number[] = 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;
}

impl Solution {
    pub fn count_largest_group(n: i32) -> i32 {
        let mut cnt = vec![0; 40];
        let mut ans = 0;
        let mut mx = 0;

        for i in 1..=n {
            let mut s = 0;
            let mut x = i;
            while x > 0 {
                s += x % 10;
                x /= 10;
            }
            cnt[s as usize] += 1;
            if mx < cnt[s as usize] {
                mx = cnt[s as usize];
                ans = 1;
            } else if mx == cnt[s as usize] {
                ans += 1;
            }
        }

        ans
    }
}

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