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Formatted question description: https://leetcode.ca/all/1399.html

# 1399. Count Largest Group (Easy)

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

Return how many groups 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.


Example 3:

Input: n = 15
Output: 6


Example 4:

Input: n = 24
Output: 5


Constraints:

• 1 <= n <= 10^4

Related Topics:
Array

## Solution 1.

• class Solution {
public int countLargestGroup(int n) {
int maxCount = 0;
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 1; i <= n; i++) {
int sum = sumDigits(i);
int count = map.getOrDefault(sum, 0) + 1;
map.put(sum, count);
maxCount = Math.max(maxCount, count);
}
int groups = 0;
Set<Integer> keySet = map.keySet();
for (int key : keySet) {
if (map.get(key) == maxCount)
groups++;
}
return groups;
}

public int sumDigits(int num) {
char[] array = String.valueOf(num).toCharArray();
int sum = 0;
for (char c : array)
sum += c - '0';
return sum;
}
}

############

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

• // OJ: https://leetcode.com/problems/count-largest-group/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int countLargestGroup(int n) {
unordered_map<int, int> m;
for (int i = 1; i <= n; ++i) {
int sum = 0, x = i;
while (x) {
sum += x % 10;
x /= 10;
}
m[sum]++;
}
int maxSize = 0;
for (auto &p : m) {
maxSize = max(maxSize, p.second);
}
int ans = 0;
for (auto &p : m) {
if (p.second == maxSize) ++ans;
}
return ans;
}
};

• class Solution:
def countLargestGroup(self, n: int) -> int:
cnt = Counter()
ans, mx = 0, 0
for i in range(1, n + 1):
t = sum(int(v) for v in str(i))
cnt[t] += 1
if mx < cnt[t]:
mx = cnt[t]
ans = 1
elif mx == cnt[t]:
ans += 1
return ans


• func countLargestGroup(n int) int {
cnt := make([]int, 40)
mx, ans := 0, 0
for i := 1; i <= n; i++ {
t := 0
j := i
for j != 0 {
t += j % 10
j /= 10
}
cnt[t]++
if mx < cnt[t] {
mx = cnt[t]
ans = 1
} else if mx == cnt[t] {
ans++
}
}
return ans
}

• 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;
}