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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.
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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; } }
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// 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; } };
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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
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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 }
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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; }