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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;
    }
    
    

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