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Formatted question description: https://leetcode.ca/all/1941.html

1941. Check if All Characters Have Equal Number of Occurrences

Level

Easy

Description

Given a string s, return true if s is a good string, or false otherwise.

A string s is good if all the characters that appear in s have the same number of occurrences (i.e., the same frequency).

Example 1:

Input: s = “abacbc”

Output: true

Explanation: The characters that appear in s are ‘a’, ‘b’, and ‘c’. All characters occur 2 times in s.

Example 2:

Input: s = “aaabb”

Output: false

Explanation: The characters that appear in s are ‘a’ and ‘b’. ‘a’ occurs 3 times while ‘b’ occurs 2 times, which is not the same number of times.

Constraints:

  • 1 <= s.length <= 1000
  • s consists of lowercase English letters.

Solution

Use a hash map to store each character’s number of occurrences in s. The length of s should be divisible by the hash map’s size, and each character’s number of occurrences should be the same. In this case, return true. Otherwise, return false.

  • class Solution {
    public boolean areOccurrencesEqual(String s) {
        Map<Character, Integer> map = new HashMap<Character, Integer>();
        int length = s.length();
        for (int i = 0; i < length; i++) {
            char c = s.charAt(i);
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        int size = map.size();
        if (length % size != 0)
            return false;
        int times = length / size;
        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
            int count = entry.getValue();
            if (count != times)
                return false;
        }
        return true;
    }
}

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

class Solution {
    public boolean areOccurrencesEqual(String s) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            ++cnt[c - 'a'];
        }
        int t = 0;
        for (int v : cnt) {
            if (v == 0) {
                continue;
            }
            if (t == 0) {
                t = v;
            } else if (t != v) {
                return false;
            }
        }
        return true;
    }
}

  • // OJ: https://leetcode.com/problems/check-if-all-characters-have-equal-number-of-occurrences/
// Time: O(N)
// Space: O(C) where C is the range of the characters
class Solution {
public:
    bool areOccurrencesEqual(string s) {
        int cnt[26] = {}, mx = 0;
        for (char c : s) {
            mx = max(mx, ++cnt[c - 'a']);
        }
        for (int i = 0; i < 26; ++i) {
            if (cnt[i] && cnt[i] != mx) return false;
        }
        return true;
    }
};

  • class Solution:
    def areOccurrencesEqual(self, s: str) -> bool:
        cnt = Counter(s)
        return len(set(cnt.values())) == 1

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

# 1941. Check if All Characters Have Equal Number of Occurrences
# https://leetcode.com/problems/check-if-all-characters-have-equal-number-of-occurrences/

class Solution:
    def areOccurrencesEqual(self, s: str) -> bool:
        counter = collections.Counter(s)
        values = list(counter.values())
        
        for v in values:
            if v != values[0]: return False
        
        return True


  • func areOccurrencesEqual(s string) bool {
	cnt := make([]int, 26)
	for _, c := range s {
		cnt[c-'a']++
	}
	ss := map[int]bool{}
	for _, v := range cnt {
		if v == 0 {
			continue
		}
		ss[v] = true
	}
	return len(ss) == 1
}

  • function areOccurrencesEqual(s: string): boolean {
    const cnt: number[] = new Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
    }
    let x = 0;
    for (const v of cnt) {
        if (v) {
            if (!x) {
                x = v;
            } else if (x !== v) {
                return false;
            }
        }
    }
    return true;
}


  • class Solution {
    /**
     * @param String $s
     * @return Boolean
     */
    function areOccurrencesEqual($s) {
        for ($i = 0; $i < strlen($s); $i++) {
            $hashtable[$s[$i]] += 1;
        }
        $rs = array_unique($hashtable);
        return count($rs) === 1;
    }
}

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