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1941. Check if All Characters Have Equal Number of Occurrences

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.

Solutions

Solution 1: Counting

We use a hash table or an array of length \(26\) called \(\textit{cnt}\) to record the number of occurrences of each character in the string \(s\).

Next, we traverse each value in \(\textit{cnt}\) and check if all non-zero values are equal.

The time complexity is \(O(n)\), and the space complexity is \(O(|\Sigma|)\). Here, \(n\) is the length of the string \(s\), and \(\Sigma\) is the size of the character set. In this problem, the character set consists of lowercase English letters, so \(|\Sigma|=26\).

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

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class Solution {
    public boolean areOccurrencesEqual(String s) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            ++cnt[c - 'a'];
        }
        int v = 0;
        for (int x : cnt) {
            if (x == 0) {
                continue;
            }
            if (v > 0 && v != x) {
                return false;
            }
            v = x;
        }
        return true;
    }
}

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class Solution {
public:
    bool areOccurrencesEqual(string s) {
        vector<int> cnt(26);
        for (char c : s) {
            ++cnt[c - 'a'];
        }
        int v = 0;
        for (int x : cnt) {
            if (x == 0) {
                continue;
            }
            if (v && v != x) {
                return false;
            }
            v = x;
        }
        return true;
    }
};

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func areOccurrencesEqual(s string) bool {
    cnt := [26]int{}
    for _, c := range s {
        cnt[c-'a']++
    }
    v := 0
    for _, x := range cnt {
        if x == 0 {
            continue
        }
        if v > 0 && v != x {
            return false
        }
        v = x
    }
    return true
}

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function areOccurrencesEqual(s: string): boolean {
    const cnt: number[] = Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0) - 'a'.charCodeAt(0)];
    }
    const v = cnt.find(v => v);
    return cnt.every(x => !x || v === x);
}

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class Solution {
    /**
     * @param String $s
     * @return Boolean
     */
    function areOccurrencesEqual($s) {
        $cnt = array_fill(0, 26, 0);
        for ($i = 0; $i < strlen($s); $i++) {
            $cnt[ord($s[$i]) - ord('a')]++;
        }
        $v = 0;
        foreach ($cnt as $x) {
            if ($x == 0) {
                continue;
            }
            if ($v && $v != $x) {
                return false;
            }
            $v = $x;
        }
        return true;
    }
}

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