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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

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

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class Solution {
    public boolean areOccurrencesEqual(String s) {
        int[] cnt = new int[26];
        for (int i = 0; i < s.length(); ++i) {
            ++cnt[s.charAt(i) - 'a'];
        }
        int x = 0;
        for (int v : cnt) {
            if (v > 0) {
                if (x == 0) {
                    x = v;
                } else if (x != v) {
                    return false;
                }
            }
        }
        return true;
    }
}

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

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

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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;
}

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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;
    }
}

Solution 2

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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)];
    }
    const x = cnt.find(v => v);
    return cnt.every(v => !v || v === x);
}

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