Check if All Characters Have Equal Number of Occurrences — Easy Problem & Solution

Problem statement

A string is good if every character that appears in it appears the same number of times.

Given s, return true if it is good.

Example 1

Input: s = "abacbc"
Output: true
Explanation: a, b and c each appear twice.

Example 2

Input: s = "aaabb"
Output: false
Explanation: a appears three times and b twice.

Example 3

Input: s = "codekairo"
Output: false
Explanation: o appears twice, everything else once.

Constraints

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

How to solve Check if All Characters Have Equal Number of Occurrences

Tally the letters, then check that all non-zero tallies agree. The first non-zero tally becomes the reference value.

Approach

1. Build count[26] from s.
2. Scan the tallies, skipping zeros.
3. Remember the first non-zero tally and reject as soon as another differs.

Why it works

The property is exactly 'the multiset of non-zero counts has one distinct value', and comparing every count against the first is the cheapest way to test that in one pass.

Complexity

• Time — O(n)
• Space — O(1)

Pitfalls

• Including zero counts in the comparison rejects every string that does not use all 26 letters.
• Comparing count[c] against count[0] rather than the first non-zero tally fails whenever 'a' is absent.

Reference solution

Python

def areOccurrencesEqual(s: str) -> bool:
    count = [0] * 26
    for c in s:
        count[ord(c) - 97] += 1
    seen = [c for c in count if c > 0]
    return all(c == seen[0] for c in seen)

JavaScript

var areOccurrencesEqual = function(s) {
    var count = [];
    for (var t = 0; t < 26; t++) count.push(0);
    for (var i = 0; i < s.length; i++) count[s.charCodeAt(i) - 97]++;
    var first = -1;
    for (var c = 0; c < 26; c++) {
        if (count[c] === 0) continue;
        if (first < 0) first = count[c];
        else if (count[c] !== first) return false;
    }
    return true;
};

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