LeetCode 2423. Remove Letter To Equalize Frequency Solution in Java, C++, Python & More | Explanation + Code

Description

You are given a 0-indexed string word, consisting of lowercase English letters. You need to select one index and remove the letter at that index from word so that the frequency of every letter present in word is equal.

Return true if it is possible to remove one letter so that the frequency of all letters in word are equal, and false otherwise.

Note:

• The frequency of a letter x is the number of times it occurs in the string.
• You must remove exactly one letter and cannot choose to do nothing.

 

Example 1:

Input: word = "abcc"
Output: true
Explanation: Select index 3 and delete it: word becomes "abc" and each character has a frequency of 1.

Example 2:

Input: word = "aazz"
Output: false
Explanation: We must delete a character, so either the frequency of "a" is 1 and the frequency of "z" is 2, or vice versa. It is impossible to make all present letters have equal frequency.

 

Constraints:

• 2 <= word.length <= 100
• word consists of lowercase English letters only.

Solutions

Solution 1: Counting + Enumeration

First, we use a hash table or an array of length 26 named cnt to count the number of occurrences of each letter in the string.

Next, we enumerate the 26 letters. If letter c appears in the string, we decrement its count by one, then check whether the counts of the remaining letters are the same. If they are, return true. Otherwise, increment the count of c by one and continue to enumerate the next letter.

If the enumeration ends, it means that it is impossible to make the counts of the remaining letters the same by deleting one letter, so return false.

The time complexity is O(n + C²), and the space complexity is O(C). Here, n is the length of the string word, and C is the size of the character set. In this problem, C = 26.

PythonJavaC++GoTypeScript

class Solution: def equalFrequency(self, word: str) -> bool: cnt = Counter(word) for c in cnt.keys(): cnt[c] -= 1 if len(set(v for v in cnt.values() if v)) == 1: return True cnt[c] += 1 return False(code-box)

class Solution { public boolean equalFrequency(String word) { int[] cnt = new int[26]; for (int i = 0; i < word.length(); ++i) { ++cnt[word.charAt(i) - 'a']; } for (int i = 0; i < 26; ++i) { if (cnt[i] > 0) { --cnt[i]; int x = 0; boolean ok = true; for (int v : cnt) { if (v == 0) { continue; } if (x > 0 && v != x) { ok = false; break; } x = v; } if (ok) { return true; } ++cnt[i]; } } return false; } }(code-box)

class Solution { public: bool equalFrequency(string word) { int cnt[26]{}; for (char& c : word) { ++cnt[c - 'a']; } for (int i = 0; i < 26; ++i) { if (cnt[i]) { --cnt[i]; int x = 0; bool ok = true; for (int v : cnt) { if (v == 0) { continue; } if (x && v != x) { ok = false; break; } x = v; } if (ok) { return true; } ++cnt[i]; } } return false; } };(code-box)

func equalFrequency(word string) bool { cnt := [26]int{} for _, c := range word { cnt[c-'a']++ } for i := range cnt { if cnt[i] > 0 { cnt[i]-- x := 0 ok := true for _, v := range cnt { if v == 0 { continue } if x > 0 && v != x { ok = false break } x = v } if ok { return true } cnt[i]++ } } return false }(code-box)

function equalFrequency(word: string): boolean { const cnt: number[] = new Array(26).fill(0); for (const c of word) { cnt[c.charCodeAt(0) - 97]++; } for (let i = 0; i < 26; ++i) { if (cnt[i]) { cnt[i]--; let x = 0; let ok = true; for (const v of cnt) { if (v === 0) { continue; } if (x && v !== x) { ok = false; break; } x = v; } if (ok) { return true; } cnt[i]++; } } return false; }(code-box)