2423. Remove Letter To Equalize Frequency LeetCode Solution
In this guide, you will get 2423. Remove Letter To Equalize Frequency LeetCode Solution with the best time and space complexity. The solution to Remove Letter To Equalize Frequency problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Remove Letter To Equalize Frequency
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.
2 <= word.length <= 100
word consists of lowercase English letters only.
Complexity Analysis
Time Complexity: O(26n) = O(n)
Space Complexity: O(26) = O(1)
2423. Remove Letter To Equalize Frequency LeetCode Solution in C++
class Solution {
public:
bool equalFrequency(string word) {
vector<int> count(26);
for (const char c : word)
++count[c - 'a'];
// Try to remove each letter, then check if the frequency of all the letters
// in `word` are equal.
for (const char c : word) {
--count[c - 'a'];
if (equalFreq(count))
return true;
++count[c - 'a'];
}
return false;
}
private:
static constexpr int kMax = 101;
bool equalFreq(const vector<int>& count) {
int minfreq = kMax;
int maxfreq = 0;
for (const int freq : count)
if (freq > 0) {
minfreq = min(minfreq, freq);
maxfreq = max(maxfreq, freq);
}
return minfreq == maxfreq;
}
}; /* code provided by PROGIEZ */
2423. Remove Letter To Equalize Frequency LeetCode Solution in Java
class Solution {
public boolean equalFrequency(String word) {
int[] count = new int[26];
for (final char c : word.toCharArray())
++count[c - 'a'];
// Try to remove each letter, then check if the frequency of all the letters
// in `word` are equal.
for (final char c : word.toCharArray()) {
--count[c - 'a'];
if (equalFreq(count))
return true;
++count[c - 'a'];
}
return false;
}
private static final int kMax = 101;
private boolean equalFreq(int[] count) {
int minfreq = kMax;
int maxfreq = 0;
for (final int freq : count)
if (freq > 0) {
minfreq = Math.min(minfreq, freq);
maxfreq = Math.max(maxfreq, freq);
}
return minfreq == maxfreq;
}
} // code provided by PROGIEZ
2423. Remove Letter To Equalize Frequency LeetCode Solution in Python
class Solution:
def equalFrequency(self, word: str) -> bool:
count = collections.Counter(word)
# Try to remove each letter, then check if the frequency of all the letters
# in `word` are equal.
for c in word:
count[c] -= 1
if count[c] == 0:
del count[c]
if min(count.values()) == max(count.values()):
return True
count[c] += 1
return False # code by PROGIEZ
