Two strings word1
and word2
are considered almost equivalent if the differences between the frequencies of each letter from 'a'
to 'z'
between word1
and word2
is at most 3
.
Given two strings word1
and word2
, each of length n
, return true
if word1
and word2
are almost equivalent, or false
otherwise.
The frequency of a letter x
is the number of times it occurs in the string.
Example 1:
Input: word1 = "aaaa", word2 = "bccb" Output: false Explanation: There are 4 'a's in "aaaa" but 0 'a's in "bccb". The difference is 4, which is more than the allowed 3.
Example 2:
Input: word1 = "abcdeef", word2 = "abaaacc" Output: true Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3: - 'a' appears 1 time in word1 and 4 times in word2. The difference is 3. - 'b' appears 1 time in word1 and 1 time in word2. The difference is 0. - 'c' appears 1 time in word1 and 2 times in word2. The difference is 1. - 'd' appears 1 time in word1 and 0 times in word2. The difference is 1. - 'e' appears 2 times in word1 and 0 times in word2. The difference is 2. - 'f' appears 1 time in word1 and 0 times in word2. The difference is 1.
Example 3:
Input: word1 = "cccddabba", word2 = "babababab" Output: true Explanation: The differences between the frequencies of each letter in word1 and word2 are at most 3: - 'a' appears 2 times in word1 and 4 times in word2. The difference is 2. - 'b' appears 2 times in word1 and 5 times in word2. The difference is 3. - 'c' appears 3 times in word1 and 0 times in word2. The difference is 3. - 'd' appears 2 times in word1 and 0 times in word2. The difference is 2.
Constraints:
n == word1.length == word2.length
1 <= n <= 100
word1
andword2
consist only of lowercase English letters.
// OJ: https://leetcode.com/problems/check-whether-two-strings-are-almost-equivalent/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool checkAlmostEquivalent(string s, string t) {
int cnt[26] = {};
for (char c : s) cnt[c - 'a']++;
for (char c : t) cnt[c - 'a']--;
for (int i = 0; i < 26; ++i) {
if (abs(cnt[i]) > 3) return false;
}
return true;
}
};