# 2957. Remove Adjacent Almost-Equal Characters

## Description

You are given a 0-indexed string word.

In one operation, you can pick any index i of word and change word[i] to any lowercase English letter.

Return the minimum number of operations needed to remove all adjacent almost-equal characters from word.

Two characters a and b are almost-equal if a == b or a and b are adjacent in the alphabet.

Example 1:

Input: word = "aaaaa"
Output: 2
Explanation: We can change word into "acaca" which does not have any adjacent almost-equal characters.
It can be shown that the minimum number of operations needed to remove all adjacent almost-equal characters from word is 2.


Example 2:

Input: word = "abddez"
Output: 2
Explanation: We can change word into "ybdoez" which does not have any adjacent almost-equal characters.
It can be shown that the minimum number of operations needed to remove all adjacent almost-equal characters from word is 2.

Example 3:

Input: word = "zyxyxyz"
Output: 3
Explanation: We can change word into "zaxaxaz" which does not have any adjacent almost-equal characters.
It can be shown that the minimum number of operations needed to remove all adjacent almost-equal characters from word is 3.


Constraints:

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

## Solutions

Solution 1: Greedy

We start traversing the string word from index $1$. If word[i] and word[i - 1] are approximately equal, we greedily replace word[i] with a character that is not equal to both word[i - 1] and word[i + 1] (we can choose not to perform the replacement operation, just record the number of operations). Then, we skip word[i + 1] and continue to traverse the string word.

Finally, we return the recorded number of operations.

The time complexity is $O(n)$, where $n$ is the length of the string word. The space complexity is $O(1)$.

• class Solution {
public int removeAlmostEqualCharacters(String word) {
int ans = 0, n = word.length();
for (int i = 1; i < n; ++i) {
if (Math.abs(word.charAt(i) - word.charAt(i - 1)) < 2) {
++ans;
++i;
}
}
return ans;
}
}

• class Solution {
public:
int removeAlmostEqualCharacters(string word) {
int ans = 0, n = word.size();
for (int i = 1; i < n; ++i) {
if (abs(word[i] - word[i - 1]) < 2) {
++ans;
++i;
}
}
return ans;
}
};

• class Solution:
def removeAlmostEqualCharacters(self, word: str) -> int:
ans = 0
i, n = 1, len(word)
while i < n:
if abs(ord(word[i]) - ord(word[i - 1])) < 2:
ans += 1
i += 2
else:
i += 1
return ans


• func removeAlmostEqualCharacters(word string) (ans int) {
for i := 1; i < len(word); i++ {
if abs(int(word[i])-int(word[i-1])) < 2 {
ans++
i++
}
}
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function removeAlmostEqualCharacters(word: string): number {
let ans = 0;
for (let i = 1; i < word.length; ++i) {
if (Math.abs(word.charCodeAt(i) - word.charCodeAt(i - 1)) < 2) {
++ans;
++i;
}
}
return ans;
}