2645. Minimum Additions to Make Valid String

Description

Given a string word to which you can insert letters "a", "b" or "c" anywhere and any number of times, return the minimum number of letters that must be inserted so that word becomes valid.

A string is called valid if it can be formed by concatenating the string "abc" several times.

Example 1:

Input: word = "b"
Output: 2
Explanation: Insert the letter "a" right before "b", and the letter "c" right next to "a" to obtain the valid string "abc".


Example 2:

Input: word = "aaa"
Output: 6
Explanation: Insert letters "b" and "c" next to each "a" to obtain the valid string "abcabcabc".


Example 3:

Input: word = "abc"
Output: 0
Explanation: word is already valid. No modifications are needed.


Constraints:

• 1 <= word.length <= 50
• word consists of letters "a", "b" and "c" only.

Solutions

Solution 1: Greedy + Two Pointers

We define the string $s$ as "abc", and use pointers $i$ and $j$ to point to $s$ and $word$ respectively.

If $word[j] \neq s[i]$, we need to insert $s[i]$, and we add $1$ to the answer; otherwise, it means that $word[j]$ can match with $s[i]$, and we move $j$ one step to the right.

Then, we move $i$ one step to the right, i.e., $i = (i + 1) \bmod 3$. We continue the above operations until $j$ reaches the end of the string $word$.

Finally, we check whether the last character of $word$ is 'b' or 'a'. If it is, we need to insert 'c' or 'bc', and we add $1$ or $2$ to the answer and return it.

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

• class Solution {
String s = "abc";
int ans = 0, n = word.length();
for (int i = 0, j = 0; j < n; i = (i + 1) % 3) {
if (word.charAt(j) != s.charAt(i)) {
++ans;
} else {
++j;
}
}
if (word.charAt(n - 1) != 'c') {
ans += word.charAt(n - 1) == 'b' ? 1 : 2;
}
return ans;
}
}

• class Solution {
public:
string s = "abc";
int ans = 0, n = word.size();
for (int i = 0, j = 0; j < n; i = (i + 1) % 3) {
if (word[j] != s[i]) {
++ans;
} else {
++j;
}
}
if (word[n - 1] != 'c') {
ans += word[n - 1] == 'b' ? 1 : 2;
}
return ans;
}
};

• class Solution:
def addMinimum(self, word: str) -> int:
s = 'abc'
ans, n = 0, len(word)
i = j = 0
while j < n:
if word[j] != s[i]:
ans += 1
else:
j += 1
i = (i + 1) % 3
if word[-1] != 'c':
ans += 1 if word[-1] == 'b' else 2
return ans


• func addMinimum(word string) (ans int) {
s := "abc"
n := len(word)
for i, j := 0, 0; j < n; i = (i + 1) % 3 {
if word[j] != s[i] {
ans++
} else {
j++
}
}
if word[n-1] == 'b' {
ans++
} else if word[n-1] == 'a' {
ans += 2
}
return
}

• function addMinimum(word: string): number {
const s: string = 'abc';
let ans: number = 0;
const n: number = word.length;
for (let i = 0, j = 0; j < n; i = (i + 1) % 3) {
if (word[j] !== s[i]) {
++ans;
} else {
++j;
}
}
if (word[n - 1] === 'b') {
++ans;
} else if (word[n - 1] === 'a') {
ans += 2;
}
return ans;
}