2207. Maximize Number of Subsequences in a String
Description
You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both of which consist of only lowercase English letters.
You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the character can be added even at the beginning or at the end of text.
Return the maximum number of times pattern can occur as a subsequence of the modified text.
A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.
Example 1:
Input: text = "abdcdbc", pattern = "ac" Output: 4 Explanation: If we add pattern[0] = 'a' in between text[1] and text[2], we get "abadcdbc". Now, the number of times "ac" occurs as a subsequence is 4. Some other strings which have 4 subsequences "ac" after adding a character to text are "aabdcdbc" and "abdacdbc". However, strings such as "abdcadbc", "abdccdbc", and "abdcdbcc", although obtainable, have only 3 subsequences "ac" and are thus suboptimal. It can be shown that it is not possible to get more than 4 subsequences "ac" by adding only one character.
Example 2:
Input: text = "aabb", pattern = "ab" Output: 6 Explanation: Some of the strings which can be obtained from text and have 6 subsequences "ab" are "aaabb", "aaabb", and "aabbb".
Constraints:
1 <= text.length <= 105pattern.length == 2textandpatternconsist only of lowercase English letters.
Solutions
Solution 1: Traversal + Counting
We can use two variables $x$ and $y$ to record the current counts of $\textit{pattern}[0]$ and $\textit{pattern}[1]$ in the string, respectively.
Then, traverse the string $\textit{text}$. For the current character $c$:
If $c$ equals $\textit{pattern}[1]$, increment $y$ by one. At this point, all previously encountered $\textit{pattern}[0]$ can form a $\textit{pattern}$ subsequence with the current $c$, so add $x$ to the answer.
If $c$ equals $\textit{pattern}[0]$, increment $x$ by one.
After the traversal, since we can insert one character, if we add $\textit{pattern}[0]$ at the beginning of the string, we can get $y$ $\textit{pattern}$ subsequences. If we add $\textit{pattern}[1]$ at the end of the string, we can get $x$ $\textit{pattern}$ subsequences. Therefore, we add the larger value of $x$ and $y$ to the answer.
The time complexity is $O(n)$, where $n$ is the length of the string $\textit{text}$. The space complexity is $O(1)$.
Python3
class Solution:
def maximumSubsequenceCount(self, text: str, pattern: str) -> int:
ans = x = y = 0
for c in text:
if c == pattern[1]:
y += 1
ans += x
if c == pattern[0]:
x += 1
ans += max(x, y)
return ans
Java
class Solution {
public long maximumSubsequenceCount(String text, String pattern) {
long ans = 0;
int x = 0, y = 0;
for (int i = 0; i < text.length(); ++i) {
if (text.charAt(i) == pattern.charAt(1)) {
++y;
ans += x;
}
if (text.charAt(i) == pattern.charAt(0)) {
++x;
}
}
ans += Math.max(x, y);
return ans;
}
}
C++
class Solution {
public:
long long maximumSubsequenceCount(string text, string pattern) {
long long ans = 0;
int x = 0, y = 0;
for (char& c : text) {
if (c == pattern[1]) {
++y;
ans += x;
}
if (c == pattern[0]) {
++x;
}
}
ans += max(x, y);
return ans;
}
};
Go
func maximumSubsequenceCount(text string, pattern string) (ans int64) {
x, y := 0, 0
for _, c := range text {
if byte(c) == pattern[1] {
y++
ans += int64(x)
}
if byte(c) == pattern[0] {
x++
}
}
ans += int64(max(x, y))
return
}
TypeScript
function maximumSubsequenceCount(text: string, pattern: string): number {
let ans = 0;
let [x, y] = [0, 0];
for (const c of text) {
if (c === pattern[1]) {
++y;
ans += x;
}
if (c === pattern[0]) {
++x;
}
}
ans += Math.max(x, y);
return ans;
}