Leetcode 2207: Maximize Number of Subsequences in a String
You are given a string text and another string pattern of length 2, consisting of lowercase English letters. Your task is to insert one of the characters from pattern into text exactly once, and then return the maximum number of times pattern appears as a subsequence in the modified text.
Problem
Approach
Steps
Complexity
Input: The input consists of a string text and a string pattern of length 2.
Example: text = 'abdcdbc', pattern = 'ac'
Constraints:
• 1 <= text.length <= 10^5
• pattern.length == 2
• text and pattern consist only of lowercase English letters.
Output: Return the maximum number of times pattern can appear as a subsequence of the modified text after adding one of the characters from pattern to text.
Example: For input text = 'abdcdbc', pattern = 'ac', the output is 4.
Constraints:
• The output should be an integer.
Goal: Maximize the number of subsequences of the pattern in the modified text after one character is inserted.
Steps:
• Initialize counters to track subsequences of pattern[0] and pattern[1].
• Iterate over the text and count occurrences of pattern[0] and pattern[1].
• Add the maximum possible count of subsequences after inserting either pattern[0] or pattern[1].
• Return the final result.
Goal: The length of text and pattern are constrained by the given limits.
Steps:
• 1 <= text.length <= 10^5
• pattern.length == 2
• The characters in text and pattern are lowercase English letters.
Assumptions:
• The length of text is at least 1.
• The length of pattern is fixed at 2 characters.
• Input: text = 'abdcdbc', pattern = 'ac'
• Explanation: By inserting 'a' in the correct position, we can form up to 4 subsequences of 'ac'.
• Input: text = 'aabb', pattern = 'ab'
• Explanation: Adding 'a' to the string 'aabb' results in 6 subsequences of 'ab'.
Approach: To solve this problem, we can iterate through the string and keep track of how many subsequences of the pattern appear. We then calculate the effect of adding either character from pattern and maximize the count of subsequences.
Observations:
• The insertion of a character can be done at any position, and the goal is to find the optimal position.
• By counting the occurrences of pattern[0] and pattern[1] in the original string, we can determine the best place to insert one of these characters.
Steps:
• Count the number of pattern[0] and pattern[1] subsequences in the string.
• Calculate the effect of inserting each character of the pattern at different positions.
• Maximize the subsequences after each possible insertion.
Empty Inputs:
• The input string text cannot be empty.
Large Inputs:
• For very large text lengths, efficient iteration and counting are needed.
Special Values:
• If the text already contains multiple subsequences of pattern, inserting the other character might not increase the count.
Constraints:
• The solution should handle text strings with lengths up to 10^5.
long long maximumSubsequenceCount(string text, string pattern) {
long long cnt1 = 0, cnt2 = 0, res = 0, n = text.length();
for(int i = 0; i < n; i++) {
if(text[i] == pattern[1]) {
cnt2++;
res += cnt1;
}
if(text[i] == pattern[0]) cnt1++;
}
res += max(cnt1, cnt2);
return res;
}
1 : Function Definition
long long maximumSubsequenceCount(string text, string pattern) {
Define the function that calculates the number of subsequences of a given pattern in a string.
2 : Variable Declaration
long long cnt1 = 0, cnt2 = 0, res = 0, n = text.length();
Declare and initialize variables: cnt1 and cnt2 for counting occurrences of characters in the pattern, res for storing the result, and n for the length of the input text.
3 : For Loop
for(int i = 0; i < n; i++) {
Start the loop to iterate through each character in the text.
4 : Condition Check 1
if(text[i] == pattern[1]) {
Check if the current character in the text matches the second character of the pattern.
5 : Increment Counter 2
cnt2++;
Increment the counter for the second character of the pattern.
6 : Update Result
res += cnt1;
Add the count of the first character occurrences to the result.
7 : Condition Check 2
if(text[i] == pattern[0]) cnt1++;
Check if the current character in the text matches the first character of the pattern and increment the counter.
8 : Final Calculation
res += max(cnt1, cnt2);
Add the maximum count of the first or second character to the result to account for any remaining subsequences.
9 : Return Statement
return res;
Return the computed result of the maximum subsequence count.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in the length of the input text.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant since only a few variables are used for counting.
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