LeetCode 2405: Optimal Partition of String
LeetCode 2405 Solution Explanation
Explanation:
To solve this problem, we can use a greedy approach. We will iterate through each character in the string and maintain a set to keep track of unique characters we have encountered so far in the current substring. Whenever we encounter a character that is already in the set, it means we need to start a new substring from that character.
Algorithm:
- Initialize an empty set to store unique characters.
- Initialize a variable
countto keep track of the number of substrings. - Iterate through each character in the string.
- If the character is not in the set, add it to the set.
- If the character is already in the set, increment
countand clear the set. - Return the final value of
count.
Time Complexity:
The time complexity of this algorithm is O(n), where n is the length of the input string s.
Space Complexity:
The space complexity of this algorithm is O(26) = O(1) since we are using a set to store unique characters.
:
LeetCode 2405 Solutions in Java, C++, Python
class Solution {
public int minPartitions(String s) {
Set<Character> uniqueChars = new HashSet<>();
int count = 0;
for (char c : s.toCharArray()) {
if (!uniqueChars.contains(c)) {
uniqueChars.add(c);
} else {
count++;
uniqueChars.clear();
uniqueChars.add(c);
}
}
return count + 1;
}
}Interactive Code Editor for LeetCode 2405
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