Leetcode Problem 2405. Optimal Partition of String

2405. Optimal Partition of String

Leetcode Solutions

Initialize an array lastSeen of size 26 to -1 to keep track of the most recent position of each character.
Initialize count to 1 as we start with the first substring.
Initialize substringStart to 0 to mark the beginning of the current substring.
Iterate over the string s and for each character at index i:
- Check if the character has been seen in the current substring by comparing lastSeen[s[i] - 'a'] with substringStart.
- If it has been seen, increment count, start a new substring, and update substringStart to i.
- Update lastSeen[s[i] - 'a'] to i.
Return count.

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