花花酱 LeetCode 1653. Minimum Deletions to Make String Balanced

You are given a string s consisting only of characters 'a' and 'b'​​​​.

You can delete any number of characters in s to make s balanced. s is balanced if there is no pair of indices (i,j) such that i < j and s[i] = 'b' and s[j]= 'a'.

Return the minimum number of deletions needed to make s balanced.

Example 1:

Input: s = "aababbab"
Output: 2
Explanation: You can either:
Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").

Example 2:

Input: s = "bbaaaaabb"
Output: 2
Explanation: The only solution is to delete the first two characters.

Constraints:

• 1 <= s.length <= 105
• s[i] is 'a' or 'b'​​.

Solution: DP

dp[i][0] := min # of dels to make s[0:i] all ‘a’s;
dp[i][1] := min # of dels to make s[0:i] ends with 0 or mode ‘b’s

if s[i-1] == ‘a’:
dp[i + 1][0] = dp[i][0], dp[i + 1][1] = min(dp[i + 1][0], dp[i][1] + 1)
else:
dp[i + 1][0] = dp[i][0] + 1, dp[i + 1][1] = dp[i][1]

Time complexity: O(n)
Space complexity: O(n) -> O(1)