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)

// Author: Huahua
class Solution {
public:
  int minimumDeletions(string s) {
    // const int n = s.length();
    // 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+ 'b's
    // vector<vector<int>> dp(n + 1, vector<int>(2));
    // for (int i = 0; i < n; ++i) {
    //   if (s[i] == '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];
    //   }
    // }
    // return min(dp[n][0], dp[n][1]);
    int a = 0, b = 0;
    for (char c : s) {
      if (c == 'a') b = min(a, b + 1); 
      else ++a;
    }
    return min(a, b);
  }
};

The post 花花酱 LeetCode 1653. Minimum Deletions to Make String Balanced appeared first on Huahua's Tech Road.