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'​​.

Approach 01:

#include <string>
#include <algorithm>

using namespace std;

class Solution {
public:
    int minimumDeletions(string s) {
        int minDeletions = 0;   // Minimum deletions to make the string balanced
        int countB = 0;         // Count of 'b's encountered so far

        for (const char c : s) {
            if (c == 'a') {
                // Either delete this 'a' or keep it and delete some 'b's seen before
                minDeletions = min(minDeletions + 1, countB);
            } else {
                ++countB;  // Increment count of 'b's
            }
        }

        return minDeletions;
    }
};

class Solution:
    def minimumDeletions(self, s: str) -> int:
        minDeletions = 0  # Minimum deletions to make the string balanced
        countB = 0        # Count of 'b's encountered so far

        for c in s:
            if c == 'a':
                # Either delete this 'a' or keep it and delete some 'b's seen before
                minDeletions = min(minDeletions + 1, countB)
            else:
                countB += 1  # Increment count of 'b's

        return minDeletions

Time Complexity

• Initialization:

  Initializing minDeletions and countB takes \( O(1) \) time.

• Traversal of String:

  We traverse the string s once, which takes \( O(n) \) time, where \( n \) is the length of the string.

• Operations within Loop:

  Inside the loop, the operations (checking characters and updating variables) are all \( O(1) \) operations.

• Overall Time Complexity:

  The overall time complexity is \( O(n) \).

Space Complexity

• Space for Variables:

  The space used by the variables minDeletions and countB is \( O(1) \).

• Input String:

  The input string s is provided as input, so it does not count towards the space complexity.

• Overall Space Complexity:

  The overall space complexity is \( O(1) \).