1653. Minimum Deletions to Make String Balanced

/**
 * Problem: LeetCode 1653 - Minimum Deletions to Make String Balanced
 * Difficulty: Medium
 * Language: Java
 * Author: CodeByTushu
 */
class Solution {
    public int minimumDeletions(String s) {
        int bCount = 0;
        int deletions = 0;

        for (char c : s.toCharArray()) {
            if (c == 'b') {
                bCount++;
            } else {
                // For character 'a', we either delete this 'a' (deletions + 1)
                // or keep it by deleting all prior 'b' characters (bCount).
                deletions = Math.min(deletions + 1, bCount);
            }
        }

        return deletions;
    }
}

"""
Problem: LeetCode 1653 - Minimum Deletions to Make String Balanced
Difficulty: Medium
Language: Python
Author: CodeByTushu
"""
class Solution:
    def minimumDeletions(self, s: str) -> int:
        b_count = 0
        deletions = 0

        for ch in s:
            if ch == 'b':
                b_count += 1
            else:
                # Either delete this 'a' or delete all previous 'b's
                deletions = min(deletions + 1, b_count)

        return deletions

/**
 * Problem: LeetCode 1653 - Minimum Deletions to Make String Balanced
 * Difficulty: Medium
 * Language: C++
 * Author: CodeByTushu
 */
#include <string>
#include <algorithm>

using namespace std;

class Solution {
public:
    int minimumDeletions(string s) {
        int bCount = 0;
        int deletions = 0;

        for (char c : s) {
            if (c == 'b') {
                bCount++;
            } else {
                // Either delete this 'a' or delete all previous 'b's
                deletions = min(deletions + 1, bCount);
            }
        }

        return deletions;
    }
};

/**
 * Problem: LeetCode 1653 - Minimum Deletions to Make String Balanced
 * Difficulty: Medium
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {string} s
 * @return {number}
 */
var minimumDeletions = function(s) {
    let bCount = 0;
    let deletions = 0;

    for (let i = 0; i < s.length; i++) {
        if (s[i] === 'b') {
            bCount++;
        } else {
            // Either delete this 'a' or delete all previous 'b's
            deletions = Math.min(deletions + 1, bCount);
        }
    }

    return deletions;
};