LeetCode in Kotlin

1653. Minimum Deletions to Make String Balanced

Medium

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

class Solution {
    fun minimumDeletions(s: String): Int {
        var a = 0
        var b = 0
        for (ch in s.toCharArray()) {
            if (ch == 'a') {
                a++
            } else {
                b = Math.max(a, b) + 1
            }
        }
        return s.length - Math.max(a, b)
    }
}

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