1544 - Make The String Great (Easy)

Problem Link

https://leetcode.com/problems/make-the-string-great/

Problem Statement

Given a string s of lower and upper case English letters.

A good string is a string which doesn't have two adjacent characters s[i] and s[i + 1] where:

• 0 <= i <= s.length - 2
• s[i] is a lower-case letter and s[i + 1] is the same letter but in upper-case or vice-versa.

To make the string good, you can choose two adjacent characters that make the string bad and remove them. You can keep doing this until the string becomes good.

Return the string after making it good. The answer is guaranteed to be unique under the given constraints.

Notice that an empty string is also good.

Example 1:

Input: s = "leEeetcode"
Output: "leetcode"
Explanation: In the first step, either you choose i = 1 or i = 2, both will result "leEeetcode" to be reduced to "leetcode".

Example 2:

Input: s = "abBAcC"
Output: ""
Explanation: We have many possible scenarios, and all lead to the same answer. For example:
"abBAcC" --> "aAcC" --> "cC" --> ""
"abBAcC" --> "abBA" --> "aA" --> ""

Example 3:

Input: s = "s"
Output: "s"

Constraints:

• 1 <= s.length <= 100
• s contains only lower and upper case English letters.

Approach 1: String as Stack

C++

Written by @wingkwong

class Solution {
public:
    string makeGood(string s) {
        // we can use stack as well but we need to build the final string at the end
        // hence, using string is enough
        string t;
        // for each character
        for (auto c : s) {
            // as long as `t` has a character, we check if the last character is same letter but in upper-case or vice-versa
            // here we can use XOR and 1 << 5 to convert a lower character to a upper one and vice-versa
            // A: 01[0]00001
            // a: 01[1]00001
            // Z: 01[0]11010
            // z: 01[1]11010
            // a -> A / A -> a
            if (t.size() && (t.back() ^ (1 << 5)) == c) {
                // if it matches the requirement, we remove the last character in `t`
                t.pop_back();
            } else {
                // otherweise, we add the current char to `t`
                t.push_back(c);
            }
        }
        return t;
    }
};

Java

Written by @wingkwong

class Solution {
    public String makeGood(String s) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            // as long as `sb` has a character, 
            // we check if the last character is same letter but in upper-case or vice-versa
            // here we can use XOR and 1 << 5 to convert a lower character to a upper one and vice-versa
            // A: 01[0]00001
            // a: 01[1]00001
            // Z: 01[0]11010
            // z: 01[1]11010
            // a -> A / A -> a
            if (sb.length() > 0  && ((sb.charAt(sb.length() - 1) ^ (1 << 5)) == s.charAt(i))) {
                // if it matches the requirement, we remove the last character in `sb`
                sb.deleteCharAt(sb.length() - 1);
            } else {
                // otherwise, we add the current char to `sb`
                sb.append(s.charAt(i));
            }
        }
        return sb.toString();
    }
}

Kotlin

Written by @wingkwong

class Solution {
    fun makeGood(s: String): String {
        var t : String = ""
        // for each character
        for (i in 0 .. s.length - 1) {
            // as long as `t` has a character, we check if the last character is same letter but in upper-case or vice-versa
            // here we can use XOR and 1 << 5 to convert a lower character to a upper one and vice-versa
            // A: 01[0]00001
            // a: 01[1]00001
            // Z: 01[0]11010
            // z: 01[1]11010
            // a -> A / A -> a
            if (t.length > 0 && (t.last() ^ s[i] == (1 shl 5)) {
                // if it matches the requirement, we remove the last character in `t`
                t = t.dropLast(1)
            } else {
                // otherweise, we add the current char to `t`
                t += s[i]
            }
        }
        return t
    }
}