1544. Make The String Great

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";

Approach

Initialize a Stack

Create an empty stack to store characters.

Iterate Through String

Loop through each character in the input string.

Check Stack Status

If the stack is empty, push the current character onto the stack.
If the stack is not empty, proceed to the next step.

Check for Case-Insensitive Match

Calculate the absolute difference between the ASCII values of the current character and the character at the top of the stack.
If the difference is 32 (indicating a case-insensitive match), pop the character from the stack.
If there is no match, push the current character onto the stack.

Construct Final String

After processing all characters in the input string, pop characters from the stack and construct the final string.

Return Final String

Return the constructed final string as the result.

Solution

class Solution {
public:
    string makeGood(string s) {
        stack<char> st;
        for (int i = 0; i < s.size(); i++) {
            if (st.empty())
                st.push(s[i]);
            else if (abs(st.top() - s[i]) == 32) // 32 is the difference between the ASCII values of the upper and lower case letters
                st.pop();
            else
                st.push(s[i]);
        }
 
        string ans = "";
        while (!st.empty()) {
            ans = st.top() + ans;
            st.pop();
        }
        return ans;
    }
};

Time Complexity: O(N), Space Complexity: O(N)

Patterns Path Sum