tags: leetcode easy string stack

1544. Make The String Great

Description

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:

• $0 \leq i \leq s.length - 2$
• 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.

Examples

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:

• $1 \leq s.length \leq 100$
• s contains only lower and upper case English letters

Code

char * makeGood(char * s){
    int length = strlen(s);
    //If input is too short, just return
    if( (length <= 1) )
        return s;
    //Linear search for bad case and modify it
    restart:
    length = strlen(s);
    for(int i = 0 ; i < length - 1 ; i++){
        if ( (s[i] == s[i+1] + 32) || (s[i] == s[i+1] - 32) ){
            for(; i < length - 2; i++)
                s[i] = s[i + 2];
            s[length - 2] = '\0';
            s = (char *)realloc(s, (length - 1) * sizeof(char));
            goto restart;
        }
    }
    return s;
}

Complexity

Space	Time
$O(1)$	$O(N^2)$

Result

• Runtime: 7 ms, faster than 20.97%
• Memory Usage: 6.1 MB, less than 7.73%

Notes