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Leetcode 1963: Minimum Number of Swaps to Make the String Balanced

DevGod
DevGod
Elf Vtuber

About

This leetcode problem is basically us to find the minimum number of swap operations are needed to make a string of ”[]” symbols balanced. Balanced being if each opening ”["" symbol also has a closing ”]” symbol. For example, the following would be examples of balanced strings.

  1. []
  2. [[]]
  3. [][][]

Thoughts

One thing to notice off the bat is when considering certain testcases, it is already possible that a string already contains a balanced string. For example, testcase ]][][[ already has a balanced string in the middle, so it is the same as ]][[. For example, all testcases either eventually become testcases such as ][, ]][[, ]]][[[ ect.

After we see this, it is now about finding a remainder inside our testcase string of extra unmatched chars. For example, in ][ the remained would be one, which can be viewed either as having one unmatched opening char, or one unmatched closed char.

After drawing out these certain testcases, I saw a certain pattern involving the remaining total of unmatched charecters in the sring.

  1. ][ = 1 Swaps needed
  2. ]][[ = 1 Swaps needed
  3. ]]][[[ = 2 Swaps needed
  4. ]]]][[[[ = 2 Swaps needed
  5. ]]]]][[[[[ = 3 Swaps needed

The pattern being, that for the remaining amount of unmatched symbols at the end of the string is then Math Ceil (remaining amount of unmatched chars / 2) is the number of swaps needed.

Implementation

There are two ways you can solve this problem. One way is to use a stack data structure to keep track of previous symbols, and then pushing and popping accordingly. However, what I ended up doing to save memory usage was to simulate the stack size instead. Since we only need to return the minimal amount of swaps needed, we don’t need to have a stack, only the state of the would be stack’s size / length. Once I for loop to simulate the stack size, I simply just return the Math.ceil(remainder / 2) to return the minimal amount of swaps needed.

Leetcode 2696: Minimum String Length After Removing Substrings

DevGod
DevGod
Elf Vtuber

About

This problem is basically asking after you replace every occurrence of a substring “AB” or “CD” in your testcase string, return the length of the new string. For example:

  1. “ABCDZ” => “Z” => return 1;
  2. “QABABABQ” => “QQ” => return 2;
  3. “ACDB” => "" return 0;

Thoughts

The first two example testcases are pretty straight forward. Just remove “AB” substrings, then remove all “CD” substrings. However, the third testcase gets interesting where we remove “AB”, then remove “CD”. However, after removing the “CD” we are left with “AB”, which can then be removed again to return a blank string "" / a return string length of 0. After seeing this I immediately just wanted to use regular expressions, as that would be perfect for finding substrings present in my main string, as well as for preforming replaceAll operations to remove “AB” and “CD” substrings from the main testcase string.

Implementation

I simply just made a while loop that only ends once my given String s no longer has a “AB” or a “CD” substring present inside of itself. This is done by using the regular expression /AB|CD/, which just checks for any occurrence of an “AB” or “CD”. The .test(s) then just takes my regular expressions, searches the string using said regular expression, then returns true or false.

Inside my while loop, I then use s.repalceAll and /AB|CD/g to replace all occurrences of ABs and CDs from my current string. The regular expression is pretty similar to the top, I just use G to make sure its global since it’s a replace all.

Lastly, once my String S no longer has any occurrences of substrings “AB” or “CD”, I exit my while loop and then just return the length of my now AB and CD substring freed string.