Easy Solution with 100% beats in Java
The problem is asking us to find the number of pairs of strings in the array words such that one string is the reverse of another. The brute-force approach woul
The problem is asking us to find the number of pairs of strings in the array words such that one string is the reverse of another. The brute-force approach would involve checking every string and comparing it with the reversed form of every other string in the list. However, we can optimize this solution.

Approach
We can use a List<String> (or a Set for better lookup time) to track the words we've seen so far. For each word, we check if its reverse already exists in the list. If it does, we count it as a valid pair. Otherwise, we add the word to the list. This ensures that we're counting each valid pair exactly once.

Time Complexity:O(n)
O(n) time, and we need to check the reversed string in the list of previously seen words.

Space Complexity: O(n)
O(n), because we store each word in a list. In the worst case, all words are distinct, so we will store all the words.

# Code
```java []
class Solution {
    public int maximumNumberOfStringPairs(String[] words) {
        List<String> li =new ArrayList<>();
        int count=0;
        for(int i=0;i<words.length;i++){
            StringBuilder sb=new StringBuilder(words[i]);
            if(li.contains(sb.reverse().toString())){
                count++;
            }else{
                li.add(words[i]);
            }
        }
        return count;    
    }
}
```
['Array', 'Java']