feat: add solutions to lc problem: No.2744~2746 (doocs#2207)

* No.2744.Find Maximum Number of String Pairs
* No.2745.Construct the Longest New String
* No.2746.Decremental String Concatenation

Author: yanglbme

solution/2700-2799/2744.Find Maximum Number of String Pairs/README_EN.md

@@ -58,6 +58,16 @@ It can be proven that 1 is the maximum number of pairs that can be formed.
 
 ## Solutions
 
+**Solution 1: Hash Table**
+
+We can use a hash table $cnt$ to store the number of occurrences of each string's reversed string in the array $words$.
+
+Traverse the array $words$, for each string $w$, we directly add the value of $cnt[w]$ to the answer, then increase the occurrence count of $w$'s reversed string by $1$.
+
+After the traversal ends, we can get the maximum number of matches.
+
+The time complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the sum of the lengths of the strings in the array $words$.
+
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 ### **Python3**

solution/2700-2799/2745.Construct the Longest New String/README_EN.md

@@ -40,6 +40,16 @@ That string has length 14, and we can show that it is impossible to construct a
 
 ## Solutions
 
+**Solution 1: Case Discussion**
+
+We observe that the string 'AA' can only be followed by 'BB', and the string 'AB' can be placed at the beginning or end of the string. Therefore:
+
+-   If $x < y$, we can first alternately place 'BBAABBAA..BB', placing a total of $x$ 'AA' and $x+1$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x \times 2 + z + 1) \times 2$;
+-   If $x > y$, we can first alternately place 'AABBAABB..AA', placing a total of $y$ 'BB' and $y+1$ 'AA', then place the remaining $z$ 'AB', with a total length of $(y \times 2 + z + 1) \times 2$;
+-   If $x = y$, we only need to alternately place 'AABB', placing a total of $x$ 'AA' and $y$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x + y + z) \times 2$.
+
+The time complexity is $O(1)$, and the space complexity is $O(1)$.
+
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 ### **Python3**

solution/2700-2799/2746.Decremental String Concatenation/README_EN.md

@@ -68,6 +68,16 @@ It can be shown that the minimum possible length of str<sub>2</sub> is 6.
 
 ## Solutions
 
+**Solution 1: Case Discussion**
+
+We observe that the string 'AA' can only be followed by 'BB', and the string 'AB' can be placed at the beginning or end of the string. Therefore:
+
+-   If $x < y$, we can first alternately place 'BBAABBAA..BB', placing a total of $x$ 'AA' and $x+1$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x \times 2 + z + 1) \times 2$;
+-   If $x > y$, we can first alternately place 'AABBAABB..AA', placing a total of $y$ 'BB' and $y+1$ 'AA', then place the remaining $z$ 'AB', with a total length of $(y \times 2 + z + 1) \times 2$;
+-   If $x = y$, we only need to alternately place 'AABB', placing a total of $x$ 'AA' and $y$ 'BB', then place the remaining $z$ 'AB', with a total length of $(x + y + z) \times 2$.
+
+The time complexity is $O(1)$, and the space complexity is $O(1)$.
+
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 ### **Python3**