feat: add solutions to lc problem: No.0383

No.0383.Ransom Note

Author: yanglbme

solution/0000-0099/0076.Minimum Window Substring/README_EN.md

@@ -49,7 +49,7 @@ Since the largest window of s only has one 'a', return empty string.
 
 ## Solutions
 
-**Approach 1: Counting + Double Pointers**
+**Approach 1: Counting + Two Pointers**
 
 We use a hash table or array $need$ to count the number of characters in string $t$, and use another hash table or array $window$ to count the number of characters in the sliding window. In addition, define two pointers $j$ and $i$ pointing to the left and right boundaries of the window, and the variable $cnt$ represents the number of characters in $t$ that are already included in the window. The variables $k$ and $mi$ represent the starting position and length of the minimum cover substring respectively.
 

solution/0300-0399/0383.Ransom Note/README.md

@@ -149,6 +149,25 @@ function canConstruct(ransomNote: string, magazine: string): boolean {
 }
 ```
 
+### **C#**
+
+```cs
+public class Solution {
+    public bool CanConstruct(string ransomNote, string magazine) {
+        int[] cnt = new int[26];
+        foreach (var c in magazine) {
+            ++cnt[c - 'a'];
+        }
+        foreach (var c in ransomNote) {
+            if (--cnt[c - 'a'] < 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
 ### **PHP**
 
 ```php

solution/0300-0399/0383.Ransom Note/README_EN.md

@@ -29,6 +29,14 @@
 
 ## Solutions
 
+**Approach 1: Hash Table or Array**
+
+We can use a hash table or an array $cnt$ of length $26$ to record the number of times each character appears in the string `magazine`. Then traverse the string `ransomNote`, for each character $c$ in it, we decrease the number of $c$ by $1$ in $cnt$. If the number of $c$ is less than $0$ after the decrease, it means that the number of $c$ in `magazine` is not enough, so it cannot be composed of `ransomNote`, just return $false$.
+
+Otherwise, after the traversal, it means that each character in `ransomNote` can be found in `magazine`. Therefore, return $true$.
+
+The time complexity is $O(m + n)$, and the space complexity is $O(C)$. Where $m$ and $n$ are the lengths of the strings `ransomNote` and `magazine` respectively; and $C$ is the size of the character set, which is $26$ in this question.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -118,6 +126,25 @@ function canConstruct(ransomNote: string, magazine: string): boolean {
 }
 ```
 
+### **C#**
+
+```cs
+public class Solution {
+    public bool CanConstruct(string ransomNote, string magazine) {
+        int[] cnt = new int[26];
+        foreach (var c in magazine) {
+            ++cnt[c - 'a'];
+        }
+        foreach (var c in ransomNote) {
+            if (--cnt[c - 'a'] < 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
 ### **PHP**
 
 ```php

solution/0300-0399/0383.Ransom Note/Solution.cs

@@ -0,0 +1,14 @@
+public class Solution {
+    public bool CanConstruct(string ransomNote, string magazine) {
+        int[] cnt = new int[26];
+        foreach (var c in magazine) {
+            ++cnt[c - 'a'];
+        }
+        foreach (var c in ransomNote) {
+            if (--cnt[c - 'a'] < 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}