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+# 2486. Append Characters to String to Make Subsequence
+
+## **Date**: June 3, 2024
+**Difficulty**: ![Medium](https://img.shields.io/badge/Medium-Yellow)  
+**Related Topics**: ![String](https://img.shields.io/badge/String-blue) ![TwoPointer](https://img.shields.io/badge/TwoPointer-blue)
+
+<p>
+  <a href="https://github.com/Hasheditz/Leetcode-CSES-GFG-Codeforces-Coding-Solutions?tab=readme-ov-file#append-characters-to-string-to-make-subsequence" style="margin-right: 5px;">
+    <img src="https://img.shields.io/badge/All%20Problem%20Solutions-green" alt="All Problem Solutions">
+  </a>
+  <a href="https://leetcode.com/problems/append-characters-to-string-to-make-subsequence/">
+    <img src="https://img.shields.io/badge/Link%20To%20The%20Question-blue" alt="Link To The Question">
+  </a>
+</p>
+
+## Editorial
+
+This problem requires us to determine the minimum number of characters that need to be appended to the end of a string `s` so that another string `t` becomes a subsequence of `s`.
+
+### Solution Explanation
+
+To solve this problem efficiently, we can use a two-pointer technique. The idea is to iterate through both strings `s` and `t` simultaneously to find the number of characters in `t` that are not in `s`. This will tell us the number of characters we need to append to `s`.
+
+#### Key Steps:
+1. **Initialize Pointers**: Use two pointers, `temp1` for iterating through `s` and `temp2` for iterating through `t`.
+2. **Iterate through the Strings**:
+   - If the current character in `t` (`t[temp2]`) matches the current character in `s` (`s[temp1]`), increment both pointers.
+   - If they do not match, only increment the pointer for `s` (`temp1`).
+3. **Calculate the Result**: After the loop, the difference between the length of `t` and the pointer `temp2` gives the number of characters that need to be appended.
+
+### Code
+
+```cpp
+class Solution {
+public:
+    int appendCharacters(string s, string t) {
+        int n = s.size(), m = t.size();
+        int temp1 = 0, temp2 = 0;
+
+        while (temp1 < n && temp2 < m) {
+            if (t[temp2] == s[temp1]) {
+                temp1++;
+                temp2++;
+            } 
+			else {
+                temp1++;
+            }
+        }
+
+        return m - temp2;
+    }
+};
+```
+### Complexity Analysis
+- **Time Complexity**: O(n + m), where `n` is the length of `s` and `m` is the length of `t`. This is because we are iterating through both strings once.
+- **Space Complexity**: O(1), as we are using a constant amount of extra space.
+
+## Like and Upvote
+
+If you found this solution helpful, please consider liking 👍 and upvoting ⬆️. Your support helps me to keep providing high-quality solutions.
+
+<p align="center">
+  <img src="https://preview.redd.it/petition-to-change-the-upvote-and-downvote-button-to-like-v0-jbrdq402054c1.jpg?width=640&crop=smart&auto=webp&s=8225d21c98a245f44fd6c1f74a4c6c67f0061f25" width="280">
+</p>