Merge pull request #3583 from sjain1909/t10

Adding solution of levenshtein distance algorithm

Author: ajay-dhangar

dsa/Advance/levenshtein-distance-algorithm.md

@@ -0,0 +1,108 @@
+---
+id: levenshtein-distance
+title: Levenshtein Distance Algorithm
+sidebar_label: 0007 - Levenshtein Distance Algorithm
+tags: [Levenshtein Distance, String Matching, Algorithm, C++, Problem Solving]
+description: This is a solution for implementing the Levenshtein Distance Algorithm for measuring the difference between two sequences.
+---
+
+## Problem Statement 
+
+### Problem Description
+
+The Levenshtein Distance algorithm calculates the minimum number of single-character edits (insertions, deletions, or substitutions) required to change one string into another. It is widely used in applications like spell checking, DNA sequencing, and natural language processing.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input: 
+String1: "kitten"
+String2: "sitting"
+Output: 
+3
+Explanation: The Levenshtein Distance between "kitten" and "sitting" is 3 (kitten -> sitten -> sittin -> sitting).
+```
+
+### Constraints
+- The length of both strings can be up to 10^3.
+
+## Solution of Given Problem
+
+### Intuition and Approach
+
+The Levenshtein Distance algorithm uses dynamic programming to efficiently compute the edit distance between two strings. It constructs a matrix where the cell at position (i, j) contains the Levenshtein Distance between the first i characters of the first string and the first j characters of the second string.
+
+### Approaches
+
+#### Codes in Different Languages
+
+<Tabs>
+  <TabItem value="cpp" label="C++">
+  <SolutionAuthor name="sjain1909"/>
+   ```cpp
+    #include <bits/stdc++.h>
+    using namespace std;
+
+    int levenshteinDistance(const string& str1, const string& str2) {
+    int len1 = str1.size();
+    int len2 = str2.size();
+    vector<vector<int>> dp(len1 + 1, vector<int>(len2 + 1));
+
+    for (int i = 0; i <= len1; ++i) {
+        for (int j = 0; j <= len2; ++j) {
+            if (i == 0) {
+                dp[i][j] = j;
+            } else if (j == 0) {
+                dp[i][j] = i;
+            } else if (str1[i - 1] == str2[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1];
+            } else {
+                dp[i][j] = 1 + min({dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]});
+            }
+        }
+    }
+
+    return dp[len1][len2];
+}
+
+int main() {
+    string str1, str2;
+    cout << "Enter the first string: ";
+    cin >> str1;
+    cout << "Enter the second string: ";
+    cin >> str2;
+
+    int distance = levenshteinDistance(str1, str2);
+    cout << "The Levenshtein Distance between \"" << str1 << "\" and \"" << str2 << "\" is " << distance << ".\n";
+
+    return 0;
+}
+    ```
+  </TabItem>  
+</Tabs>
+
+### Complexity Analysis
+
+- **Time Complexity:** $O(N * M)$ where N is the length of the first string and M is the length of the second string.
+- **Space Complexity:** $O(N * M)$ for the dynamic programming matrix.
+
+## Video Explanation of Given Problem
+
+    <LiteYouTubeEmbed
+      id="obWXjtg0L64"
+      params="autoplay=1&autohide=1&showinfo=0&rel=0"
+      title="Problem Explanation | Solution | Approach"
+      poster="maxresdefault"
+      webp 
+    />
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['sjain1909'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>