Added: DP algorithm - Edit Distance

dsa/Algorithms/Dynamic Programming/09-Edit-Distance.md

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+---
+id: edit-distance
+title: Edit Distance using Dynamic Programming
+sidebar_label: Edit Distance
+tags: [python, java, c++, javascript, programming, algorithms, dynamic programming, tutorial, in-depth]
+description: In this tutorial, we will learn about the Edit Distance problem and its solution using Dynamic Programming in Python, Java, C++, and JavaScript with detailed explanations and examples.
+---
+
+# Edit Distance using Dynamic Programming
+
+Edit Distance, also known as Levenshtein distance, is a classic problem in computer science that measures the minimum number of operations required to transform one string into another. The allowed operations are insertion, deletion, and substitution.
+
+## Problem Statement
+
+Given two strings, find the minimum number of operations required to transform one string into the other.
+
+### Intuition
+
+The problem can be solved efficiently using dynamic programming by breaking it down into subproblems. The idea is to create a DP table where `dp[i][j]` represents the edit distance between the first `i` characters of the first string and the first `j` characters of the second string.
+
+## Dynamic Programming Approach
+
+Using dynamic programming, we fill the table based on the recurrence relation:
+- If the characters are the same, no new operation is needed.
+- If the characters are different, consider the minimum cost of the three operations: insertion, deletion, and substitution.
+
+## Pseudocode for Edit Distance using DP
+
+#### Initialize:
+
+```markdown
+for i from 0 to m:
+    dp[i][0] = i
+for j from 0 to n:
+    dp[0][j] = j
+
+for i from 1 to m:
+    for j from 1 to n:
+        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[m][n]
+```
+
+### Example Output:
+
+Given the strings:
+- str1 = "kitten"
+- str2 = "sitting"
+
+The minimum edit distance between `str1` and `str2` is `3`.
+
+### Output Explanation:
+
+The edit distance of `3` can be achieved by the following operations:
+
+- Replace 'k' with 's': kitten -> sitten
+- Replace 'e' with 'i': sitten -> sittin
+- Insert 'g' at the end: sittin -> sitting
+
+Therefore, the output is: `Minimum edit distance is: 3`
+
+## Implementing Edit Distance using DP
+
+### Python Implementation
+
+```python
+def edit_distance(str1, str2):
+    m, n = len(str1), len(str2)
+    dp = [[0 for _ in range(n + 1)] for _ in range(m + 1)]  # DP table initialization
+
+    for i in range(m + 1):
+        for j in range(n + 1):
+            if i == 0:
+                dp[i][j] = j  # If first string is empty, insert all characters of second string
+            elif j == 0:
+                dp[i][j] = i  # If second string is empty, remove all characters of first string
+            elif str1[i - 1] == str2[j - 1]:
+                dp[i][j] = dp[i - 1][j - 1]
+            else:
+                dp[i][j] = 1 + min(dp[i - 1][j],    # Remove
+                                   dp[i][j - 1],    # Insert
+                                   dp[i - 1][j - 1])  # Replace
+
+    return dp[m][n]
+
+str1 = "kitten"
+str2 = "sitting"
+print("Minimum edit distance is:", edit_distance(str1, str2))
+
+```
+
+### Java Implementation
+
+```java
+public class EditDistance {
+    public static int editDistance(String str1, String str2) {
+        int m = str1.length();
+        int n = str2.length();
+        int[][] dp = new int[m + 1][n + 1];  // DP table initialization
+
+        for (int i = 0; i <= m; i++) {
+            for (int j = 0; j <= n; j++) {
+                if (i == 0) {
+                    dp[i][j] = j;  // If first string is empty, insert all characters of second string
+                } else if (j == 0) {
+                    dp[i][j] = i;  // If second string is empty, remove all characters of first string
+                } else if (str1.charAt(i - 1) == str2.charAt(j - 1)) {
+                    dp[i][j] = dp[i - 1][j - 1];
+                } else {
+                    dp[i][j] = 1 + Math.min(dp[i - 1][j],    // Remove
+                                            Math.min(dp[i][j - 1],    // Insert
+                                                     dp[i - 1][j - 1]));  // Replace
+                }
+            }
+        }
+
+        return dp[m][n];
+    }
+
+    public static void main(String[] args) {
+        String str1 = "kitten";
+        String str2 = "sitting";
+        System.out.println("Minimum edit distance is: " + editDistance(str1, str2));
+    }
+}
+
+```
+### C++ Implementation
+
+```cpp
+#include <iostream>
+#include <vector>
+#include <algorithm>
+using namespace std;
+
+int editDistance(string str1, string str2) {
+    int m = str1.length();
+    int n = str2.length();
+    vector<vector<int>> dp(m + 1, vector<int>(n + 1));  // DP table initialization
+
+    for (int i = 0; i <= m; i++) {
+        for (int j = 0; j <= n; j++) {
+            if (i == 0) {
+                dp[i][j] = j;  // If first string is empty, insert all characters of second string
+            } else if (j == 0) {
+                dp[i][j] = i;  // If second string is empty, remove all characters of first string
+            } 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],    // Remove
+                                   dp[i][j - 1],    // Insert
+                                   dp[i - 1][j - 1]});  // Replace
+            }
+        }
+    }
+
+    return dp[m][n];
+}
+
+int main() {
+    string str1 = "kitten";
+    string str2 = "sitting";
+    cout << "Minimum edit distance is: " << editDistance(str1, str2) << endl;
+    return 0;
+}
+
+```
+
+### JavaScript Implementation
+
+```javascript
+function editDistance(str1, str2) {
+    let m = str1.length;
+    let n = str2.length;
+    let dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));  // DP table initialization
+
+    for (let i = 0; i <= m; i++) {
+        for (let j = 0; j <= n; j++) {
+            if (i === 0) {
+                dp[i][j] = j;  // If first string is empty, insert all characters of second string
+            } else if (j === 0) {
+                dp[i][j] = i;  // If second string is empty, remove all characters of first string
+            } else if (str1[i - 1] === str2[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1];
+            } else {
+                dp[i][j] = 1 + Math.min(dp[i - 1][j],    // Remove
+                                        dp[i][j - 1],    // Insert
+                                        dp[i - 1][j - 1]);  // Replace
+            }
+        }
+    }
+
+    return dp[m][n];
+}
+
+let str1 = "kitten";
+let str2 = "sitting";
+console.log("Minimum edit distance is:", editDistance(str1, str2));
+
+```
+
+## Complexity Analysis
+
+- Time Complexity: $O(m \times n)$, where m and n are the lengths of the two strings.
+- Space Complexity: $O(m \times n)$, for storing the DP table.
+
+## Conclusion
+
+Dynamic programming provides an efficient solution for the Edit Distance problem by breaking it into subproblems and storing intermediate results. This technique can be extended to solve other problems with overlapping subproblems and optimal substructure properties.