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Edit Distance

Certification: Advanced Level Accuracy: 0% Submissions: 0 Points: 15

Write a C# program to implement the EditDistance(string word1, string word2) function, which calculates the minimum number of operations required to convert word1 to word2. The allowed operations are:

  1. Insert a character
  2. Delete a character
  3. Replace a character
Example 1
    
  
  • Input: word1 = "horse", word2 = "ros"
    • 
  
  • Output: 3
    • 
  
  • Explanation: 
    
      
      
    • horse → rorse (replace 'h' with 'r')
      • 
      
    • rorse → rose (remove 'r')
      • 
      
    • rose → ros (remove 'e')
      • 
    
    
  
    •
Example 2
    
  
  • Input: word1 = "intention", word2 = "execution"
    • 
  
  • Output: 5
    • 
  
  • Explanation: 
    
      
      
    • intention → inention (remove 't')
      • 
      
    • inention → enention (replace 'i' with 'e')
      • 
      
    • enention → exention (replace 'n' with 'x')
      • 
      
    • exention → exection (replace 'n' with 'c')
      • 
      
    • exection → execution (insert 'u')
      • 
    
    
  
    •
Constraints
    
  
  • 0 <= word1.length, word2.length <= 500
    • 
  
  • word1 and word2 consist of lowercase English letters
    • 
  
  • Time Complexity: O(m * n) where m and n are lengths of the two strings
    • 
  
  • Space Complexity: O(m * n)
    •
ArraysDynamic Programming IBMGoldman Sachs
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Hint
Solution

Solution Hints

    
  
  • Use dynamic programming to solve this problem efficiently
    • 
  
  • Create a 2D array to store intermediate results
    • 
  
  • Consider all three operations at each step and choose the minimum
    • 
  
  • The value at dp[i][j] represents the minimum edit distance between word1[0...i-1] and word2[0...j-1]
    • 
  
  • Base cases: converting an empty string to a non-empty string requires insertions equal to the length
    •

Steps to solve by this approach:

 Step 1: Create a 2D array dp where dp[i][j] represents the minimum edit distance between word1[0...i-1] and word2[0...j-1].
 Step 2: Initialize base cases - converting an empty string to a non-empty string requires insertions equal to the length.
 Step 3: For each character in word1 and word2, compare the characters.
 Step 4: If characters match, no operation is needed, so dp[i][j] = dp[i-1][j-1].
 Step 5: If characters don't match, take the minimum of three operations: insertion (dp[i][j-1]), deletion (dp[i-1][j]), or replacement (dp[i-1][j-1]), and add 1.
 Step 6: Continue filling the dp table until we process all characters.
 Step 7: The value at dp[m][n] is the minimum edit distance between the two strings.

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