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

Write a C program to find the minimum number of operations required to convert string word1 to word2. You have the following three operations permitted on a word: insert a character, delete a character, or replace a character. This is also known as the Levenshtein distance.

Example 1
    
  
  • Input: word1 = "horse", word2 = "ros"
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  • Output: 3
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  • Explanation: 
    
      
      
    • Step 1: horse -> rorse (replace 'h' with 'r')
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    • Step 2: rorse -> rose (remove 'r')
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    • Step 3: rose -> ros (remove 'e')
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    • Step 4: Total operations = 3
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    •
Example 2
    
  
  • Input: word1 = "intention", word2 = "execution"
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  • Output: 5
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  • Explanation: 
    
      
      
    • Step 1: intention -> inention (remove 't')
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    • Step 2: inention -> enention (replace 'i' with 'e')
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    • Step 3: enention -> exention (replace 'n' with 'x')
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    • Step 4: exention -> exection (replace 'n' with 'c')
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    • Step 5: exection -> execution (insert 'u')
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    • Step 6: Total operations = 5
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    •
Constraints
    
  
  • 0 ≤ word1.length, word2.length ≤ 500
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  • word1 and word2 consist of lowercase English letters
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  • Time Complexity: O(m * n)
    • 
  
  • Space Complexity: O(m * n)
    •
StringsDynamic Programming HCL TechnologiesLTIMindtree
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Hint
Solution

Solution Hints

    
  
  • Use dynamic programming with a 2D array dp[i][j].
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  • dp[i][j] represents minimum operations to convert word1[0...i-1] to word2[0...j-1].
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  • Initialize base cases: dp[0][j] = j and dp[i][0] = i.
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  • If characters match, dp[i][j] = dp[i-1][j-1].
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  • If characters don't match, take minimum of three operations plus 1.
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  • The three operations are: insert, delete, replace.
    •

Steps to solve by this approach:

 Step 1: Create a 2D DP array where dp[i][j] represents minimum operations to convert word1[0...i-1] to word2[0...j-1].
 Step 2: Initialize base cases - dp[i][0] = i (delete all) and dp[0][j] = j (insert all).
 Step 3: For each cell, check if current characters match.
 Step 4: If characters match, copy value from dp[i-1][j-1] (no operation needed).
 Step 5: If characters don't match, take minimum of three operations (insert, delete, replace) and add 1.
 Step 6: Fill the entire DP table using the recurrence relation.
 Step 7: Return dp[m][n] which contains the minimum edit distance.

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