#400 edit_distance problem

Author: AbhiC7721

Dynamic Programming/Edit_distance.cpp

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+/* problem statement -> https://leetcode.com/problems/edit-distance/
+
+Description ->
+Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
+
+You have the following three operations permitted on a word:
+
+Insert a character
+Delete a character
+Replace a character
+
+
+Example -> word1 = "horse", word2 = "ros"
+Output -> 3
+horse -> rorse (replace 'h' with 'r')
+rorse -> rose (remove 'r')
+rose -> ros (remove 'e')
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+int minDistance(string word1, string word2) {
+        int n = word1.length();
+        int m = word2.length();
+        // dp array 
+        int dp[n+1][m+1];    
+        // empty string to empty -> no change
+        dp[0][0]=0;
+
+        //To convert a string to a empty string, or 
+        //To convert empty string to string use (length => i or j) operations
+        for(int j=1;j<m+1;j++)
+            dp[0][j]=j;
+        for(int i=1;i<n+1;i++)
+            dp[i][0]=i;
+        
+        for(int i=1;i<n+1;i++)
+        {
+            for(int j=1;j<m+1;j++)
+            {
+                // if word[i-1]==word[j-1]; dp[i][j] is same as dp[i-1][j-1]
+                if(word1[i-1]==word2[j-1])
+                    dp[i][j] = dp[i-1][j-1];
+                else
+                // 3 cases of insert(dp[i-1][j]), delete(dp[i][j-1]), replace(dp[i-1][j-1])+1
+                    dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1]))+1;
+            }
+        }
+        // ans is last element
+        return dp[n][m];
+    }
+
+    int main()
+    {
+        string w1;
+        string w2;
+        cin>>w1>>w2;
+        // calling the minDistance function
+        int ans = minDistance(w1, w2);
+        cout<<ans<<"\n";
+        return 0;
+    }