Create minDistance.cpp

Author: SteffiKeranJ

Dynamic-Programming/minDistance.cpp

@@ -0,0 +1,58 @@
+/** 
+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 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')
+**/
+
+class Solution {
+public:
+    int minDistance(string word1, string word2) {
+        int m = word1.size(); // source
+        int n = word2.size(); // destination
+        
+        if(m == 0 && n == 0) return 0;
+        int dp[501][501] = {0};
+        
+        int i, j; // iterators
+        for(i = 1; i <=m; ++i) {
+            dp[0][i] = i;
+        } 
+        for(i = 1; i <=n;++i) {
+            dp[i][0] = i;
+        }
+        for(i = 1; i <=n; ++i) {
+            for(int j = 1; j <= m; ++j) {
+                if(word1[j-1] == word2[i-1]) 
+                    dp[i][j] = dp[i-1][j-1];
+                else {
+                    dp[i][j] = 1 + min(dp[i-1][j-1], min(dp[i][j-1], dp[i-1][j]));
+                }
+            }
+        }
+        return dp[n][m];
+    }
+};