MinimumNumberOfDeletionsAndInsertions using c++

Author: raunakpandey1

Dynamic Programming/MinimumNumberOfDeletionsAndInsertions.cpp

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+
+/*
+Given two strings str1 and str2. The task is to remove or insert the minimum number of characters from/in str1 so as to transform it into str2. It could be possible that the same character needs to be removed/deleted from one point of str1 and inserted to some another point.
+
+Example 1:
+
+Input: str1 = "heap", str2 = "pea"
+Output: 3
+Explanation: 2 deletions and 1 insertion
+p and h deleted from heap. Then, p is 
+inserted at the beginning One thing to 
+note, though p was required yet it was 
+removed/deleted first from its position 
+and then it is inserted to some other 
+position. Thus, p contributes one to the 
+deletion_count and one to the 
+insertion_count.*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+  
+class Solution{
+		
+
+	public:
+	int LCS(string X, string Y, int m, int n){
+        int dp[m+1][n+1];
+        memset(dp,-1,sizeof(dp));
+        for(int i=0;i<m+1;i++)
+        {
+            for(int j=0;j<n+1;j++)
+            {
+                if(i==0 || j==0) dp[i][j]=0;
+                else if(X[i-1] == Y[j-1]){
+                    dp[i][j] = 1+dp[i-1][j-1];
+                }
+                else{
+                    dp[i][j] = max(dp[i-1][j] , dp[i][j-1]);
+                }
+            }
+        }
+        return dp[m][n];
+    }
+	int minOperations(string str1, string str2) 
+	{ 
+	    // Your code goes 
+	    int len1 = str1.length();
+	    int len2 = str2.length();
+	    int lengthLCS = LCS(str1 , str2, len1, len2);
+	    return len1+ len2 - (2*lengthLCS);
+	    
+	} 
+};
+ 
+int main() 
+{
+   	
+   
+    int t;
+    cin >> t;
+    while (t--)
+    {
+        string s1, s2;
+        cin >> s1 >> s2;
+
+	    Solution ob;
+	    cout << ob.minOperations(s1, s2) << "\n";
+	     
+    }
+    return 0;
+}
+