From 48053e440944afd6d40644aad3c6f700d8cd72c3 Mon Sep 17 00:00:00 2001
From: Srishti Dubey <56927445+srieeee@users.noreply.github.com>
Date: Wed, 23 Oct 2019 21:03:12 +0530
Subject: [PATCH] adding edit distance problem

---
 .../dynamic programming/edist.cpp             | 50 +++++++++++++++++++
 1 file changed, 50 insertions(+)
 create mode 100644 Data Structure & Algorithm/dynamic programming/edist.cpp

diff --git a/Data Structure & Algorithm/dynamic programming/edist.cpp b/Data Structure & Algorithm/dynamic programming/edist.cpp
new file mode 100644
index 0000000..8b6d287
--- /dev/null
+++ b/Data Structure & Algorithm/dynamic programming/edist.cpp	
@@ -0,0 +1,50 @@
+// A Naive recursive C++ program to find minimum number 
+// operations to convert str1 to str2 
+#include <bits/stdc++.h> 
+using namespace std; 
+
+// Utility function to find minimum of three numbers 
+int min(int x, int y, int z) 
+{ 
+	return min(min(x, y), z); 
+} 
+
+int editDist(string str1, string str2, int m, int n) 
+{ 
+	// If first string is empty, the only option is to 
+	// insert all characters of second string into first 
+	if (m == 0) 
+		return n; 
+
+	// If second string is empty, the only option is to 
+	// remove all characters of first string 
+	if (n == 0) 
+		return m; 
+
+	// If last characters of two strings are same, nothing 
+	// much to do. Ignore last characters and get count for 
+	// remaining strings. 
+	if (str1[m - 1] == str2[n - 1]) 
+		return editDist(str1, str2, m - 1, n - 1); 
+
+	// If last characters are not same, consider all three 
+	// operations on last character of first string, recursively 
+	// compute minimum cost for all three operations and take 
+	// minimum of three values. 
+	return 1 + min(editDist(str1, str2, m, n - 1), // Insert 
+				editDist(str1, str2, m - 1, n), // Remove 
+				editDist(str1, str2, m - 1, n - 1) // Replace 
+				); 
+} 
+
+// Driver program 
+int main() 
+{ 
+	// your code goes here 
+	string str1 = "sunday"; 
+	string str2 = "saturday"; 
+
+	cout << editDist(str1, str2, str1.length(), str2.length()); 
+
+	return 0; 
+}