From 0607b7cbdc59d13503da55b094139c2130456dcb Mon Sep 17 00:00:00 2001
From: sakssingh <85640472+sakssingh@users.noreply.github.com>
Date: Tue, 14 Sep 2021 00:31:10 +0530
Subject: [PATCH] chore(CPlusPlus): Add Edit Distance (#458)

Co-authored-by: Arsenic <54987647+Arsenic-ATG@users.noreply.github.com>
---
 .../Dynamic-Programming/edit-distance.cpp     | 72 +++++++++++++++++++
 algorithms/CPlusPlus/README.md                |  1 +
 2 files changed, 73 insertions(+)
 create mode 100644 algorithms/CPlusPlus/Dynamic-Programming/edit-distance.cpp

diff --git a/algorithms/CPlusPlus/Dynamic-Programming/edit-distance.cpp b/algorithms/CPlusPlus/Dynamic-Programming/edit-distance.cpp
new file mode 100644
index 00000000..6eeb8996
--- /dev/null
+++ b/algorithms/CPlusPlus/Dynamic-Programming/edit-distance.cpp
@@ -0,0 +1,72 @@
+// Given two strings 's1' and 's2' and below operations that can performed on 's1'.
+// 1.Insert
+// 2.Remove
+// 3.Replace
+//Find minimum number of edits (operations) required to convert ‘s1’ into ‘s2’.  
+
+// NOTE: All above operations are of equal cost
+// Input:   s1 = "car", s2 = "cars"
+// Output:  1
+// We can convert s1 into s2 by inserting a 's'.
+
+//Program using Dynamic Programming
+//TIME COMPLEXITY: O(m*n) 
+
+#include <iostream>
+#include <vector>
+using namespace std;
+ 
+// Utility function to find the minimum of three numbers
+int min(int x, int y, int z) { return min(min(x, y), z); }
+ 
+int editDistDP(string s1, string s2, int m, int n)
+{
+    vector<vector<int>> dp(m + 1, vector<int>(n + 1));
+ 
+    for (int i = 0; i <= m; i++) {
+        for (int j = 0; j <= n; j++) {
+            // If first string is empty, only option is to
+            // insert all characters of second string
+            if (i == 0)
+                dp[i][j] = j; // Min. operations = j
+ 
+            // If second string is empty, only option is to
+            // remove all characters of second string
+            else if (j == 0)
+                dp[i][j] = i; // Min. operations = i
+ 
+            // If last characters are same, ignore last char
+            // and recur for remaining string
+            else if (s1[i - 1] == s2[j - 1])
+                dp[i][j] = dp[i - 1][j - 1];
+ 
+            // If the last character is different, consider
+            // all possibilities and find the minimum
+            else
+                dp[i][j]
+                    = 1
+                      + min(dp[i][j - 1], // Insert
+                            dp[i - 1][j], // Remove
+                            dp[i - 1][j - 1]); // Replace
+        }
+    }
+ 
+    return dp[m][n];
+}
+ 
+int main()
+{
+    string s1, s2;
+    cout << "Enter 1st string: " << '\n'; 
+    cin >> s1;
+    cout << "Enter 2nd string: " << '\n';
+    cin >> s2;
+    int n = s1.length();
+    int m = s2.length();
+    cout << editDistDP(s1, s2, n, m);
+    return 0;
+}
+
+// s1: food
+// s2: money
+// output: 4
diff --git a/algorithms/CPlusPlus/README.md b/algorithms/CPlusPlus/README.md
index 393d9be7..bae45a72 100644
--- a/algorithms/CPlusPlus/README.md
+++ b/algorithms/CPlusPlus/README.md
@@ -21,6 +21,7 @@
 2. [Longest Common Substring](Dynamic-Programming/longest-common-substring.cpp)
 3. [0/1-knapsack](Dynamic-Programming/01-knapsack.cpp)
 4. [Matrix chain Multiplication](Dynamic-Programming/matrix-chain-multiplication.cpp)
+5. [Edit Distance](Dynamic-Programming/edit-distance.cpp)
 
 ## Graphs