cpp code for day 7 and 8 (#98)

Author: profgrammer

day7/C++/profgrammer_oneedit.cpp

@@ -0,0 +1,42 @@
+/*
+  *@author: profgrammer
+  *@date: 31-12-2018
+*/
+
+#include <bits/stdc++.h>
+#define inf 100000000
+using namespace std;
+
+string s1, s2;
+
+int dp[1500][1500];
+
+// function that returns minimum edits required to convert s1[0 .. i-1] into s2[0 .. j-1]
+int minEdit(int i, int j){
+  // base cases. if s1 finishes we need j insertions, if s2 finishes we need i insertions
+  if(i == 0) return j;
+  if(j == 0) return i;
+  // housekeeping for dp with memoisation
+  if(dp[i][j] != inf) return dp[i][j];
+  // if the last characters are the same, no need to change anything and move both pointers by 1 unit
+  if(s1[i-1] == s2[j-1]) {
+    return dp[i][j] = minEdit(i-1, j-1);
+  }
+  // replace s1[i] to be s2[j]. the strings to be checked are still s1[0 .. i-1] and s2[0 .. j-1] but the cost is +1
+  int minReplace = 1 + minEdit(i-1, j-1);
+  // delete s1[i], cost is +1 and the strings are now s1[0 .. i-1] and s2[0 .. j] because we need to compare the i-1th character with the jth character
+  int minDelete = 1 + minEdit(i-1, j); 
+  // insert a character into s1 at index i. cost is +1 and now the strings to be checked are s1[0 .. i] because the ith character isn't compared yet and s2[0 .. j-1]
+  int minInsert = 1 + minEdit(i, j-1);
+  // return min of these 3 values
+  return dp[i][j] = min(minReplace, min(minDelete, minInsert));
+}
+
+int main() {
+  for(int i = 0;i < 1500;i++){
+    for(int j = 0;j < 1500;j++) dp[i][j] = inf;
+  }
+  cin>>s1>>s2;
+  if(minEdit(s1.size(), s2.size()) <= 1) cout<<"The strings are 1 edit away\n";
+  else cout<<"The strings are not 1 edit away\n";
+}

day7/README.md

@@ -498,6 +498,52 @@ int main(){
 }
 ```
 
+### [C++ Solution](./C++/profgrammer_oneedit.cpp)
+```cpp
+/*
+  *@author: profgrammer
+  *@date: 31-12-2018
+*/
+
+#include <bits/stdc++.h>
+#define inf 100000000
+using namespace std;
+
+string s1, s2;
+
+int dp[1500][1500];
+
+// function that returns minimum edits required to convert s1[0 .. i-1] into s2[0 .. j-1]
+int minEdit(int i, int j){
+  // base cases. if s1 finishes we need j insertions, if s2 finishes we need i insertions
+  if(i == 0) return j;
+  if(j == 0) return i;
+  // housekeeping for dp with memoisation
+  if(dp[i][j] != inf) return dp[i][j];
+  // if the last characters are the same, no need to change anything and move both pointers by 1 unit
+  if(s1[i-1] == s2[j-1]) {
+    return dp[i][j] = minEdit(i-1, j-1);
+  }
+  // replace s1[i] to be s2[j]. the strings to be checked are still s1[0 .. i-1] and s2[0 .. j-1] but the cost is +1
+  int minReplace = 1 + minEdit(i-1, j-1);
+  // delete s1[i], cost is +1 and the strings are now s1[0 .. i-1] and s2[0 .. j] because we need to compare the i-1th character with the jth character
+  int minDelete = 1 + minEdit(i-1, j); 
+  // insert a character into s1 at index i. cost is +1 and now the strings to be checked are s1[0 .. i] because the ith character isn't compared yet and s2[0 .. j-1]
+  int minInsert = 1 + minEdit(i, j-1);
+  // return min of these 3 values
+  return dp[i][j] = min(minReplace, min(minDelete, minInsert));
+}
+
+int main() {
+  for(int i = 0;i < 1500;i++){
+    for(int j = 0;j < 1500;j++) dp[i][j] = inf;
+  }
+  cin>>s1>>s2;
+  if(minEdit(s1.size(), s2.size()) <= 1) cout<<"The strings are 1 edit away\n";
+  else cout<<"The strings are not 1 edit away\n";
+}
+```
+
 
 ## C Implementation
 

day8/Cpp/profgrammer_editdistance.cpp

@@ -0,0 +1,42 @@
+/*
+  *@author: profgrammer
+  *@date: 31-12-2018
+*/
+
+#include <bits/stdc++.h>
+#define inf 100000000
+using namespace std;
+
+string s1, s2;
+
+int dp[1500][1500];
+
+// function that returns minimum edits required to convert s1[0 .. i-1] into s2[0 .. j-1]
+int minEdit(int i, int j){
+  // base cases. if s1 finishes we need j insertions, if s2 finishes we need i insertions
+  if(i == 0) return j;
+  if(j == 0) return i;
+  // housekeeping for dp with memoisation
+  if(dp[i][j] != inf) return dp[i][j];
+  // if the last characters are the same, no need to change anything and move both pointers by 1 unit
+  if(s1[i-1] == s2[j-1]) {
+    return dp[i][j] = minEdit(i-1, j-1);
+  }
+  // replace s1[i] to be s2[j]. the strings to be checked are still s1[0 .. i-1] and s2[0 .. j-1] but the cost is +1
+  int minReplace = 1 + minEdit(i-1, j-1);
+  // delete s1[i], cost is +1 and the strings are now s1[0 .. i-1] and s2[0 .. j] because we need to compare the i-1th character with the jth character
+  int minDelete = 1 + minEdit(i-1, j); 
+  // insert a character into s1 at index i. cost is +1 and now the strings to be checked are s1[0 .. i] because the ith character isn't compared yet and s2[0 .. j-1]
+  int minInsert = 1 + minEdit(i, j-1);
+  // return min of these 3 values
+  return dp[i][j] = min(minReplace, min(minDelete, minInsert));
+}
+
+int main() {
+  for(int i = 0;i < 1500;i++){
+    for(int j = 0;j < 1500;j++) dp[i][j] = inf;
+  }
+  cin>>s1>>s2;
+  cout<<"The minimum edit distance is: ";
+  cout<<minEdit(s1.size(), s2.size())<<endl;
+}

day8/README.md

@@ -101,7 +101,7 @@ console.log(minEditDist('kitten', 'sitting'));
 
 ## C++ Implementation
 
-### [Solution 1](./Cpp/day8.cpp)
+### [Solution 1 by @dhruv-gupta14](./Cpp/day8.cpp)
 
 ```cpp
 /*
@@ -154,7 +154,53 @@ int main()
 }
 ``` 
 
-### [Solution 2 by @divyakhetan](./Cpp/EditDistanceday8.cpp)
+### [Solution 2 by @profgrammer](./Cpp/profgrammer_editdistance.cpp)
+```cpp
+/*
+  *@author: profgrammer
+  *@date: 31-12-2018
+*/
+
+#include <bits/stdc++.h>
+#define inf 100000000
+using namespace std;
+
+string s1, s2;
+
+int dp[1500][1500];
+
+// function that returns minimum edits required to convert s1[0 .. i-1] into s2[0 .. j-1]
+int minEdit(int i, int j){
+  // base cases. if s1 finishes we need j insertions, if s2 finishes we need i insertions
+  if(i == 0) return j;
+  if(j == 0) return i;
+  // housekeeping for dp with memoisation
+  if(dp[i][j] != inf) return dp[i][j];
+  // if the last characters are the same, no need to change anything and move both pointers by 1 unit
+  if(s1[i-1] == s2[j-1]) {
+    return dp[i][j] = minEdit(i-1, j-1);
+  }
+  // replace s1[i] to be s2[j]. the strings to be checked are still s1[0 .. i-1] and s2[0 .. j-1] but the cost is +1
+  int minReplace = 1 + minEdit(i-1, j-1);
+  // delete s1[i], cost is +1 and the strings are now s1[0 .. i-1] and s2[0 .. j] because we need to compare the i-1th character with the jth character
+  int minDelete = 1 + minEdit(i-1, j); 
+  // insert a character into s1 at index i. cost is +1 and now the strings to be checked are s1[0 .. i] because the ith character isn't compared yet and s2[0 .. j-1]
+  int minInsert = 1 + minEdit(i, j-1);
+  // return min of these 3 values
+  return dp[i][j] = min(minReplace, min(minDelete, minInsert));
+}
+
+int main() {
+  for(int i = 0;i < 1500;i++){
+    for(int j = 0;j < 1500;j++) dp[i][j] = inf;
+  }
+  cin>>s1>>s2;
+  cout<<"The minimum edit distance is: ";
+  cout<<minEdit(s1.size(), s2.size())<<endl;
+}
+```
+
+### [Solution 3 by @divyakhetan](./Cpp/EditDistanceday8.cpp)
 
 ```cpp
 /**
@@ -200,7 +246,7 @@ int main(){
 }
 ``` 
 
-### [Solution 3 by @aaditkamat] (./C++/levenshtein_distance.cpp)
+### [Solution 4 by @aaditkamat] (./C++/levenshtein_distance.cpp)
 
 ```cpp
 /**