Edit Distance.cpp
/* Given two strings str1 & str2⏎
* and below operations that can⏎
* be performed on str1. Find⏎
* minimum number of edits⏎
* (operations) required to convert⏎
* 'str1' into 'str2'/⏎
* a. Insert⏎
* b. Remove⏎
* c. Replace⏎
* All of the above operations are ⏎
* of equal cost⏎
*/⏎
⏎
#include <iostream>⏎
#include <string>⏎
using namespace std;⏎
⏎
int min(int x, int y, int z)⏎
{⏎
return min(min(x, y), z);⏎
}⏎
⏎
/* A Naive recursive C++ program to find⏎
* minimum number of operations to convert⏎
* str1 to str2.⏎
* O(3^m)⏎
*/⏎
int editDist(string str1, string str2, int m, int n)⏎
{⏎
if (m == 0)⏎
return n;⏎
if (n == 0)⏎
return m;⏎
⏎
//If last characters are same then continue⏎
//for the rest of them.⏎
if (str1[m - 1] == str2[n - 1])⏎
return editDist(str1, str2, m - 1, n - 1);⏎
⏎
//If last not same, then 3 possibilities⏎
//a.Insert b.Remove c. Replace⏎
//Get min of three and continue for rest.⏎
return 1 + min(editDist(str1, str2, m, n - 1),⏎
editDist(str1, str2, m - 1, n),⏎
editDist(str1, str2, m - 1, n - 1));⏎
}⏎
⏎
/* A DP based program⏎
* O(m x n)⏎
*/⏎
int editDistDP(string str1, string str2, int m, int n)⏎
{⏎
⏎
//Create Table for SubProblems⏎
int dp[m + 1][n + 1];⏎
⏎
//Fill d[][] in bottom up manner⏎
for (int i = 0; i <= m; i++)⏎
{⏎
for (int j = 0; j <= n; j++)⏎
{⏎
//If str1 empty. Then add all of str2⏎
if (i == 0)⏎
dp[i][j] = j;⏎
⏎
//If str2 empty. Then add all of str1⏎
else if (j == 0)⏎
dp[i][j] = i;⏎
⏎
//If character same. Recur for remaining⏎
else if (str1[i - 1] == str2[j - 1])⏎
dp[i][j] = dp[i - 1][j - 1];⏎
⏎
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 str1 = "sunday";⏎
string str2 = "saturday";⏎
⏎
cout << editDist(str1, str2, str1.length(), str2.length()) << endl;⏎
cout << editDistDP(str1, str2, str1.length(), str2.length()) << endl;⏎
⏎
return 0;⏎
}⏎