TheAlgorithms-C-Plus-Plus/Dynamic Programming/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-2])
            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, str1, str1.length(), str2.length()) << endl;
    cout << editDistDP(str1, str1, str1.length(), str2.length()) << endl;

    return 0;
}