Program to find the minimum edit distance between two strings in C++

Suppose we have two words S and T, we have to find the minimum number of operations needed to convert from S to T. The operations can be of three types, these are

insert a character,
delete a character
replace a character.

So if the input strings are “evaluate” and “fluctuate”, then the result will be 5.

To solve this, we will follow these steps −

n := size of s, m := size of t,
create an array dp of size n + 1
for i in range 0 to n
- dp[i] := new array of size m + 1
- for j in range 0 to m:
  - dp[i, j] := 0
  - if i = 0, then dp[i,j] = j
  - otherwise when j = 0, then dp[i, j] := i
s := blank space and concatenate s, t := blank space and concatenate t
for i in range 1 to n
- for j in range 1 to m
  - if s[i] is not t[j], then dp[i, j] := 1 + min of dp[i – 1, j], dp[i, j - 1], dp[i – 1, j – 1]
  - otherwise dp[i, j] := dp[i – 1, j – 1]
return dp[n, m]

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int minDistance(string s, string t) {
      int n = s.size();
      int m =t.size();
      int** dp = new int*[n+1];
      for(int i =0;i<=n;i++){
         dp[i] = new int[m+1];
         for(int j=0;j<=m;j++){
            dp[i][j]=0;
            if(i==0)dp[i][j]=j;
            else if(j==0)dp[i][j] = i;
         }
      }
      s = " " + s;
      t = " " + t;
      for(int i =1;i<=n;i++){
         for(int j = 1;j<=m;j++){
            if(s[i] !=t[j]){
               dp[i][j] = 1+min({dp[i-1][j],dp[i][j-1],dp[i-1][j-1]});
            }else{
               dp[i][j] = dp[i-1][j-1];
            }
         }
      }
      return dp[n][m];
   }
};
main(){
   Solution ob;
   cout << (ob.minDistance("fluctuate", "evaluate"));
}