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

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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"));
}

## Input

"fluctuate"
"evaluate"

## Output

5