C - Dynamic Programming - Edit Distance

Question

(I have simplified my code after reading advice)

I am writing a program to take 2 strings and return the smallest editing distance as int.

eg. str1 = ab, str2 = ab ; //distance will be 0. 
(when both char of str1 and str2 are the same, distance will be 0)

eg. str1 = abc, str2 = c ; distance will be 2.

In my code,I have used below strings.

str1 = editing
str2 = distance

correct answer should be 5 (e, s, i&a, g&c, e), but my program returns 6. I have tried other strings eg, (short & ports) which gives the answer what it is supposed to be.

I have tried to debug, but I do not understand why sometimes those 3 options (ans1, ans2, ans3) in the function "check" give correct outcome, sometimes they dont.

 ans1 is the result when index of str1 +1
 ans2 is the result when index of str2 +1
 ans3 is the result when index of both str1 and str2 +1

I have read my code many times and I cant find what went wrong. please help me to understand my problems.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int check(const char* str1, const char* str2, int i, int j, int dp[10][10])
{
  int steps = 0;
  while (str1[i] != '\0' || str2[j] != '\0')
  { //while one side has not reach '\0'
        while (str1[i] != '\0' && str2[j] != '\0')
        { //while both sides has not reach '\0'
            if (dp[i][j] != -1){ return dp[i][j];} //if there is stored answer, return that
             if (str1[i] == str2[j]) 
            { // if both char are the same, both move forward
                i++;
                j++;
            }
            else  //if both char are not the same
            { 
              int ans1 = check(str1, str2, i+1, j, dp);  //ans1 = when str1 move forward
              int ans2 = check(str1, str2, i, j+1, dp);  //ans2 = when str2 move forward
              int ans3 = check(str1, str2, i+1, j+1, dp);//ans3 = when both move forward

                //compare the result below to find the smallest steps(distance)
            
                if (ans1 <= ans2)
                { //ans1 is smaller than ans2
                    if (ans1 <= ans3)
                    { //ans1 is smallest
                        dp[i][j] = ans1 + 1; //store the answer to dp array
                        i++;                        //move forward
                        steps++;                    //steps +1
                    }
                    else //ans3 is smallest
                    { 
                        dp[i][j] = ans3 + 1;
                        i++;
                        j++;
                        steps++;
                    }
                }
                else //ans2 is smaller than ans1
                { 
                    if (ans2 <= ans3)
                    { //ans2 is smallest
                        dp[i][j] = ans2 + 1;
                        j++;
                        steps++;
                    }
                    else //ans3 is smallest
                    { 
                        dp[i][j] = ans3 + 1;
                        i++;
                        j++;
                        steps++;
                    }
                }
            }//else ends here
        }//while loop ends (both sides are not \0) 
        
        //when str1 or str2 reaches the end '\0'
        if (str1[i] != '\0')
        {
            i++;
            steps++;
        }
        else if (str2[j] != '\0')
        {
            j++;
            steps++;
        }
  }//while loop ends(one side is not \0)
    
  //printf("i=%d\nj=%d\nsteps=%d\n", i, j, steps);
  return steps;
  
}//function ends here


int main() {
  char str1[10] = "editing";
  char str2[10] = "distance";
    int min_step[10][10]; //with index [i][j]
  for (int i = 0; i < 10; i++)
    for (int j = 0; j < 10; j++)
      min_step[i][j] = -1; 
  //scanf("%s %s", str1, str2);
  printf("%d\n", check(str1, str2, 0, 0, min_step));
  return 0;
}