• We will write a recursive approach and memoize it. So first make a 2D array of size (N + 1) x (M + 1) where N and M are lengths of the two strings and fill it with -1. -1 means that we don’t know the answer for that state.
• The base case would be if the length of the first string is 0 then we need to add all the characters of the second string to the first string hence return the length of the second string. Similarly, if the length of the second string is 0 return the length of the first string. Else check the value of the dp array at index i, j. If it's not equal to -1, simply return its value. Otherwise, go the following step
• The recurrence relation is as follows:
  • If the last characters of two strings are not the same, try all operations and it will cost 1. Here i is initially the last index of s1 and j is initially the last index of s2. f(i, j) is the edit distance of str1(0, i) and str2(0, j).
  • str(i, j) denotes the substring formed by taking the characters at index i through index j.
  • f(i, j) = 1 + min(f(i - 1, j), f(i, j - 1), f(i - 1, j - 1)) 
  • Here the three recursive calls are for the insert, remove and replace respectively and s1, s2 will be parameters of all function calls. If the last characters of two strings are the same (s1[i] == s2[j]) then the relationship would be:
    • f(i, j) = f(i - 1, j - 1)
• Before returning the result, store it in the array at index i, j.  This will save the overlapping subproblems.

We will use the dynamic programming approach here and will have a 2-dimensional array, dp where

dp[i][j] stores the edit distance of the (i+1)th length substring of str1 and (j+1)th length substring of str2 starting from index 0

The base case would be when the size of first-string is 0 then edit distance will be equal to the size of other string i.e when i = 0 then dp[i][j] = j and similarly vice versa, so if j = 0 then dp[i][j] = i.

Now we will use our recurrence relation  

// If str1[i - 1] == str2[j - 1]
dp[i][j] = dp[i-1][j-1]
// Otherwise
dp[i][j] = 1 + min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1])