Dynamic Programming solution to edit distance problem

EditDistance.py

######################################################################################################################
#   Edit Distance
#   Find the minimum number of operations (replace, delete, insert) to convert string a to string b. Uses a bottom up
#   dynamic programming approach. Based on example from YouTube channel 'Back to Back SWE': https://www.youtube.com/watch?v=MiqoA-yF-0M
#
#                                         replace       delete      insert
#   Recursive Equation: ED(a, b) = min( ED(a-1, b-1), ED(a-1, b), ED(a, b-1) ) where a-1 means removing 1 character from string a
#
######################################################################################################################

def EditDistance(astring, bstring):
    #Initialize matrix values
    matrix = [[0 for i in range(0, len(astring)+1)] for j in range(0, len(bstring)+1)]
    #For the empty string, number of operations will always be equal to number of characters
    for i in range(len(astring)+1): matrix[0][i] = i 
    for i in range(len(bstring)+1): matrix[i][0] = i

    #Bottom up dynamic programming
    for i in range(1, len(bstring)+1):
        for j in range(1, len(astring)+1):
            #If same character there should be no change
            if bstring[i-1] == astring[j-1]: 
                matrix[i][j] = matrix[i-1][j-1]
            #Takes minimum of replace [i-1][j-1], delete[i-1][j], or insert [i][j-1] and adds the current character
            else: 
                matrix[i][j] = min(matrix[i-1][j-1], matrix[i-1][j], matrix[i][j-1])+1

    return matrix[len(bstring)][len(astring)]


def printMat(matrix):
    for i in matrix:
        print(i)

a = "benyam"
b = "ephrem"     
ans = EditDistance(a, b)
print(ans)


a = "string"
b = "bring"
ans = EditDistance(a, b)
print(ans)