input: Write a function to compute the edit distance between two strings in Python.
  
  def editDistance(string1, string2, m, n):
      # Create an empty matrix
      dp = [[0 for x in range(n+1)] for x in range(m+1)] 
    
      # Filling the first column of the matrix
      for i in range(m+1): 
          dp[i][0] = i 
    
      # Filling the second column of the matrix 
      for j in range(n+1): 
          dp[0][j] = j 
    
      # Populate the matrix
      for i in range(1, m+1): 
          for j in range(1, n+1): 
    
              if string1[i-1] == string2[j-1]: 
                  dp[i][j] = dp[i-1][j-1] 
              else: 
                  dp[i][j] = 1 + min(dp[i][j-1],        # Insert
                                     dp[i-1][j],        # Remove 
                                     dp[i-1][j-1])      # Replace
    
      return dp[m][n] 
    
    
  # Driver program 
  string1 = "kitten"
  string2 = "sitting"
  m = len(string1) 
  n = len(string2) 
  assert editDistance(string1, string2, m, n) == 3
  
  string1 = "horse"
  string2 = "ros"
  m = len(string1) 
  n = len(string2) 
  assert editDistance(string1, string2, m, n) == 3
  
  string1 = "intention"
  string2 = "execution"
  m = len(string1) 
  n = len(string2) 
  assert editDistance(string1, string2, m, n) == 5
  
  string1 = ""
  string2 = "a"
  m = len(string1) 
  n = len(string2) 
  assert editDistance(string1, string2, m, n) == 1
  
  string1 = "a"
  string2 = ""
  m = len(string1) 
  n = len(string2) 
  assert editDistance(string1, string2, m, n) == 1
output: def editDistance(string1, string2, m, n):
      dp = [[0 for x in range(n+1)] for x in range(m+1)] 
    
      for i in range(m+1): 
          dp[i][0] = i 
    
      for j in range(n+1): 
          dp[0][j] = j 
    
      for i in range(1, m+1): 
          for j in range(1, n+1): 
    
              if string1[i-1] == string2[j-1]: 
                  dp[i][j] = dp[i-1][j-1] 
              else: 
                  dp[i][j] = 1 + min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) 
    
      return dp[m][n]