deletion_distance.py

"""
Deletion Distance

The deletion distance of two strings is the minimum number of characters you need to delete in the two strings in order to get the same string. 
For instance, the deletion distance between "heat" and "hit" is 3:
By deleting 'e' and 'a' in "heat", and 'i' in "hit", we get the string "ht" in both cases.

Given the strings str1 and str2, write an efficient function deletionDistance that returns the deletion distance between them. 


input:  str1 = "dog", str2 = "frog"
output: 3

input:  str1 = "some", str2 = "some"
output: 0

input:  str1 = "some", str2 = "thing"
output: 9

input:  str1 = "", str2 = ""
output: 0

"""

# O(mn) time
# O(mn) space
def deletion_distance(str1, str2):
  if str1 == str2:
    return 0
  
  m, n = len(str1), len(str2)
  dp = [[0 for j in range(n + 1)] for i in range(m + 1)]

  # Update first row
  for i in range(m + 1):
    dp[i][0] = i
  
  # Update first col
  for j in range(n + 1):
    dp[0][j] = j
  
  # Update the rest of the matrix
  for i in range(1, m + 1):
    for j in range(1, n + 1):
      if str1[i - 1] != str2[j - 1]:
        dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1
      else:
        dp[i][j] = dp[i - 1][j - 1]
        
  return dp[-1][-1]


# Same as previous
def deletion_distance(str1, str2):
  if str1 == str2:
    return 0
  
  m, n = len(str1), len(str2)
  dp = [[0 for j in range(n + 1)] for i in range(m + 1)]

  for i in range(m + 1):
    for j in range(n + 1):
      if i == 0:
        dp[i][j] = j
      elif j == 0:
        dp[i][j] = i
      elif str1[i - 1] != str2[j - 1]:
        dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + 1
      else:
        dp[i][j] = dp[i - 1][j - 1]
        
  return dp[-1][-1]


# O(mn) time
# O(min(n,m)) space
"""
Pseudocode:

function deletionDistance(str1, str2):
    # make sure the length of str2 isn't
    # longer than the length of str1
    if (str1.length < str2.length)
        tmpStr = str1
        str1 = str2
        str2 = tmpStr

    str1Len = str1.length
    str2Len = str2.length
    prevMemo = new Array(str2Len  + 1)
    currMemo = new Array(str2Len  + 1)

    for i from 0 to str1Len:
        for j from 0 to str2Len:
            if (i == 0):
                currMemo[j] = j
            else if (j == 0):
                currMemo[j] = i
            else if (str1[i-1] == str2[j-1]):
                currMemo[j] = prevMemo[j-1]
            else:
                currMemo[j] = 1 + min(prevMemo[j], currMemo[j-1])

        prevMemo = currMemo
        currMemo = new Array(str2Len + 1);  
                                           
    return prevMemo[str2Len]
"""