Deletion Distance between 2 strings












1












$begingroup$


I was solving this problem at Pramp and I have trouble figuring out the algorithm for this problem. I'll paste the problem description and how I kind of solved it. It's the correct solution. It is similar to the edit distance algorithm and I used the same approach. I just wanted to see what are other ways to solve this problem.




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.
We cannot get the same string from both strings by deleting 2 letters or fewer.
Given the strings str1 and str2, write an efficient function deletionDistance that returns the deletion distance between them. Explain how your function works, and analyze its time and space complexities.



Examples:

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




What I want to do in this solution, is to use dynamic programming in order to build a function that calculates opt(str1Len, str2Len). Notice the following:
I use dynamic programming methods to calculate opt(str1Len, str2Len), i.e. the deletion distance for the two strings, by calculating opt(i,j) for all 0 ≤ i ≤ str1Len, 0 ≤ j ≤ str2Len, and saving previous values



  def deletion_distance(s1, s2):
m = [[0 for j in range(len(s2) +1)] for i in range(len(s1)+1)]
for i in range(len(s1)+1):
for j in range(len(s2)+1):
if i == 0:
m[i][j] = j
elif j == 0:
m[i][j] = i
elif s1[i-1] == s2[j-1]:
m[i][j] = m[i-1][j-1]
else:
m[i][j] = 1 + min(m[i-1][j], m[i][j-1])
return m[len(s1)][len(s2)]









share|improve this question







New contributor




Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    1












    $begingroup$


    I was solving this problem at Pramp and I have trouble figuring out the algorithm for this problem. I'll paste the problem description and how I kind of solved it. It's the correct solution. It is similar to the edit distance algorithm and I used the same approach. I just wanted to see what are other ways to solve this problem.




    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.
    We cannot get the same string from both strings by deleting 2 letters or fewer.
    Given the strings str1 and str2, write an efficient function deletionDistance that returns the deletion distance between them. Explain how your function works, and analyze its time and space complexities.



    Examples:

    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




    What I want to do in this solution, is to use dynamic programming in order to build a function that calculates opt(str1Len, str2Len). Notice the following:
    I use dynamic programming methods to calculate opt(str1Len, str2Len), i.e. the deletion distance for the two strings, by calculating opt(i,j) for all 0 ≤ i ≤ str1Len, 0 ≤ j ≤ str2Len, and saving previous values



      def deletion_distance(s1, s2):
    m = [[0 for j in range(len(s2) +1)] for i in range(len(s1)+1)]
    for i in range(len(s1)+1):
    for j in range(len(s2)+1):
    if i == 0:
    m[i][j] = j
    elif j == 0:
    m[i][j] = i
    elif s1[i-1] == s2[j-1]:
    m[i][j] = m[i-1][j-1]
    else:
    m[i][j] = 1 + min(m[i-1][j], m[i][j-1])
    return m[len(s1)][len(s2)]









    share|improve this question







    New contributor




    Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$


      I was solving this problem at Pramp and I have trouble figuring out the algorithm for this problem. I'll paste the problem description and how I kind of solved it. It's the correct solution. It is similar to the edit distance algorithm and I used the same approach. I just wanted to see what are other ways to solve this problem.




      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.
      We cannot get the same string from both strings by deleting 2 letters or fewer.
      Given the strings str1 and str2, write an efficient function deletionDistance that returns the deletion distance between them. Explain how your function works, and analyze its time and space complexities.



      Examples:

      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




      What I want to do in this solution, is to use dynamic programming in order to build a function that calculates opt(str1Len, str2Len). Notice the following:
      I use dynamic programming methods to calculate opt(str1Len, str2Len), i.e. the deletion distance for the two strings, by calculating opt(i,j) for all 0 ≤ i ≤ str1Len, 0 ≤ j ≤ str2Len, and saving previous values



        def deletion_distance(s1, s2):
      m = [[0 for j in range(len(s2) +1)] for i in range(len(s1)+1)]
      for i in range(len(s1)+1):
      for j in range(len(s2)+1):
      if i == 0:
      m[i][j] = j
      elif j == 0:
      m[i][j] = i
      elif s1[i-1] == s2[j-1]:
      m[i][j] = m[i-1][j-1]
      else:
      m[i][j] = 1 + min(m[i-1][j], m[i][j-1])
      return m[len(s1)][len(s2)]









      share|improve this question







      New contributor




      Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I was solving this problem at Pramp and I have trouble figuring out the algorithm for this problem. I'll paste the problem description and how I kind of solved it. It's the correct solution. It is similar to the edit distance algorithm and I used the same approach. I just wanted to see what are other ways to solve this problem.




      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.
      We cannot get the same string from both strings by deleting 2 letters or fewer.
      Given the strings str1 and str2, write an efficient function deletionDistance that returns the deletion distance between them. Explain how your function works, and analyze its time and space complexities.



      Examples:

      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




      What I want to do in this solution, is to use dynamic programming in order to build a function that calculates opt(str1Len, str2Len). Notice the following:
      I use dynamic programming methods to calculate opt(str1Len, str2Len), i.e. the deletion distance for the two strings, by calculating opt(i,j) for all 0 ≤ i ≤ str1Len, 0 ≤ j ≤ str2Len, and saving previous values



        def deletion_distance(s1, s2):
      m = [[0 for j in range(len(s2) +1)] for i in range(len(s1)+1)]
      for i in range(len(s1)+1):
      for j in range(len(s2)+1):
      if i == 0:
      m[i][j] = j
      elif j == 0:
      m[i][j] = i
      elif s1[i-1] == s2[j-1]:
      m[i][j] = m[i-1][j-1]
      else:
      m[i][j] = 1 + min(m[i-1][j], m[i][j-1])
      return m[len(s1)][len(s2)]






      python






      share|improve this question







      New contributor




      Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question







      New contributor




      Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question






      New contributor




      Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 24 mins ago









      GideoGideo

      62




      62




      New contributor




      Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Gideo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          0






          active

          oldest

          votes











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ifUsing("editor", function () {
          StackExchange.using("externalEditor", function () {
          StackExchange.using("snippets", function () {
          StackExchange.snippets.init();
          });
          });
          }, "code-snippets");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "196"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });






          Gideo is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212607%2fdeletion-distance-between-2-strings%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          0






          active

          oldest

          votes








          0






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          Gideo is a new contributor. Be nice, and check out our Code of Conduct.










          draft saved

          draft discarded


















          Gideo is a new contributor. Be nice, and check out our Code of Conduct.













          Gideo is a new contributor. Be nice, and check out our Code of Conduct.












          Gideo is a new contributor. Be nice, and check out our Code of Conduct.
















          Thanks for contributing an answer to Code Review Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212607%2fdeletion-distance-between-2-strings%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Список кардиналов, возведённых папой римским Каликстом III

          Deduzione

          Mysql.sock missing - “Can't connect to local MySQL server through socket”