Deletion Distance between 2 strings
$begingroup$
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)]
python
asked 10 mins ago
GideoGideo
11
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.
$endgroup$
add a comment |
$begingroup$
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)]
python
asked 10 mins ago
GideoGideo
11
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.
$endgroup$
add a comment |
$begingroup$
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)]
python
asked 10 mins ago
GideoGideo
11
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
python
asked 10 mins ago
GideoGideo
11
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 10 mins ago
GideoGideo
11
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 10 mins ago
GideoGideo
11
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 10 mins ago
GideoGideo
11
asked 10 mins ago
GideoGideo
11
11
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.
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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.
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
