# Append and Delete

# Append and Delete

ngtzeyang1994 + 32 comments Let me do my best to explain the different cases that could happen:

int commonLength = 0; for (int i=0; i<java.lang.Math.min(s.length(),t.length());i++){ if (s.charAt(i)==t.charAt(i)) commonLength++; else break; } //CASE A if((s.length()+t.length()-2*commonLength)>k){ System.out.println("No"); } //CASE B else if((s.length()+t.length()-2*commonLength)%2==k%2){ System.out.println("Yes"); } //CASE C else if((s.length()+t.length()-k)<0){ System.out.println("Yes"); } //CASE D else{ System.out.println("No"); }

CASE A:

What this case is finding is if k is bigger than the difference in length of the two strings.

example: s = "123456789", t = "1", k = 5

in this case, you definitely need a bigger k to transform s to t or vice versa, therefore you reject any such cases.

Case B:

now that the case has passed A, we can say that the total number of letters required to change is less than or equal to k.

however the next problem comes because the question specified that exactly k moves must be done, no more no less. this leads to an example whereby:

s = "010101", t = "01010", k = any EVEN number

OR

s = "010101", t = "010101", k = any ODD number

from these two examples you can see that only if k is odd/even matches the odd/even of number of different digits will such cases pass.

Case C:

However there is a way to overcome this odd/even problem if you are able to completely delete away one string as a deletion action on an empty string results in another empty string.

Example: s = '1' t = '101' k = 5

in this case, to get a S from T you could do delete-delete-delete-delete-add(1) and you will satisfy the condition.

Case D:

all other cases will fail the test

Hope i helped

withasingle_t + 0 comments Nice algo!

tanuj_kapoor15 + 1 comment I'm sorry I might be wrong but what if I append by a null character? I mean in case B where t="01010" let's assume k=7 what if I just append t by a null character and then append it with a 1, it would pass the test case, right?

lemuel_dulfo + 1 comment You are not allowed to append a null character. The problem clearly states that you can perform two operations per "step", and the first operation is:

- Append a lowercase English alphabetic letter to the end of the string.

tanuj_kapoor15 + 1 comment Thanks for clearing it out.

madhu703 + 0 comments Can you explain case B with one more example . I didn't get it exactly how it works.

alexermashev + 4 comments Yor explanation is hard to understand, check a look at me code. I think it easy to understand

function isItPossibleToModify($s, $t, $k) { $sLength = strlen($s); $tLength = strlen($t); if ($sLength < $tLength) { return ($tLength - $sLength) % 2 == 0; } $commonPrefix = 0; // find a largest common prefix between s and t for ($i = 0; $i < min($sLength, $tLength); $i++) { if ($s[$i] != $t[$i]) { break; } $commonPrefix++; } $needToProcess = ($tLength - $commonPrefix) + ($sLength - $commonPrefix); return $needToProcess <= $k; }

Vishnu_Vijayan + 7 comments What about for this testcase ?

abc abc 1

It gives Yes. Shouldn't the answer be No.

riyadali + 0 comments Along the lines of what was previously mentioned ...

---> note prefixLen contains the length of the matching prefix in both strings

`int minOps=(schar.length-prefixLen)+(tchar.length-prefixLen); if (k>=minOps) { // check if you have sufficient operations to // remove first string entirely and the replace it // with the second string if (k>=(schar.length+tchar.length)) System.out.println("Yes"); else { // cannot replace first string totally so extra available // ops must be an even multiple allowing you to // replace portions of the prefix that both strings have // in common if ((k-minOps)%2==0) System.out.println("Yes"); else System.out.println("No"); } } else { System.out.println("No"); }`

jaywhy13 + 2 comments I'm not clear as to why the answer for this test case isn't Yes and then 0 ops.

rjain90 + 1 comment It's No because it has to be exactly K moves, neither less nor more.

mohammedjunaid + 0 comments [deleted]

edrouwendaal + 0 comments you can use an unlimited number of deletions on an empty string

ashitpanda2107 + 0 comments Of course the answer should be NO :)

tarunkarang + 0 comments you have one step and its nessecary to be performed, so if you add a char it wont be the same, similarly if you remove it wont be same

kininge007 + 0 comments it can solve in 0 steps so, Yes

sakshamgoyalswm + 0 comments int i=0,j=0; while(

*(s+i)==*(t+i) && *(s+i)!='\0' && *(t+i)!='\0') { i++; } int sa=strlen(s); int ta=strlen(t); j=sa-i+ta-i; static char yes[]="Yes"; static char no[]="No"; if(k`else return yes;`

neha86_arora + 0 comments ya..Answer should be No..

mukeshbhakuni590 + 0 comments didn't understood your code but it works . how did the (return (sLength) % 2 == 0;) condition works . can u please explain

_silent_ + 0 comments What about for this test case? peek seeker 3 It should give you no as answer But running your code against this test case will give Yes

if (tLength) { return (sLength) % 2 == 0; }

if(4<6){ return(6-4)%2==0 gives yes But can't be converted }

fran_rabljenovi1 + 0 comments At k you need to check if the remaining steps can cross out each other.

sharwinsharwi + 0 comments this is the correct one CASE D: else if((s.length()+t.length()-k)>1)

ram_24 + 0 comments [deleted]jjpatel361 + 0 comments I didn't understand why you need

`2*commonLength`

in your first condition.`Kmin = commonLength + delete_bigger_word + add_small_word`

`delete_bigger_word`

= number of delete operations required in longer word.`add_small_word`

= number of addition operations required in smaller word to be equal to larger_wordRushabhPatel + 0 comments Could you explain Case B again ?

renatodepaiva + 0 comments Thank you so much for showing the solution and explaining it in such a clear way!

m4manishpushkar + 0 comments Bhaisaaaab you nailed it.

The_code_shinobi + 0 comments Brilliant solution and thanks for the lucid explanation.

nammivasant + 0 comments Can you explain case 2 one more time with different example?

Specter123 + 0 comments nice

sharmaspg + 0 comments nice explaination !but i need come to this logic myself .how do i do that ?

mukeshbhakuni590 + 0 comments nice explanation ... ! cleared my doubt

pappasbrent + 0 comments Thank you! I almost had it, but my logic was convoluted and confusing. Your method of first finding the index at which the two strings start to differ and building off of that is much cleaner

harishamarnath21 + 1 comment may i know why this test case should return "No" uoiauwrebgiwrhgiuawheirhwebvjforidkslweufgrhvjqasw vgftrheydkoslwezxcvdsqjkfhrydjwvogfheksockelsnbkeq 100 its add 50 and remove 50 right?

Lalchetaketan022 + 1 comment string length are not 50 for both. First one has 51 characters and second one has 50. so, it's required to have 51+50 = 101 not 100

eabarry + 0 comments Both strings are length 50, so k=100 should produce a "Yes".

singhal_purvi14 + 0 comments I want to know how you think of this approach. How do you start thinking this problem?

SL170031111 + 0 comments nice expalnation thank u

kishoresaldanha + 0 comments You may need to add another case if len(s)==len(t) and k=2*commonLength+1 right after case A

DarkSilence + 0 comments Example: s = 'ABCD' t = 'BCD' k = 1 return "No" ???

JJJason + 0 comments I think case A should be 'k is smaller than the difference in length of the two strings.'

jnrdn0011 + 0 comments Thank you.

Sithlords1999 + 0 comments suppose s="ash" t="ashley" ,k=10; according to algo ans is no but if we do 4 delete operations on s and 6 append operation,then we can convert s into t.

aayushsugandhi11 + 0 comments how did u paste this solution . i want to post my solution in python

[deleted] + 0 comments [deleted]ghassan_karwchan + 0 comments Nice explenation. Without your explenation, I could never got it. Their description for the problem is lacking enough explanation. They should emphacize on that the moves should match "Exactly" k.

hardik_dadga + 0 comments In the example of case C, why do you say "to get s from t you can......" When the PS stated is "Given an integer, k, and two strings, and , determine whether or not you can convert

**'s'**to**'t'**by performing exactly 'k' of the above operations on 's'. If it's possible, print Yes. Otherwise, print No."_JXD_ + 0 comments When case 3 occurs the expression "(s.length()+t.length()-k) < 0" will always be false using C++. That is because you are substracting a signed integer from an unsigned integer, which will result in an unsigned integer and thus will never be less than 0. To correct this, you can write: "(int(s.length()+t.length())-k)". I spent half an hour figuring this out...

mehul_sachdeva7 + 1 comment my js approach, quite different but almost same

function appendAndDelete(s, t, k) { var end = 0; for (var i = 0; i < Math.min(t.length, s.length); i++) { if (s[i] == t[i]) { end++; } else { break; } } var maxIter = s.length + t.length; var minIter = maxIter - (2 * end); if ((k >= minIter) & ((k - minIter) % 2 == 0)) { return "Yes"; } else if (k >= maxIter) { return "Yes"; } else { return "No"; } }

MESHKVIN + 0 comments what if (k - minIter) % 2 != 0) but (k - minIter) >= 2*end )

asad70161 + 1 comment Your codefailed at this test case y yu 2 it gives yes, answer is no

almagician + 1 comment My code also failed on this case. What is the issue? It looks like it shoud be yes.

Also this case: abcd abcdert 10

Mine is giving Yes, but it is wrong.

kurohashi + 1 comment Question asks "exactly k" moves, so we have to perform 2 moves here even if it becomes "yu" in first move. The second move will distort it again.

almagician + 0 comments thank you, i got it)

arsen_budumyan + 0 comments For Case B it should be for: s = "010101", t = "01010", k = any ODD number for: s = "010101", t = "010101", k = any EVEN number

thank you for your explanation :)

prmshaw906 + 0 comments great analysis

ryalman + 1 comment It should not be considered as an easy problem. I hardly solved it.

JamesCater + 0 comments My thoughts exactly, at first glance it appears to be simple, but it is far from Easy

apghr + 8 comments works like a charm

#include <bits/stdc++.h> using namespace std; #define L (s.size() + t.size()) int main(){ string s, t; int k, i; cin >> s >> t >> k; for(i = 0; s[i] == t[i]; i++); cout << (L <= k + i*2 && L%2 == k%2 || L < k ? "Yes" : "No"); return 0; }

(edit: cleared out some redundant code)

chandu_iitm2029 + 1 comment <= k at the end

apghr + 1 comment Not needed. When s.size() + t.size() == k, the previous part of the condition is always true.

shrads01 + 2 comments Can you explain me the need of diff%2 == k%2 ?

dekoder + 0 comments it checks whether both are even or both are odd ! because if that happens to be true then (k - diff) will always be even and string can always be build. HOW ?

by appending then deleting or vice-versa.

jadhavsaurabh191 + 0 comments check exact number of alphabates

lohit_cool + 1 comment can you please explain diff%2==k%2 condition?

Majka + 0 comments qwerasdf qwerzxcv 10 operations

substring qwer is the same in both strings, so diff = asdf+zxcv = 8, you need atleast 8 operations to create the required string(delete asdf and append zxcv).But then if they gave you exactly two more operations to work with (10), you could remove 'r' from qwer and then append it. If they gave you 12 you could do the same thing to 'e' . . .

ZeroBlankZero + 0 comments pretty decent . good job :)

mr_360 + 0 comments can u explain your logic?

mahima_goyen2015 + 0 comments can you please eplain the logic?

wohlwendk + 0 comments I'm a little late to the party, but this is undefined behaviour if s == t. Since you don't check whether i <= s.length(), s[i] might be out of bounds, and from there anything could happen. You got lucky here and there's no problems on the judge, but it might not always works

sravanthi100 + 1 comment can anyone explain the logic

wohlwendk + 1 comment Sure.

First, we try to find the minimum number of operations needed. For this, we check how large the largest common prefix of the two strings is; eg. if we have "hello" and "hellno", our largest common prefix is "hell". We compute the length of this prefix by

`for(i = 0; s[i] == t[i]; i++);`

.If we didn't have such a prefix, we'd have to first delete

`s.size()`

characters, then append`t.size()`

characters (`s.size() + t.size() = L`

steps in total). However, since we have a prefix of length`i`

, we can skip`i`

deletions and`i`

additions. Therefore, we need at least`L - 2*i`

operations in total. If`k`

is smaller than that, we can return early.`L <= k + i*2`

checks this.Now, if

`(k - (L - 2 * i)) % 2 == 0`

(which is equivalent to`k%2 == L%2`

), we can first do`L - 2*i`

operations to get the string`t`

, and then just append a character and immediately remove it. Since`(k - (L - 2 * i)) % 2 == 0`

means that the difference between`k`

and`L - 2*i`

is even, this is always possible.There is one more special case; since

`remove("") == ""`

, if we have`k >= L`

, we can just do`s.size()`

deletions, then another`k - L`

deletions, and`t.size()`

additions. In this special case, we can always find a way to convert`s`

to`t`

using`k`

operations.sravanthi100 + 0 comments thank you!

skur42 + 6 comments Don't just copy it figure out the logic

import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); String t = in.next(); int k = in.nextInt(); int i; int check=0; for(i=0;i<s.length()&&i<t.length();i++) { if(s.charAt(i)!=t.charAt(i)) { break; } } int d=s.length()-i; int a=t.length()-i; int p=k-d-a; if(p==0) { System.out.println("Yes"); } else if(p<0) { System.out.println("No"); } else { if(p%2==0) { System.out.println("Yes"); } else { if(p>=(2*i)) {System.out.println("Yes");} else {System.out.println("No");} } } } }

kangkanlahkar1 + 1 comment I didnt get

if(p>=(2*i))

iitbRaj + 1 comment Assuming you understood everything above that part, u must have realised that if the difference p is odd, two cases arise...and in that non-trivial case when the string becomes empty(like in the example "aba")during the process when u delete extra elements so that the no. of operations reaches k,realise that the string should become empty in less than or equal to p/2 operations(the other delete operations will be "just like that")(i at this point of time is denoting the no. of characters left in the string is).so u can also write --- if(p/2>=i)....

sonali007 + 0 comments thnxx...it was great help.

coolchand5725 + 0 comments [deleted]stabgan + 0 comments For the first time I enjoyed a Java code

Basu_97 + 0 comments Awesome solution!! I was trying to understand how many cases i have to consider. But your code just blew my mind. Really easy to understand. Quite a good one

weudbewiud + 6 comments Quite a compact solution in Python.

Basically, I remove characters from the end of s until t starts with s and the number of missing characters to get to t is the number of operations left. I also break if there are no more operations or if s became empty. Afterwards I simply check whether I have enough operations left to add character to s to reach t.

#!/bin/python import sys s = raw_input().strip() t = raw_input().strip() k = int(raw_input().strip()) for ops_left in reversed(range(1, k + 1)): if s == t[:len(s)] and len(t) - len(s) == ops_left or len(s) == 0: break s = s[:-1] print "Yes" if len(t) - len(s) <= ops_left else "No"

msambitkumar1991 + 0 comments @tschmorleiz can you please explain the condition of the if statement?

iamkris + 0 comments Amazing!!

pramos + 2 comments Really nice solution!

I'm not sure why you're doing:

reversed(range(1, k+1))

when you could directly do:

range(k, 1, -1)

vedikagupta + 0 comments kindly explain!!

chandan37shaw + 0 comments though this is a genuine solution but not passing all test cases

kushchoudhary8 + 0 comments no of moves left should be exactly equal to k not less then!

Shikhar47 + 0 comments The solution is elegant but the complexity of the algorithm will be "n-squared" as the list slicing operation is O(n) inside a for loop

RodneyShag + 4 comments ### Java solution - passes 100% of test cases

From my HackerRank solutions.

**Case 1:**See if we can completely erase String*s*and append String*t*. If we need to waste operations to reach*k*operations, we can do so when String*s*has no characters.**Case 2:**See if we can convert String*s*to String*t*without completely erasing String*s*. We keep erasing charcters from String*s*until it becomes a prefix of String*t*. We then add the characters needed to turn String*s*into String*t*. If we need to waste operations to reach*k*operations, we can only do so in groups of 2 by doing an append and a delete.import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner scan = new Scanner(System.in); String s = scan.next(); String t = scan.next(); int k = scan.nextInt(); scan.close(); System.out.println(canConvert(s, t, k) ? "Yes" : "No"); } private static boolean canConvert(String s, String t, int k) { /* Case 1 */ if (s.length() + t.length() <= k) { return true; } /* Case 2 */ int i = 0; // represents index of 1st non-matching character for ( ; i < s.length() && i < t.length(); i++) { if (s.charAt(i) != t.charAt(i)) { break; } } int minOperations = (s.length() - i) + (t.length() - i); return k >= minOperations && (k - minOperations) % 2 == 0; } }

Let me know if you have any questions.

abhishek_09091 + 2 comments why are you doing (k - minOperations) % 2 == 0;? please could you explain on this.

RodneyShag + 2 comments Because, if we need to waste operations to reach k operations, we can only do so in groups of 2 by doing an append and a delete. That line of code makes sure the number of operations we "waste" is divisible by 2.

abhishek_09091 + 0 comments thanx gentleman

the_night_walker + 0 comments [deleted]

psshetty15 + 1 comment How about the below case? ashla asha 4

if am not wrong it would take only take three steps to convert s to t delete l and a ,append a but the result give by your code is NO.

Please reply, i may not be correct so please help me understand.

RodneyShag + 0 comments Hi. The instructions want us to do it in

**exactly**4 steps, not less than 4. Since ashla can't be converted to asha in exactly 4 steps, the answer is*no*.

ItzMv + 0 comments Best explaination thanks bro

nishantketu903 + 0 comments [deleted]Starpallavi04 + 1 comment i want to know how( s.length()-i+t.length()-i) is the minimum number of operations...it shoild be maximum i guess

RodneyShag + 0 comments Well, maximum is actually infinite, since you can always append a character, delete a character, append a character, delete a character, repeat...

`(s.length() - i) + (t.length() - i)`

is actually how many operations you need to solve the problem, while ignoring the add/delete of same character, which does nothing.

adamxtokyo + 1 comment Here's a short and sweet JavaScript solution if anybody wants it:

function appendAndDelete (s, t, k) { let o = s.length + t.length if (k > o) return 'Yes' for (let i = 0, l = Math.min(s.length, t.length); i < l; i++, o -= 2) { if (s[i] !== t[i]) break } return o > k || (k - o) % 2 !== 0 ? 'No' : 'Yes' }

Explanation:

o is the minimum number of operations. But to deal with the one exception in this algorithm, we set it to the total length of both strings first. If the number of allowed operations is bigger than the sum of both strings, the answer will always be "Yes". So we check for that before moving along!

Now, we compare the letters in both strings and subtract each iteration from the minimum number of operations. Remember that each common letter means 2 less operations (remove a letter + add a letter). We break out of the loop once the letters aren't equal anymore.

Now, if the minimum number of operations is more than the allowed number, we return "No". Otherwise, as long as the remainder of the difference between minimum and allowed operations is even, we return "Yes".

Questions? =)

babylee2002 + 0 comments Thanks for sharing this great solution.

But..Test Case 5 has input of 'y','yu',2

Isn't this supposed to be yes? since it doesn't take any delete but one add and that is less than 2? I might be missing something. I would appreciate any insights.

KshitijSinha + 3 comments If someone have problem with test case 10 and (I think) 6 (may be). Then the solution is below.

Here is a twist, If the difference of lengths of s and t are odd and k is even, then it is not possible to convert s to t. eg,

y yu 2

let us look at all the cases for above eg:

op 1:del y from s => s=""; op 2:append y to s => s="y" => s!=t; OR

op 1:append u to s => s="yu"; op 2:del u from s => s="y" => s!=t; OR

op 1:append u to s => s="yu"; op 2:append x(say) to s => s="yux" => s!=t;

Hence, s=t can not be achieved using 2 operations. That's why ans is No

KshitijSinha + 2 comments int main(){

`char* s = (char *)malloc(512000 * sizeof(char)); scanf("%s",s); char* t = (char *)malloc(512000 * sizeof(char)); scanf("%s",t); int k,min,i; scanf("%d",&k); i=0; while(s[i]==t[i]){i++;} min=strlen(s)-i+strlen(t)-i; if((((strlen(s)-strlen(t))%2==1)||((strlen(t)-strlen(s))%2==1))&&(k%2==0)){printf("No");} else if(k>=min){printf("Yes");} else{printf("No");} return 0;`

}

Vishu40585 + 0 comments i dint understand ur if statement can u explain it for me?

_HELLO_WORd + 0 comments # include

# include

# include

# include

# include

# include

# include

int main() { int k, i, j, s_size, t_size, count, br, nmod; char* s = (char *)malloc(512000 * sizeof(char)); char* t = (char *)malloc(512000 * sizeof(char)); scanf("%s", s); scanf("%s", t); scanf("%i", &k); s_size = strlen(s); t_size = strlen(t); br = 1; nmod = 0; for(i=count=0; i < s_size && br; i++){ if(s[i]==t[i]) count++; else br = 0; } if(s_size > t_size){ /* when length s greater than t

*/ nmod = s_size - count; nmod = 2*(nmod - t_size)+(s_size - t_size); }else if(s_size < t_size){ /* when length of s less than t*/ nmod = t_size - s_size; nmod = nmod + 2*(s_size - count); }else{ /* when length of s equal to t */ nmod = 2 * (s_size - count); } nmod = abs(nmod); if(k < nmod) printf("No"); else printf("Yes");`return 0;`

}

failed 2 cases

Sur02 + 0 comments [deleted]william_shen8 + 0 comments [deleted]

mitchmcdee + 4 comments Simple python3 solution:

See more solutions here, let me know if any need explaining!

s = input().strip() t = input().strip() k = int(input().strip()) numSameChars = min(len(s), len(t)) for i in range(len(t)): if s[:i] != t[:i]: numSameChars = i-1 break diff = len(s)-numSameChars + len(t)-numSameChars print('Yes' if (diff <= k and diff%2 == k%2) or len(s) + len(t) < k else 'No')

Aximm + 0 comments I came up with a very similar, if not more verbose, solution

def app_and_del(s, t, k): s_length = len(s) t_length = len(t) if s_length + t_length < k: return 'Yes' same = 0 for s_l, t_l in zip(s, t): if s_l == t_l: same += 1 else: break extra_s = s_length - same extra_t = t_length - same difference = extra_s + extra_t if difference <= k and difference % 2 == k % 2: return 'Yes' return 'No' print(app_and_del(s, t, k))

Freak_1231 + 2 comments what diff%2 ==k%2 doing here???please explain

mitchmcdee + 3 comments Essentially checks to see if the difference between the two strings are odd and that k is odd. If this is the case, it is trivial to see that an even number of appends/deletions can be made to make the strings equal :)

Freak_1231 + 0 comments thanks man :)

SebastianNielsen + 0 comments Isn't

(k - diff) % 2 == 0

the same as

k%2 == diff%2

since whenever you subtract a number from another, the following rules:

a=even, b=even a-b=even a=odd, b=odd a-b=even a=odd, b=even a-b=odd a=even, b=odd a-b=odd

So suppose: # k = 6 s = 'abcdefghij' # diff = 4 t = 'abcdef' (k - diff) % 2 = (6 - 4) % 2 = 2 % 2 == 0 True

krishp + 0 comments Does it necessarily matter?

Aximm + 1 comment i think it's easier to explain with an example. say you want to get'abcde' from'abc'. the difference is 2, which is even (2%2==0). i can only have a k value that is even to reach abcde, any odd number won't work. likewise if i wanted to reach abcd (1%2==1), only an odd k would work. hope this clears it up

Freak_1231 + 0 comments woow...thank you so much..god bless you.

SebastianNielsen + 1 comment if s[:i] != t[:i]: numSameChars = i-1 break

This is very ineffecient, compared to:

if s[i] != t[i]: numSameShars = i break

But apart from that your code is perfect, I would just prefer to break the last few lines up; instead of having it all on one line (for the sake of brevity):

a = input() b = input() k = int(input()) if len(a) + len(b) < k: print('Yes') exit() numSameChars = min(len(a), len(b)) for i in range(numSameChars): if a[i] != b[i]: numSameChars = i break diff = (len(a) - numSameChars) + (len(b) - numSameChars) if (diff <= k and (diff - k) % 2 == 0) or len(a) + len(b) < k: print('Yes') else: print('No')

krishp + 0 comments Nice one!

msuraj001 + 1 comment Test cases 4 and 8 fail in python 3 IDE, can you help

andrew_abbey + 0 comments Not the greatest algorithm... but it works in Python3

def appendAndDelete(s, t, k): x = 0

`for i in range(min([len(s),len(t)])): if (s[i] != t[i]): x = i break x = i + 1 n = len(s) - x + abs(x - len(t)) if (n <= k): if ((k-n) % 2 == 0) or (k-n > 2*x): return "Yes" return "No"`

schenjobs + 1 comment Where testcase #10 abcd abcdert 10 is No?

Should it not be 3? Append ert?

arjunmehta94 + 1 comment Had the same question, shouldn't it be Yes?

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