We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Implementation
  4. Append and Delete
  5. Discussions

Append and Delete

Sort by

recency

|

1444 Discussions

|

  • sergiosrtd
     + 0 comments

    The test case 7 is wrong

    s=aaaaaaaaaa t=aaaaa k=7

    You only need 5 delete operations on s to convert t

    But the test case expect "Yes"

  • afzal_saiyed931
     + 0 comments
    def appendAndDelete(s, t, k):
    prefix_count = 0
    len_s = len(s)
    len_t = len(t)
    while prefix_count < len(t):
        if prefix_count > (len_s - 1) or s[prefix_count] != t[prefix_count]:
            break
        else:
            prefix_count += 1
    
    delete_required = len_s - prefix_count
    addition_required = len_t - prefix_count
    total_ops_required = delete_required + addition_required
    if k >= (len_s + len_t):
        return "Yes"
        
    if k >= total_ops_required and (k - total_ops_required) % 2 == 0:
        return "Yes"
        
    return "No"

  • asadkamal1994
     + 1 comment

    There is a testcase 5 in which we have: y yu 2 Should'nt the expected output be 'Yes'? because the operation can be performed according to me.

  • yashdeora98294
     + 0 comments

    Here is problem solution in Python, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-append-and-delete-problem-solution.html

  • csiegel
     + 0 comments

    python

    def appendAndDelete(s, t, k):
    # Write your code here
    
    #all cases where the length of the two strings combined is less than or equal to k work
    if len(t) + len(s) <= k:
        return 'Yes'
    
    index=0
		
    while True:
        
        #remove the out of index range scenario
        if index > len(s) - 1 or index > len(t) - 1:
            break
        #determine up to which index is the same
        if s[index] == t[index]:
            index+=1
        else:
            break
            
    #determine remaining steps needed and if the difference between remaining and k is either 0 or even
     
    remaining_steps = len(s) + len(t) - index * 2
    if remaining_steps <= k and (k - remaining_steps) % 2 == 0:
        return 'Yes'
    else:
        return 'No'