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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Append and Delete
  5. Discussions

Append and Delete

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1446 Discussions

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  • sebaspr16
     + 0 comments

    if i´m not mistaken these tests are wrong, please if i'm wrong help me to understand my mistake

            Input (stdin):                    s:abcd, t:abcdert, k:10
        Expected Output:            No

        I think having k = 3 here is enough so the out put should be Yes

        Input (stdin):                  s:y, t:yu,  k = 2
        Expected Output:         No
        having  k = 1 is enough and the output should be Yes

  • anarod_val25
     + 0 comments

    Thing is, you need to use exactly k moves, no more, no less. Else, fail.

    string appendAndDelete(string s, string t, int k) {
    int n = s.size();
    int m = t.size();
    int i = 0;

    while (i < n && i < m && s[i] == t[i]) {
        i++;
    }

    int opsNeeded = (n - i) + (m - i);

    if (opsNeeded > k) {
        return "No";
    }
    else if ((k - opsNeeded) % 2 == 0 || k >= n + m) {
        return "Yes";
    }
    else {
        return "No";
    }
}

  • sergiosrtd
     + 1 comment

    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.