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

Append and Delete

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1453 Discussions

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  • wifivop476
     + 0 comments

    I thought some LeetCode problems were already quite counter-intuitive and frustrating. This HackerRank problem really opened my eyes, much like a guide to painting ceilings can reveal tips you never expected. I didn’t expect a problem to be designed in such a psychologically twisted way. So many traps.

  • sida9567
     + 0 comments

    I thought some LeetCode problems were already quite counter-intuitive and frustrating. This HackerRank problem really opened my eyes. I didn’t expect a problem to be designed in such a psychologically twisted way. So many traps.

  • naufalfh19
     + 0 comments

    Python3 Solution

    def appendAndDelete(s, t, k):
    if len(t) + len(s) <= k:
        return "Yes"
    
    while k > 0:     
        if s > t:
            s = s[:-1]
            k -= 1
        elif s == t:
            return "Yes"
        else:
            print(s)
            if s == t[:len(s)]:
                if (k - (len(t)-len(s))) % 2 == 0 and k >= len(t)-len(s):
                    return "Yes"
                return "No"
            else:
                s = s[:-1]
                k -= 1
            
    return "No"

  • yashdeora98294
     + 0 comments

    Here is append and delete problem solution in python, java, c++, c and javascript - https://programmingoneonone.com/hackerrank-append-and-delete-problem-solution.html

  • highsoft_soi
     + 0 comments

    C++

    string appendAndDelete(string s, string t, int k) {
    int same_letters = 0;
    int cnt1 = s.size();
    int cnt2 = t.size();
    int min_val = min(cnt1, cnt2);

    for (int i = 0; i < min_val; i++) {
        if (s[i] == t[i]) {
            same_letters++;
        } else break;
    }

    int amount = (cnt1 - same_letters) + (cnt2 - same_letters);

    if (amount == k) {
        return "Yes" ;
    } else if (amount < k) {
        if ((k - amount) % 2 == 0 || k >= cnt1 + cnt2) {
            return "Yes";
        }
    }

    return "No";
}