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

Append and Delete

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1426 Discussions

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  • suriyaskrs
     + 0 comments

    Simple program using single for loop and east to understand

    def appendAndDelete(s, t, k):

    count=0
for i,j in zip(s,t):
    if i == j:
        count+=1
    else:
        break
mini=(len(s)-count)+(len(t)-count)
if len(s)+len(t) <= k or (k >= mini and (k - mini) % 2 == 0):
    return "Yes"
return "No"

  • alban_tyrex
     + 0 comments

    Here is my O(k) c++ solution, you can watch the explanation here : https://youtu.be/8ZOmXfo6k0Y

    string appendAndDelete(string s, string t, int k) {
    if(k >= t.size() + s.size()) return "Yes";
    int same = min(t.size(), s.size());
    for(int i = 0; i< min(t.size(), s.size()); i++){
        if(s[i] != t[i]){
            same = i;
            break;
        }
    }
    k -= (s.size() - same);
    k -= (t.size() - same);
    return (k >= 0 && k%2 == 0)? "Yes":"No";
}

  • yeungsze
     + 0 comments

    My python solution:

    def appendAndDelete(s, t, k):
    # Write your code here
    equal_first_letters = 0
    while equal_first_letters < len(s) and equal_first_letters < len(t)\
    and s[equal_first_letters] == t[equal_first_letters]:
        equal_first_letters += 1
    min_moves = len(s) - equal_first_letters + len(t) - equal_first_letters
    return 'Yes' if (k >= min_moves and (k - min_moves) % 2 == 0) or k > len(s) + len(t) else 'No'

  • mustafa_rc1905
     + 0 comments

    Whoever wrote it wrote it very well.

  • gua9302
     + 0 comments

    TestCase 7 is not correct there should be K = 2 instead of K= 20