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

Append and Delete

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1409 Discussions

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  • maurizioscibilia
     + 0 comments

    python:

    def appendAndDelete(s, t, k):
    # Write your code here
    if len(s)+len(t)<=k:
        return 'Yes'
    i = 0
    while i < min(len(s), len(t)) and s[i] == t[i]:
        i += 1
    root_index = i
    delta = len(s) - root_index
    chars_to_add = len(t) - root_index
    remaining_ops = k-(delta+chars_to_add)
    if remaining_ops>=0 and remaining_ops%2==0: 
        return 'Yes'
    else: 
        return 'No'

  • maluisamrok
     + 0 comments

    Hi I tried with no more than a simple if else and I could, just my logic is kind of weird but works.

    public static String appendAndDelete(String s, String t, int k) {
    // Write your code here
        
        int uno = s.length();
        int dos = t.length();
        
        if (s.equals(t))
        {return "Yes";}
        else if (s.length()>t.length() & (uno -dos) > k ){
            
            return "No";
        }else if (s.length()<t.length() & (dos-uno) == k/2 & s.charAt(uno-1) == t.charAt(dos-1)){
            
            return "Yes";
        }
        else if (s.length()<t.length() & (dos-uno) < k){
            
            return "No";
        }
        else if (s.length()== t.length() & (uno/2) > k/2 ){
            
            return "No";
        }else if (s.charAt(0)!= t.charAt(0) && s.length()>k/2){
            return "No";
        }
        
        return "Yes";

    }

}

  • kgonz224
     + 1 comment

    Why is the test case below is expected as "No"? y yu 2

    Just remove/append the u

  • voronin_ilia
     + 0 comments

    C++ solution

    string appendAndDelete(string s, string t, int k) {
  
  int sl = s.length();
  int tl = t.length();
  
  if(k > (sl + tl))
    return "Yes";
  
  int length = 0;
  int diffLength = 0;
  if(sl >= tl) {
    length = tl;
    diffLength = sl - tl;
  }
  else {
    length = sl;
    diffLength = tl - sl;
  }
  if(diffLength > k)
    return "No";
  
  int diffIndex = length;
  for(int i = 0; i < length; i++) {
    if(s[i] != t[i]) {
      diffIndex = i;
      break;
    }
  }
  
  int operations = (sl + tl - 2*diffIndex);
  if(operations == k || (operations % 2 == k % 2 && operations < k))
    return "Yes";
    
  return "No";
}

  • bladoncgarland
     + 0 comments

    TS Solution -- optimal time complexity but not optimised:

    function appendAndDelete(s: string, t: string, k: number): string { // Write your code here

    for(let i = 0; i < Math.max(s.length, t.length); i++) {

    if(s[i] != t[i]) {

        let numCharsToDelete:number = s.length - i
        let numCharsToRebuild:number = t.length - i
        let minActionsToZeroAndBack:number = (s.length - numCharsToDelete) * 2

        k -= numCharsToDelete + numCharsToRebuild

        if (k >= 0 && (k % 2 === 0 || minActionsToZeroAndBack <= k )) {
            return "Yes"
        } else {
            return "No"
        }
    }
}

return "Yes"

    }