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

Append and Delete

  • safee7
     + 0 comments

    Step-by-step Solution in C++ :-

    1.Find the common prefix length.

    2.Calculate how many characters need to be deleted from s and added to reach t.

    3.Check if k is enough to do that and whether the leftover operations can be used meaninglessly.

    string appendAndDelete(string s, string t, int k) {
    int commonLength = 0;
    int n1 = s.size(), n2 = t.size();

    // Find common prefix
    for (int i = 0; i < min(n1, n2); i++) {
        if (s[i] == t[i])
            commonLength++;
        else
            break;
    }

    int totalOps = (n1 - commonLength) + (n2 - commonLength);

    // Case 1: Enough operations and parity matches
    if (totalOps == k)
        return "Yes";

    // Case 2: More operations than needed
    if (totalOps < k) {
        // Check if we can use extra operations meaninglessly
        if ((k - totalOps) % 2 == 0 || k >= n1 + n2)
            return "Yes";
    }

    return "No";
}