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

Append and Delete

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1449 Discussions

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  • 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";
}

  • 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";
}

  • amit_kumar_mish2
     + 0 comments

    Java ::

    public static String appendAndDelete(String s, String t, int k) { // Write your code here int n = s.length(); int m = t.length(); int i = 0; char[] a = s.toCharArray();

            char[] b = t.toCharArray();


while (i < n && i < m && a[i] == b[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";
}
}

  • sebaspr16
     + 1 comment

    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";
    }
}