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";
}

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";
}
}

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

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";
    }
}

The test case 7 is wrong

s=aaaaaaaaaa t=aaaaa k=7

You only need 5 delete operations on s to convert t

But the test case expect "Yes"