This is my solution. It's in python. It's probably a bit over complicated but I'll post it anyway

def appendAndDelete(s, t, k):

s_list = list(s)
t_list = list(t)
changes_needed = 0
last_the_same = True

if s == t:
    print("Yes")
    return "Yes"

for i in range(len(s)):
    if i < len(t):
        if s[i] == t[i]:
            if last_the_same is False:
                changes_needed += 1
            else:
                print(s[i] , " = ", t[i])
        else:
            changes_needed += 1
            last_the_same = False
    else:
        changes_needed += 1

if(changes_needed > 0):
    del s_list[-changes_needed:]

print(" removing from s = ", s_list, ", now adding t... we will add: ", t_list[-changes_needed: ])

for i in range(len(t)):

    if i < len(s_list):
        if s_list[i] != t_list[i]:
            s_list.append(t_list[i])
            changes_needed += 1
    else:
        s_list.append(t_list[i])
        changes_needed += 1


print(" s with t added = ", s_list, " the number of changes needed was :", changes_needed)

print("k - changes_needed = ", k - changes_needed, ", s_list = ",s_list, ", t_list = ", t_list )


if s_list == t_list:
    if len(s) >= len(t):
        if (k - changes_needed) == 0 or (k - changes_needed) > 1:
            print('Yes')
            return "Yes"
        else:
            print('No')
            return "No"
    else:
        if (k - changes_needed) == 0 or (k - changes_needed) % 2 == 0:
            print('Yes')
            return "Yes"
        else:
            print('No')
            return "No"
else:
    print('No')
    return "No"