# Append and Delete

### Problem Statement :

```You have two strings of lowercase English letters. You can perform two types of operations on the first string:

1. Append a lowercase English letter to the end of the string.
2. Delete the last character of the string. Performing this operation on an empty string results in an empty string.

Given an integer, k, and two strings, s and t, determine whether or not you can convert s to t by performing exactly k of the above operations on s. If it's possible, print Yes. Otherwise, print No.

Example.
s = [a, b, c]
t = [d, e, f]
k = 6

To convert s to t, we first delete all of the characters in 3 moves. Next we add each of the characters of t in order. On the 6th move, you will have the matching string. Return Yes.

If there were more moves available, they could have been eliminated by performing multiple deletions on an empty string. If there were fewer than 6 moves, we would not have succeeded in creating the new string.

Function Description

Complete the appendAndDelete function in the editor below. It should return a string, either Yes or No.

appendAndDelete has the following parameter(s):

string s: the initial string
string t: the desired string
int k: the exact number of operations that must be performed

Returns
string: either Yes or No

Input Format

The first line contains a string s, the initial string.
The second line contains a string t, the desired final string.
The third line contains an integer k, the number of operations.

Constraints
1 <= |s| <= 100
1 <= |t| <= 100
1 <= k <= 100
s and t consist of lowercase English letters, ascii[a-z].```

### Solution :

```                            ```Solution in C :

C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
char* s = (char *)malloc(512000 * sizeof(char));
scanf("%s",s);
char* t = (char *)malloc(512000 * sizeof(char));
scanf("%s",t);
int k,i=0,l1,l2,del,append,same;
scanf("%d",&k);
l1=strlen(s),l2=strlen(t);
if(strcmp(s,t)==0)
{
if(k%2==0 || k>=2*l1)
printf("Yes");
else
printf("No");
}
else
{
if(k>=2*l2)
printf("Yes");
else
{
while(i<l1 && i<l2)
{
if(s[i]==t[i])
i++;
else
break;
}
same=i;
del=l1-same;
append=l2-same;
if(del+append > k)
printf("No");
else
{
if((del+append)%2==0)
{
if(k%2==0)
printf("Yes");
else
printf("No");
}
else
{
if(k%2==0)
printf("No");
else
printf("Yes");
}
}
}
}
return 0;
}

C ++  :

#include <iostream>
#include <vector>
#include <string>
#include <string.h>
#include <cmath>

using namespace std;

int main() {
string s, t;
cin >> s >> t;
int k;
cin >> k;

int i = 0, j = 0;
for (; i < (int)s.size() && j < (int)t.size(); ++i,++j) {
if (s[i] != t[j])
break;
}

int need = ((int)s.size() - i) + ((int)t.size() - j);
if ((need <= k && (k-need) % 2 == 0) || k >= (int)s.size() + (int)t.size()) {
cout << "Yes";
} else {
cout << "No";
}

return 0;
}

Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String s = in.next();
String t = in.next();
int k = in.nextInt();
int toDelete = 0;
int i = 0;
while (i < s.length() && i < t.length() && s.charAt(i) == t.charAt(i)) {
i++;
}
toDelete = s.length() - i;
int ops = toDelete + (t.length() - i);
if (ops <= k && ((k - ops) % 2 == 0 || (k - ops) > 2 * i)) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}

python 3:

#!/bin/python3

s = input().strip()
t = input().strip()
k = int(input().strip())

ls = len(s)
lt = len(t)

lcp = 0 # Length of common prefix
while lcp <= ls-1 and lcp <= lt-1 and s[lcp] == t[lcp]:
lcp += 1

if k >= ls + lt:
print ("Yes")
elif k >= ls + lt - 2*lcp and (k - ls - lt + 2*lcp)%2 == 0:
print ("Yes")
else:
print ("No")```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink