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")
View More Similar Problems
Kindergarten Adventures
Meera teaches a class of n students, and every day in her classroom is an adventure. Today is drawing day! The students are sitting around a round table, and they are numbered from 1 to n in the clockwise direction. This means that the students are numbered 1, 2, 3, . . . , n-1, n, and students 1 and n are sitting next to each other. After letting the students draw for a certain period of ti
View Solution →Mr. X and His Shots
A cricket match is going to be held. The field is represented by a 1D plane. A cricketer, Mr. X has N favorite shots. Each shot has a particular range. The range of the ith shot is from Ai to Bi. That means his favorite shot can be anywhere in this range. Each player on the opposite team can field only in a particular range. Player i can field from Ci to Di. You are given the N favorite shots of M
View Solution →Jim and the Skyscrapers
Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently. Let us describe the problem in one dimensional space
View Solution →Palindromic Subsets
Consider a lowercase English alphabetic letter character denoted by c. A shift operation on some c turns it into the next letter in the alphabet. For example, and ,shift(a) = b , shift(e) = f, shift(z) = a . Given a zero-indexed string, s, of n lowercase letters, perform q queries on s where each query takes one of the following two forms: 1 i j t: All letters in the inclusive range from i t
View Solution →Counting On a Tree
Taylor loves trees, and this new challenge has him stumped! Consider a tree, t, consisting of n nodes. Each node is numbered from 1 to n, and each node i has an integer, ci, attached to it. A query on tree t takes the form w x y z. To process a query, you must print the count of ordered pairs of integers ( i , j ) such that the following four conditions are all satisfied: the path from n
View Solution →Polynomial Division
Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie
View Solution →