Abbreviation

Problem Statement :

```You can perform the following operations on the string, :

1. Capitalize zero or more of a's lowercase letters.
2. Delete all of the remaining lowercase letters in a.

Given two strings, a and b, determine if it's possible to make a equal to b as described. If so, print YES on a new line. Otherwise, print NO.

Function Description

Complete the function abbrevation in the editor below. It must return either YES or NO.

abbreviation has the following parameter(s):

a: the string to modify
b: the string to match
Input Format

The first line contains a single integer q, the number of queries.

Each of the next q pairs of lines is as follows:
- The first line of each query contains a single string, a.
- The second line of each query contains a single string, b.

Constraints

1   <=   q  <=  10
1   <=   | a |, | b |   <=  1000
String a consists only of uppercase and lowercase English letters, ascii[A-Za-z].
String b consists only of uppercase English letters, ascii[A-Z].

Output Format

For each query, print YES on a new line if it's possible to make string a equal to string b. Otherwise, print NO.

Sample Input

1
daBcd
ABC
Sample Output

YES```

Solution :

```                            ```Solution in C :

In    C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include<ctype.h>

int main() {

int q,i,j,n,m;
char a[1005],s[1005];
scanf("%d",&q);
while(q--)
{
scanf("%s",s);
scanf("%s",a);
n=strlen(s);
m=strlen(a);
for(i=0,j=0;i<n && j<m;i++)
{
if(s[i]<94)
{
while(s[i]!=a[j] && j<m)
j++;
if(j==m)
break;
else
{
s[i]=a[j]=91;
}
}
}
if(i!=n && j==m)
{
printf("NO\n");
continue;
}
i=0;
while(s[i]==91 && i<n)
i++;
for(j=i;i<n&&j<m;i++)
{
if(s[i]==91 && a[j]!=91 )
break;
else if(s[i]==91 && a[j]==91)
j++;
else
{
if(s[i]==a[j]+32)
j++;
}
}
if(j==m)
printf("YES\n");
else
printf("NO\n");

}
return 0;
}```
```

```                        ```Solution in C++ :

In   C ++  :

#include <cstdio>
#include <algorithm>
#include <vector>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <numeric>
#include <functional>

using namespace std;

char a[2048], b[2048];

int main() {
int T;
scanf("%d", &T);
for (int testcase = 1; testcase <= T; testcase++) {
scanf("%s%s", a, b);
int n = strlen(a), m = strlen(b);
vector<vector<int>> dp(n+1, vector<int>(m+1));
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
if (i == 0 && j == 0) {
dp[i][j] = 1;
continue;
}
bool ok = false;
if (i > 0 && j > 0 && toupper(a[i-1]) == b[j-1] && dp[i-1][j-1]) {
ok = true;
}
if (i > 0 && islower(a[i-1]) && dp[i-1][j]) {
ok = true;
}
dp[i][j] = ok ? 1 : 0;
}
}
if (dp[n][m]) {
printf("YES\n");
}
else {
printf("NO\n");
}
}
return 0;
}```
```

```                        ```Solution in Java :

In   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 sc = new Scanner(System.in);
int q = sc.nextInt();
sc.nextLine();
for (int z = 0; z < q; z++) {
char[] a = sc.nextLine().toCharArray();
char[] b = sc.nextLine().toCharArray();
boolean[][] dp = new boolean[a.length+1][b.length+1];
for (int i = 0; i <= a.length; i++)
dp[i][0] = true;
for (int i = 1; i <= a.length; i++) {
if (a[i-1]>='A'&&a[i-1]<='Z') {
for (int j = 1; j <= b.length; j++) {
if (b[j-1]==a[i-1])
dp[i][j] = dp[i-1][j-1];
}
} else {
char c = (char)(a[i-1]-32);
for (int j = 1; j <= b.length; j++) {
if (b[j-1]==c)
dp[i][j] = dp[i-1][j-1];
dp[i][j] |= dp[i-1][j];
}
}
}
System.out.println(dp[a.length][b.length]?"YES":"NO");
}
}
}```
```

```                        ```Solution in Python :

In   Python3  :

#!/usr/bin/env python3

def search(needle, haystack):
if not haystack:
return False
if not needle:
for ch in haystack:
if ch.isupper():
return False
return True
if needle == haystack.upper():
return True
if haystack[0].isupper() and needle[0] != haystack[0]:
return False
elif needle[0] == haystack[0]:
return search(needle[1:], haystack[1:])
if needle[0] == haystack[0].upper():
if search(needle[1:], haystack[1:]):
return True
return search(needle, haystack[1:])

def main():
queries = int(input().strip())
for i in range(queries):
src = input().strip()
tgt = input().strip()
if search(tgt, src):
print('YES')
else:
print('NO')

if __name__ == '__main__':
main()```
```

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