# Abbreviation

### Problem Statement :

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

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.

For example, given a = AbcDE and b = ABDE, in a we can convert b and delete c to match b. If a = AbcDE and b = AFDE, matching is not possible because letters may only be capitalized or discarded, not changed.

Function Description

Complete the function abbreviation 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.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <bits/stdc++.h>
using namespace std;

const int maxN = 1000 + 100;

bool dp[maxN][maxN];

inline void solve() {
string a,b; cin >> a >> b;
memset( dp , 0 , sizeof dp );
int n = a.size() , m = b.size();
dp[0][0] = true;
for( int i = 1 ; i <= n ; i++ )
for( int j = 0 ; j <= m ; j++ ) {
if( j && dp[i-1][j-1] && toupper(a[i-1]) == b[j-1] )
dp[i][j] = true;
if( dp[i-1][j] && islower(a[i-1]) )
dp[i][j] = true;
}
if( dp[n][m] )
cout << "YES" << endl;
else
cout << "NO" << endl;
}

int main() {
int t; cin >> t;
while( t-- )
solve();
}

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");
}
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int change(char *A,int n, char *B,int m){
if( A[n]=='\0' && B[m]=='\0' )
return 1;
if(A[n]=='\0')
return 0;
if( B[m]=='\0' ){
while( A[n]!= '\0' ){
if(A[n]<='Z' && A[n]>='A')
return 0;
n++;
}
return 1;
}
if( (A[n]<='Z' && A[n]>='A') && (A[n] != B[m]) )
return 0;
if( (A[n]==B[m]) )
return change(A,n+1,B,m+1);
if ( (A[n]-'a' +'A')==B[m] ){
if(change(A,n+1,B,m+1) )
return 1;
else
return change(A,n+1,B,m);
}
return change(A,n+1,B,m);
}
int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
char A[1010],B[1010];
int q,n;
scanf("%d",&q);
while(q--){
scanf("%s",A);
scanf("%s",B);
if( change(A,0,B,0) ){
printf("YES\n");

}else{
printf("NO\n");
}
}
return 0;
}

In Python3 :

def recurse(a,b):
if len(a) < len(b):
return False
if len(b) == 0:
return a.islower()
if a.upper() == b:
return True
if a[0].isupper():
if a[0] == b[0]:
return recurse(a[1:],b[1:])
else:
return False
else:
if a[0].upper() != b[0]:
return recurse(a[1:],b)
else:
return (recurse(a[1:],b[1:]) or recurse(a[1:],b))

def solveProblem():
a = input()
b = input()
x = recurse(a,b)
if x:
print('YES')
else:
print('NO')

T = int(input())
for t in range(T):
solveProblem()```
```

## Find Merge Point of Two Lists

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