Abbreviation

                            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()