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 :



title-img


                            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()
                        




View More Similar Problems

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 →

Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →

The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →