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()
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