Your Name is Mine

• Problem Description

  
  In an attempt to control the rise in population, Archer was asked to come up with a plan. This time he is targeting marriages. Archer, being as intelligent as he is, came up with the following plan:
  
  A man with name M is allowed to marry a woman with name W, only if M is a subsequence of W or W is a subsequence of M.
  
  A is said to be a subsequence of B, if A can be obtained by deleting some elements of B without changing the order of the remaining elements.
  
  Your task is to determine whether a couple is allowed to marry or not, according to Archers rule.
  Input
  
  The first line contains an integer T, the number of test cases. T test cases follow. Each test case contains two space separated strings M and W.
  Output
  
  For each test case print ""YES"" if they are allowed to marry, else print ""NO"". (quotes are meant for clarity, please dont print them)
  Constraints
  
  1 <= T<=100
  1<= |M|, |W| <=25000 (|A| denotes the length of the string A.)
  All names consist of lowercase English letters only.

• Test Case 1

  
  Input (stdin)

  3
  
  john johanna
  
  ira ira
  
  kayla jayla
  
  

  Expected Output

  YES
  
  YES
  
  NO

• Test Case 2

  
  Input (stdin)

  3
  
  nivi pavi
  
  tifu tifk
  
  vishu nisha
  
  

  Expected Output

  NO
  
  NO
  
  NO

• Program

• #include <stdio.h>
  #include<string.h>
  #define p 25000
  int main()
  {
    char s1[p],s2[p];
    int g=0,i,t,k,h=0;
    scanf("%d",&t);
    for(k=1;k<=t;k++)
    {
      g=0;
      h=0;
      scanf("%s",s1);
      scanf("%s",s2);
      for(i=0;s2[i]!='\0' && s1[g]!='\0';i++)
      {
        if(s2[i]==s1[g])
        {
          g++;
        }
      }
      for(i=0;s1[i]!='\0' && s2[h]!='\0';i++)
      {
        if(s1[i]==s2[h])
          h++;
      }
      int l=strlen(s1);
      int l1=strlen(s2);
      if(l==g || l1==h)
        printf("YES\n");
      else
        printf("NO\n");
    }
  
   return 0;
  }