CareerCup

int check (char[] s1 , char[] s2) { 	
	int i,j,s,ss;
	char cur_char;
	s1 = strlen(s1);
	ss = strlen(s2);
	cur_char = s1[0];	i=0;
	for(j=0;j<ss;++j) {
		if ( s2[j] == cur_char) {
			++i;
			cur_char = s[i]
		}
	}
	if ( i == s) return 1;
	return 0;
}

Ex: String2 = "ABC"; String1="AXBYCZ". Answer = true;

Simple Algo:
First look for 1st char, then 2nd char in remaining string and so on..

boolean f(String str1, String str2)
{
	int begIndex = 0;
	for(int i=0; i<str2.length; ++i){
		int pos = str1.indexOf(str2.charAt(i), begIndex);
		if(pos == -1)	return false;
		begIndex = pos+1;
	}

	return true;
}

Need more details in problem def, for more sophisticated solution.

1.) Put all characters string2 in a stack.
2.) Then start traversing string1 in reverse order
2.1) If currect character matches with top of stack then pop it from stack.
3.) After traversing string1 fully, If stack is empty then return true else return false

just traverse the strings character by character

Here's my idea in Python:

# Using the test case provided above
	s1 = 'AXBYCZ'
	s2 = 'ABC'
	
	common_str = [c for c in s2 if c in s1]  # Filter out unwanted chars
	common_str2 = [c for c in s1 if c in common_str] 
        
	return common_str == common_str2

for(i=0,j=0;j<strlen(str1)&&i<strlen(str2);i++)
{
if(str2[i]==str1[j])
j++;
}
if(j==strlen(str1))
printf("\n Found the code");
else
printf("\nsorry!");

for(i=0,j=0;j<strlen(str1)&&i<strlen(str2);i++)
    {
        if(str2[i]==str1[j])
            j++;
    }
    if(j==strlen(str1))
        printf("\n Found the code");
    else
        printf("\nsorry!");

Compare the two strings character by character