str1 = "spqrstrupvqw"
str2 = "sprt"
str3 = "q"
We're looking for the minimum sub-string from str1
that contain all str2
characters (assume ordered) and no characters from str3
..
i = 1 .. str1.length
cursor = 1 .. str2.length
The solution must be on the form:
str2.first X X .. X X str2.last
So to check for that sub-string we use a cursor over str2
, but we also have the constraint of avoiding str3
characters, so we have:
if str3.contain(str1[i])
cursor = 1
else
if str1[i] == str2[cursor]
cursor++
Goal check is:
if cursor > str2.length
return solution
else
if i >= str1.length
return not-found
And for optimization, you can skip to the next look-ahead which is:
look-ahead = { str2[cursor] or { X | X in str3 }}
In case str2
is not ordered:
i = 1 .. str1.length
lookup = { X | X in str2 }
The solution must be on the form:
str2[x] X X .. X X str2[x]
So to check for that sub-string we use a check-list str2
, but we also have the constraint of avoiding str3
characters, so we have:
if str3.contain(str1[i])
lookup = { X | X in str2 }
else
if lookup.contain(str1[i])
lookup.remove(str1[i])
Goal check is:
if lookup is empty
return solution
else
if i >= str1.length
return not-found
And for optimization, you can skip to the next look-ahead which is:
look-ahead = {{ X | X in lookup } or { X | X in str3 }}
Code
class Solution
{
private static ArrayList<Character> getCharList (String str)
{
return Arrays.asList(str.getCharArray());
}
private static void findFirst (String a, String b, String c)
{
int cursor = 0;
int start = -1;
int end = -1;
ArrayList<Character> stream = getCharList(a);
ArrayList<Character> lookup = getCharList(b);
ArrayList<Character> avoid = getCharList(c);
for(Character ch : stream)
{
if (avoid.contains(ch))
{
lookup = getCharList(b);
start = -1;
end = -1;
}
else
{
if (lookup.contains(ch))
{
lookup.remove(ch)
if (start == -1) start = cursor;
end = cursor;
}
}
if (lookup.isEmpty())
break;
cursor++;
}
if (lookup.isEmpty())
{
System.out.println(" found at ("+start+":"+end+") ");
}
else
{
System.out.println(" not found ");
}
}
}