Find the lexicographically largest unique string - algorithm

Question

I need an algorithm to find the largest unique (no duplicate characters) substring from a string by removing character (no rearranging).
String A is greater than String B if it satisfies these two conditions.
1. Has more characters than String B
Or
2. Is lexicographically greater than String B if equal length
For example, if the input string is dedede, then the possible unique combinations are de, ed, d, and e.
Of these combinations, the largest one is therefore ed since it has more characters than d and e and is lexicographically greater than de.
The algorithm must more efficient than generating all possible unique strings and sorting them to find the largest one.
Note: this is not a homework assignment.

Answer 1

How about this
string getLargest(string s)
{
int largerest_char_pos=0;
string result="";
if(s.length() == 1) return s;
for(int i=0;i<s.length();)
{
p=i;
for(int j=i+1;j<s.length();j++)
{
if(s[largerest_char_pos]< s[j]) largerest_char_pos =j;
}
res+=s[largerest_char_pos];
i=largerest_char_pos+1;
}
return result;
}
This is code snipet just gives you the lexicigraphically larger string. If you dont want duplicates you can just keep track of already added characters .

Answer 2

Let me state the rules for ordering in a way that I think is more clear.
String A is greater than string B if
- A is longer than B
OR
- A and B are the same length and A is lexicographically greater than B
If my restatement of the rules is correct then I believe I have a solution that runs in O(n^2) time and O(n) space. My solution is a greedy algorithm based on the observation that there are as many characters in the longest valid subsequence as there are unique characters in the input string. I wrote this in Go, and hopefully the comments are sufficient enough to describe the algorithm.
func findIt(str string) string {
// exc keeps track of characters that we cannot use because they have
// already been used in an earlier part of the subsequence
exc := make(map[byte]bool)
// ret is where we will store the characters of the final solution as we
// find them
var ret []byte
for len(str) > 0 {
// inc keeps track of unique characters as we scan from right to left so
// that we don't take a character until we know that we can still make the
// longest possible subsequence.
inc := make(map[byte]bool, len(str))
fmt.Printf("-%s\n", str)
// best is the largest character we have found that can also get us the
// longest possible subsequence.
var best byte
// best_pos is the lowest index that we were able to find best at, we
// always want the lowest index so that we keep as many options open to us
// later if we take this character.
best_pos := -1
// Scan through the input string from right to left
for i := len(str) - 1; i >= 0; i-- {
// Ignore characters we've already used
if _, ok := exc[str[i]]; ok { continue }
if _, ok := inc[str[i]]; !ok {
// If we haven't seen this character already then it means that we can
// make a longer subsequence by including it, so it must be our best
// option so far
inc[str[i]] = true
best = str[i]
best_pos = i
} else {
// If we've already seen this character it might still be our best
// option if it is a lexicographically larger or equal to our current
// best. If it is equal we want it because it is at a lower index,
// which keeps more options open in the future.
if str[i] >= best {
best = str[i]
best_pos = i
}
}
}
if best_pos == -1 {
// If we didn't find any valid characters on this pass then we are done
break
} else {
// include our best character in our solution, and exclude it for
// consideration in any future passes.
ret = append(ret, best)
exc[best] = true
// run the same algorithm again on the substring that is to the right of
// best_pos
str = str[best_pos+1:]
}
}
return string(ret)
}
I am fairly certain you can do this in O(n) time, but I wasn't sure of my solution so I posted this one instead.