1593. Split a String Into the Max Number of Unique Substrings

Difficulty:
Related Topics:
Similar Questions:

    Problem

    Given a string s, return the maximum number of unique substrings that the given string can be split into.

    You can split string s into any list of non-empty substrings, where the concatenation of the substrings forms the original string. However, you must split the substrings such that all of them are unique.

    A substring is a contiguous sequence of characters within a string.

      Example 1:

    Input: s = "ababccc"
    Output: 5
    Explanation: One way to split maximally is ['a', 'b', 'ab', 'c', 'cc']. Splitting like ['a', 'b', 'a', 'b', 'c', 'cc'] is not valid as you have 'a' and 'b' multiple times.
    

    Example 2:

    Input: s = "aba"
    Output: 2
    Explanation: One way to split maximally is ['a', 'ba'].
    

    Example 3:

    Input: s = "aa"
    Output: 1
    Explanation: It is impossible to split the string any further.
    

      Constraints:

    1 <= s.length&nbsp;<= 16
    
    
    s contains&nbsp;only lower case English letters.
    

    Solution (Java)

    class Solution {
        public int maxUniqueSplit(String s) {
            int lo = 1;
            int hi = s.length();
            // binary search
            while (lo < hi) {
                int mid = (lo + hi + 1) >> 1;
                if (ok(0, mid, 0, s, new HashSet<>())) {
                    lo = mid;
                } else {
                    hi = mid - 1;
                }
            }
            return lo;
        }
    
        private boolean ok(int depth, int end, int curLen, String s, Set<String> seen) {
            if (depth == end) {
                return true;
            }
            for (int j = curLen; j < s.length(); j++) {
                // not enough length remains to reach the end.
                if (s.length() - j < end - depth) {
                    break;
                }
                String cur = s.substring(curLen, j + 1);
                if (seen.add(cur)) {
                    if (ok(depth + 1, end, j + 1, s, seen)) {
                        return true;
                    }
                    seen.remove(cur);
                }
            }
            return false;
        }
    }
    

    Explain:

    nope.

    Complexity: