1715. Split A String Into The Max Number Of Unique Substrings¶

Difficulty: Medium

1715. Split a String Into the Max Number of Unique Substrings

Medium

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 <= 16
s contains only lower case English letters.

Solution¶

class Solution {
    public int maxUniqueSplit(String s) {
        return solve(s, 0, new HashSet<>());
    }

    private int solve(String s, int left, HashSet<String> set) {
        if (left >= s.length()) return 0;
        int max = 0;
        for (int right = left + 1; right <= s.length(); right++) {
            String current = s.substring(left, right);
            if (!set.contains(current)) {
                set.add(current);
                max = Math.max(max, 1 + solve(s, right, set));
                set.remove(current);
            }
        }
        return max;
    }

    static Debug dbg = new Debug();
    static class Debug {
        public static boolean LOCAL = true;
        public static boolean getLocal() {
            try {
                return System.getProperty("LOCAL") == null;
            }catch(SecurityException e) {
                return false;
            }
        }
        public static <T> String ts(T t) {
            if(t==null) {
                return "null";
            }
            if(t instanceof Iterable) {
                return ts((Iterable<?>) t);
            }else if(t instanceof int[]) {
                String s = Arrays.toString((int[]) t);
                return "{"+s.substring(1, s.length()-1)+"}";
            }else if(t instanceof long[]) {
                String s = Arrays.toString((long[]) t);
                return "{"+s.substring(1, s.length()-1)+"}";
            }else if(t instanceof char[]) {
                String s = Arrays.toString((char[]) t);
                return "{"+s.substring(1, s.length()-1)+"}";
            }else if(t instanceof double[]) {
                String s = Arrays.toString((double[]) t);
                return "{"+s.substring(1, s.length()-1)+"}";
            }else if(t instanceof boolean[]) {
                String s = Arrays.toString((boolean[]) t);
                return "{"+s.substring(1, s.length()-1)+"}";
            }else if(t instanceof Object[]) {
                return ts((Object[]) t);
            }
            return t.toString();
        }
        private static <T> String ts(T[] arr) {
            StringBuilder ret = new StringBuilder();
            ret.append("{");
            boolean first = true;
            for(T t: arr) {
                if(!first) ret.append(", ");
                first = false;
                ret.append(ts(t));
            }
            ret.append("}");
            return ret.toString();
        }
        private static <T> String ts(Iterable<T> iter) {
            StringBuilder ret = new StringBuilder();
            ret.append("{");
            boolean first = true;
            for(T t: iter) {
                if(!first) ret.append(", ");
                first = false;
                ret.append(ts(t));
            }
            ret.append("}");
            return ret.toString();
        }
        public static void print(Object... o) {
            if(LOCAL) {
                System.out.print("Line #"+Thread.currentThread().getStackTrace()[2].getLineNumber()+": [");
                for(int i = 0; i<o.length; i++) {
                    if(i!=0) System.out.print(", ");
                    System.out.print(ts(o[i]));
                }
                System.out.println("]");
            }
        }
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here