2707. Extra Characters in a String

Problem

You are given a 0-indexed string s and a dictionary of words dictionary. You have to break s into one or more non-overlapping substrings such that each substring is present in dictionary. There may be some extra characters in s which are not present in any of the substrings.

Return **the *minimum* number of extra characters left over if you break up s optimally.**

  Example 1:

Input: s = "leetscode", dictionary = ["leet","code","leetcode"]
Output: 1
Explanation: We can break s in two substrings: "leet" from index 0 to 3 and "code" from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.

Example 2:

Input: s = "sayhelloworld", dictionary = ["hello","world"]
Output: 3
Explanation: We can break s in two substrings: "hello" from index 3 to 7 and "world" from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.

  Constraints:

• 1 <= s.length <= 50

• 1 <= dictionary.length <= 50

• 1 <= dictionary[i].length <= 50

• dictionary[i] and s consists of only lowercase English letters

• dictionary contains distinct words

Solution

/**
 * @param {string} s
 * @param {string[]} dictionary
 * @return {number}
 */
var minExtraChar = function(s, dictionary) {
    var map = dictionary.reduce((res, item) => {
        res[item] = 1;
        return res;
    }, {});
    var dp = Array(s.length);
    for (var i = s.length - 1; i >= 0; i--) {
        dp[i] = (dp[i + 1] || 0) + 1;
        var str = '';
        for (var j = i; j < s.length; j++) {
            str += s[j];
            if (map[str]) {
                dp[i] = Math.min(dp[i], dp[j + 1] || 0);
            }
        }
    }
    return dp[0];
};

Explain:

Bottom-up dynamic programming.

Complexity:

• Time complexity : O(n ^ 2).
• Space complexity : O(n).