LeetCode in Kotlin

2707. Extra Characters in a String

Medium

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

class Solution {
    fun minExtraChar(s: String, dictionary: Array<String>): Int {
        val dict: MutableSet<String> = HashSet()
        val dp = IntArray(s.length + 1)
        for (word in dictionary) dict.add(word)
        for (i in 1 until dp.size) {
            dp[i] = dp[i - 1] + 1
            for (j in i - 1 downTo 0) {
                val sub: String = s.substring(j, i)
                if (dict.contains(sub)) dp[i] = dp[i].coerceAtMost(dp[j])
            }
        }
        return dp[dp.size - 1]
    }
}

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