LeetCode in Kotlin

1416. Restore The Array

Hard

A program was supposed to print an array of integers. The program forgot to print whitespaces and the array is printed as a string of digits s and all we know is that all integers in the array were in the range [1, k] and there are no leading zeros in the array.

Given the string s and the integer k, return the number of the possible arrays that can be printed as s using the mentioned program. Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: s = “1000”, k = 10000

Output: 1

Explanation: The only possible array is [1000]

Example 2:

Input: s = “1000”, k = 10

Output: 0

Explanation: There cannot be an array that was printed this way and has all integer >= 1 and <= 10.

Example 3:

Input: s = “1317”, k = 2000

Output: 8

Explanation: Possible arrays are [1317],[131,7],[13,17],[1,317],[13,1,7],[1,31,7],[1,3,17],[1,3,1,7]

Constraints:

• 1 <= s.length <= 105
• s consists of only digits and does not contain leading zeros.
• 1 <= k <= 109

Solution

class Solution {
    fun numberOfArrays(s: String, k: Int): Int {
        // dp[i] is number of ways to print valid arrays from string s start at i
        val dp = arrayOfNulls<Int>(s.length)
        return dfs(s, k.toLong(), 0, dp)
    }

    private fun dfs(s: String, k: Long, i: Int, dp: Array<Int?>): Int {
        // base case -> Found a valid way
        if (i == s.length) return 1
        // all numbers are in range [1, k] and there are no leading zeros -> So numbers starting with 0 mean invalid!
        if (s[i] == '0') return 0
        if (dp[i] != null) return dp[i]!!
        var ans = 0
        var num: Long = 0
        for (j in i until s.length) {
            // num is the value of the substring s[i..j]
            num = num * 10 + s[j].code.toLong() - '0'.code.toLong()
            // num must be in range [1, k]
            if (num > k) break
            ans += dfs(s, k, j + 1, dp)
            ans %= 1000000007
        }
        return ans.also { dp[i] = it }
    }
}

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