g1401_1500.s1416_restore_the_array.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1401_1500.s1416_restore_the_array;

// #Hard #String #Dynamic_Programming #2022_03_26_Time_34_ms_(100.00%)_Space_42_MB_(100.00%)

/**
 * 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
**/
public class Solution {
    public int numberOfArrays(String s, int k) {
        int kMod = 1000000007;
        int n = s.length();
        int[] dp = new int[n];
        if (s.charAt(n - 1) != '0') {
            dp[n - 1] = 1;
        }
        for (int i = n - 2; i >= 0; i--) {
            if (s.charAt(i) == '0') {
                continue;
            }
            long temp = 0;
            int j = i;
            while (j < n && temp * 10 + s.charAt(j) - '0' <= k) {
                temp = temp * 10 + s.charAt(j) - '0';
                if (j == n - 1) {
                    dp[i] += 1;
                } else {
                    dp[i] += dp[j + 1];
                }
                dp[i] %= kMod;
                j++;
            }
        }
        return dp[0];
    }
}