1416.Restore-The-Array

1416. Restore The Array

题目地址

https://leetcode.com/problems/restore-the-array/

题目描述

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 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. There can be multiple ways to restore the array.

Return the number of possible array that can be printed as a string s using the mentioned program.

The number of ways could be very large so return it modulo 10^9 + 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]

Example 4:
Input: s = "2020", k = 30
Output: 1
Explanation: The only possible array is [20,20]. [2020] is invalid because 2020 > 30. [2,020] is ivalid because 020 contains leading zeros.

Example 5:
Input: s = "1234567890", k = 90
Output: 34

Constraints:
1 <= s.length <= 10^5.
s consists of only digits and doesn't contain leading zeros.
1 <= k <= 10^9.

代码

Approach #1 DFS

Time: O(n * log10(k)) && Space: O(n)

Explain: There are total n states, each state needs maximum log10(k) iteration loops to calculate the result.

class Solution {
    public int numberOfArrays(String s, int k) {
        int[] memo = new int[s.length()];
        Arrays.fill(memo, -1);
        return dfs(s, k, memo, 0);
    }

    private int dfs(String s, int k, int[] memo, int i) {
        if (i == s.length()) return 1;
        if (s.charAt(i) == '0') return 0;
        if (memo[i] != -1)   return memo[i];

        int ans = 0;
        long num = 0;
        for (int j = i; j < s.length(); j++) {
            num = num * 10 + s.charAt(j) - '0';
            if (num > k) break;
            ans += dfs(s, k, memo, j + 1);
            ans %= 1000_000_007;
        }

        memo[i] = ans;
        return ans;
    }
}

Approach #2 DP

Time: O(n log k) Space: O(n)

class Solution {
    int mod = (int)1e9 + 7;
    public int numberOfArrays(String s, int k) {
        if (s == null || s.length() == 0)     return 0;
        int n = s.length();
        int[] dp = new int[n + 1];
        dp[n] = 1; // initial 
        for (int i = n - 1; i >= 0; i--) {
            long num = s.charAt(i) - '0'; // long
            if (num != 0) {
                for (int j = i + 1; j <= n; j++) {
                    if (num > k)    break;
                    dp[i] = (dp[i] + dp[j]) % mod;
                    if (j < n) {
                        num = num * 10 + s.charAt(j) - '0';
                    }
                }
            } else {
                dp[i] = 0;
            }
        }

        return dp[0];
    }
}